192 b) Electric potential due to an electric dipole: We have studied electric and magnetic dipoles in XIth Std. Figure 8.8 shows an electric dipole AB consisting of two charges +q and -q separated by a finite distance 2l. Its dipole moment is p, of magnitude p = q × 2l, directed from -q to +q. The line joining the centres of the two charges is called dipole axis. A straight line drawn perpendicular to the axis and passing through centre O of the electric dipole is called equator of dipole. In order to determine the electric potential due to a dipole, let the origin be at the centre (O) of the dipole. V2 = –q 4 r fiff 0 2 The electrostatic potential is the work done by the electric field per unit charge, V W Q fi ff ffl ffi fl . The potential at C due to the dipole is, V V V q r r C fi ff fi ffl ffi fl 1 2 0 1 2 4 1 1 - By geometry, r r r 1 2 2 2 fiffffl 2 cos ffi r r r 2 2 2 2 fiffff 2 cos ffl r r r r 1 2 2 2 2 fi ff1 2 ffl ffi fl cos r r r r 2 2 2 2 2 fi ff1 2 ff ffl ffi fl cos For a short dipole, 2 << r and If r r >> is small ∴ 2 2 r can be neglected fi ff ffl ffi fl r r r 1 2 2 1 2 cos r r r 2 2 2 1 2 fi ff ffl ffi fl cos fi ff ffl ffi fl r r r 1 1 2 1 2 cos r r r 2 1 2 1 2 fi ff ffl ffi fl cos fi ff ffl ffi fl ff ffi fl ffl ffl 1 1 1 2 1 1 1 2 1 1 2 2 1 2 r r r r r r cos cos and fi ff ffl ff ffi fl ffi V V V q r r C 1 2 0 1 2 4 1 1 2 cos ffi ffl fl ffi 1 1 2 1 2 r r cos Using binomial expansion, ( 1 + x) n = 1 + nx, x << l and retaining terms up to the first order of r only, we get Example 8.6: A wire is bent in a circle of radius 10 cm. It is given a charge of 250µC which spreads on it uniformly. What is the electric potential at the centre ? Solution : Given : q = 250 µC = 250 × 10-6 C R = 10 cm = 10-1 m V = ? As V = 1 4π 0 q r = 9 10 250 10 10 9 6 1 fi fi fi ff ff = 2.25 × 107 volt Let C be any point near the electric dipole at a distance r from the centre O inclined at an angle θ with axis of the dipole. r1 and r2 are the distances of point C from charges +q and -q, respectively. Potential at C due to charge +q at A is, V q r 1 0 = + 4fiff 1 Potential at C due to charge -q at B is, Fig. 8.8: Electric potential due to an electric dipole.
193 V q r r r C fi ff ffl ffi fl ffl ffi fl 4 1 1 1 0 cos cos fi ff ffl ff ffi fl q r r r o 4 1 1 cos cos fi ff ffl ffi fl q r r o 4 2 cos fi ff V ffi fl ff ffl p r p q C 1 4 2 0 2 cos ∵ ff Electric potential at C, can also be expressed as, V p r r C fi 1 4 0 3 ffffl fiff ff . V p r r r C fi fi ff ffl ffi ffi fl 1 4 0 2 fiff ffl ffl ff . , r r where r is a unit vector along the position vector, OC fi ff fifi ffl = r i) Potential at an axial point,θ = 00 (towards +q) or 1800 (towards – q) V p r axial o fi ff1 4 2 fflffi i.e. This is the maximum value of the potential. ii) Potential at an equatorial point, θ = 90° and V = 0 Hence, the potential at any point on the equatorial line of a dipole is zero. This is the minimum value of the magnitude of the potential of a dipole. Thus the plane perpendicular to the line between the charges at the midpoint is an equipotential plane with potential zero. The work done to move a charge anywhere in this plane (potential difference being zero) will be zero. Example 8.7: A short electric dipole has dipole moment of 1 × 10-9 C m. Determine the electric potential due to the dipole at a point distance 0.3 m from the centre of the dipole situated a) on the axial line b) on the equatorial line c) on a line making an angle of 60° with the dipole axis. Solution: Given p = 1×10-9 Cm r = 0.3 m a) Potential at a point on the axial line V = 1 4 0 2 π p r = 9 10 1 10 0 3 9 9 2 fi fi fi ff ffl ffi . =100 volt b) Potential at a point on the equatorial line = 0 c) Potential at a point on a line making an angle of 60° with the dipole axis is V= 1 4 0 2 fi ff p r cos = 9 10 1 10 60 0 3 9 9 2 fi fi fi ff ffl ffi fl cos . = 50 volt c) Electrostatics potential due to a system of charges: We now extend the analysis to a system of charges. Fig. 8.9: System of charges. Consider a system of charges q1 , q2 ......... qn at distances r1 , r2 ...... rn respectively from point P. The potential V1 at P due to the charge q1 is V q r 1 0 = 1 4fiff 1 1 Similarly the potentials V2 , V3 ........Vn at P due to the individual charges q2 , q3 ...........qn are given by V q r V q r V q r n n n 2 0 2 2 3 0 3 3 0 1 4 1 4 1 4 fi fi fi ffffl ffffl ffffl , , By the superposition principle, the potential V at P due to the system of charges is the algebraic sum of the potentials due to the individual charges. ∴V = V1 + V2 + ... + Vn = 1 4 + + ----- + 0 1 1 2 fiff 2 q r q r q r n n ffl ffi fl
194 Use your brain power 8.5 Equipotential Surfaces: An equipotential surface is that surface, at every point of which the electric potential is the same. We know that, The potential (V) for a single charge q is given by V q r = 1 4fiff 0 If r is constant then V will be constant. Hence, equipotential surfaces of single point charge are concentric spherical surfaces centered at the charge. For a line charge, the shape of equipotential surface is cylindrical. Let the potential be zero at the point P and OP = x. For x < 0 ( i.e. to the left of O), the potentials of the two charges cannot add up to zero. Clearly, x must be positive. If x lies between O and A, then V1 + V2 = 0, where V1 and V2 are the potentials at points O and A, respectively. 1 4 0 16 0 0 1 2 fiff q x q x ffl ffi fl . 9 × 109 5 10 3 10 0 16 0 8 8 fi ff fi ff ffl ffi fl ff ff x x . ⇒9 ×109 × 10-8 5 3 0 16 0 x x . fi fi ff ffl ffi fl ⇒ 5 3 x x 0 16 . − − = 0 ∴x = 0.10 m, x = 10 cm The other possibility is that x may also lie on extended OA. Equipotential surfaces can be drawn through any region in which there is an electric field. By definition the potential difference between two points P and Q is the work done per unit positive charge displaced from Q to P. ∴VP – VQ = WQP If points P and Q lie on an equipotential surface, Vp = VQ. ∴ WQP = 0 Thus, no work is required to move a test charge along an equipotential surface. a) If dx is the small distance over the equipotential surface through which unit positive charge is carried then dW fi fi E x E dx fi fiff ff . d cos ff 0 Fig. 8.10 : Equipotential surfaces. Or, V q r i i i n fi fi ff 1 4fflffi 0 1 For a continuous charge distribution, summation should be replaced by integration. Example 8.8: Two charges 5 × 10-8 C and -3 × 10-8 C are located 16 cm apart. At what point (s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. Solution : As shown below, suppose the two point charges are placed on x- axis with the positive charge located at the origin O. Is electrostatic potential necessarily zero at a point where electric field strength is zero? Justify. As V1 + V2 = 0 1 4π 0 q x q x 1 2 0 16 fi 0 ff ffl ffi fl fl . 9 × 109 5 10 3 10 0 16 0 8 8 fi ff fi ff ffl ffi fl ff ff x x . ∴x = 0.40 m, x = 40 cm q1 = 5 × 10-8 C q2 = -3 × 10-8 C q1 = 5 × 10-8 C q2 = -3 × 10-8 C
195 Do you know? fi cos = 0 or = 900 ff ff i.e. E d x fiff ff ⊥ as shown in Fig. 8.11 Hence electric field intensity E is always normal to the equipotential surface i.e., for any charge distribution, the equipotential surface through a point is normal to the electric field at that point. Equipotential surfaces do not intersect each other as it gives two directions of electric fields at intersecting point which is not possible. Fig. 8.12: Equipotential surfaces for a uniform electric field. Fig. 8.14: Equipotential surfaces for two identical positive charges. Fig. 8.13: Equipotential surfaces for a dipole. Fig. 8.15: (a) Between 2 plane metallic sheets. Fig. 8.11: Equipotential surface ⊥ to E b) If the field is not normal, it would have a nonzero component along the surface. So to move a test charge against this component work would have to be done. But by the definition of equipotential surfaces, there is no potential difference between any two points on an equipotential surface and hence no work is required to displace the charge on the surface. Therefore, we can conclude that the electrostatic field must be normal to the equipotential surface at every point, and vice versa. Example 8.9: A small particle carrying a negative charge of 1.6 × 10-19 C is suspended in equilibrium between two horizontal metal plates 10 cm apart having a potential (b) When one of the sheet is replaced by a charged metallic sphere. Like the lines of force, the equipotential surface give a visual picture of both the direction and the magnitude of electric field in a region of space.
196 To calculate the electric potential energy of the two charge system, we assume that the two charges q1 and q2 are initially at infinity. We then determine the work done in bringing the charges to the given location by an external agency. In bringing the first charge q1 to position A r 1 fiff fi ff, no work is done because there is no external field against which work needs to be done as charge q2 is still at infinity i.e., W1 = 0. This charge produces a potential in space given by V q r 1 0 1 1 = 1 4fiff --- (8.14) Where r1 is the distance of point A from the origin. When we bring charge q2 from infinity to B r2 fiff fi ff at a distance r12, from q1 , work done is W2 = (potential at B due to charge q1 ) × q2 = 4 × 1 0 12 2 q r q fiff , (where AB = r12) --- (8.15) This work done in bringing the two charges to their respective locations is stored as the potential energy of the configuration of two charges. fiU q q r = 1 4 0 1 2 ffffl 12 --- (8.16) Equation (8.16) can be generalised for a system of any number of point charges. Example 8.10: Two charges of magnitude 5 nC and −2 nC are placed at points (2 cm, 0, 0) and (20 cm, 0, 0) in a region of space, where there is no other external field. Find the electrostatic potential energy of the system. Solution : Given q1 = 5 nC = 5 × 10-9 C q2 = -2 nC = -2 × 10-9 C r = (20 – 2) cm = 18 cm = 18 × 10-2 m U = 1 4 0 1 2 π q q r = 9 10 5 10 2 10 18 10 9 9 9 2 fi fi fi fiff fi fi ff ff ff = -5 × 10-7 J = -0.5 × 10-6 J = -0.5 µJ 8.6 Electrical Energy of Two Point Charges and of a Dipole in an Electrostatic Field: When two like charges lie infinite distance apart, their potential energy is zero because no work has to done in moving one charge at infinite distance from the other. But when they are brought closer to one another, work has to be done against the force of repulsion. As electrostatic force is conservative, this work gets stored as the potential energy of the two charges. Electrostatic potential energy of a system of point charges is defined as the total amount of work done to assemble the system of charges by bringing them from infinity to their present locations. a) Potential energy of a system of 2 point charges: Fig. 8.16: System of two point charges. difference of 4000 V across them. Find the mass of the particle. Solution: Given : q = 1.6 × 10-19 C dx = 10 cm = 10 × 10-2 m = 10-1 m dV = 4000 V E = −dV dx = − − fi4000 10 1 = - 4 × 104 Vm-1 As the charged particle remain suspended in equilibrium, F = mg = qE ∴m = qE g = ffl ffi fi ff ffl ffi fi ff fi 1 6 10 4 10 9 8 19 4 . . = 0.653 × 10-15 kg m = 6.53 × 10-16 kg Let us consider 2 charges q1 and q2 with position vectors r1 and r2 relative to some origin (O). O
197 b) Potential energy for a system of N point charges: Equation (8.16) gives an expression for potential energy for a system of two charges. We now analyse the situation for a system of N point charges. In bringing a charge q3 from ∞ to C (r3 fiff ) work has to be done against electrostatic forces of both q1 and q2 ∴ W3 = (potential at C due to q1 and q2 )× q3 = 1 4 + × 0 1 13 2 23 3 fiff q r q r q ffl ffi fl = 1 4 + 0 1 3 13 2 3 fiff 23 q q r q q r ffl ffi fl Similarly in bringing a charge q4 from ∞ to D r4 fiff fi ff work has to be done against electrostatic forces of q1 , q2 , and q3 W q q r q q r q q r 4 0 1 4 14 2 4 24 3 4 34 = 1 4 + + fiff ffl ffi fl Proceeding in the same way, we can write the electrostatic potential energy of a system of N point charges at r r r 1 2 N fiff fiff fifffi , .... as U q q r j k all pairs jk fi ff 1 4fflffi 0 (c) Potential energy of a single charge in an external field: Above, we have obtained an expression for potential energy of a system of charges when the source of the electric field, i.e., charges and their locations, were specified. In this section, we determine the potential energy of a charge (or charges) in an external field E fiff which is not produced by the given charge (or charges) whose potential energy we wish to calculate. The external sources could be known, unknown or unspecified, but what is known is the electric field E or the `electrostatic potential V due to the external sources. Here we assume that the external field is not affected by the charge q, if q is very small. The external electric field E and the corresponding external potential V may vary from point to point. If V (r ) is the external potential at any point P having position vector r , then by definition, work done in bringing a unit positive charge from ∞ to the point P is equal to V. ∴ Work done in bringing a charge q, from ∞ to the given point in the external field is qV (r ). This work is stored in the form of potential energy of a system of charge q. ∴PE of a system of a single charge q at r in an external field is given by PE fi qV rff ffl --- (8.17) (d) Potential energy of a system of two charges in an external field: In order to find the potential energy of a system of two charges q1 and q2 located at r 1 and r2 respectively in an external field, we calculate the work done in bringing the charge q1 from ∞ to r1 . From (8.17), in the said process work done = q1 V (r 1 ) --- (8.18) To bring the charge q2 to r2 , the work is done not only against the external field E but also against the field due to q1 . Example 8.11: Calculate the electrostatic potential energy of the system of charges shown in the figure. Solution : Taking zero of potential energy at ∞, we get potential energy (PE) of the system of charges PE = 1 4π 0 ∑ q q r j k jk fi ff ffi fl ffl ff ffi fl ffl ffi fl ffl ff ffi fl ffl ffi fl ff 1 4 0 q q r q q r q q r q q r ffi ffi ff ffi fl ffl ff ffi fl ffl q q r q q r ffi ffi 2 2 fi ff ffl ff ff ff ffi fl fi ffiff fl fi ff 1 4 2 2 1 4 2 2 2 0 2222 2 2 0 2 q r q r q r q r q r q r q r q r 2 0 4 ffi fl
198 ∴Work done on q2 against the external field = q2 V (r2 fiff ) and Work done on q2 against the field due to q q q r 1 1 2 0 12 = 4fiff , where r12 = distance between q1 and q2 . By the Principle of superposition for fields, we add up the work done on q2 against the two fields. ∴ Work done in bringing q2 to r2 fi ff ffl ffi fl q V r q q r 2 2 1 2 0 12 4 --- (8.19) Thus from (8.18) and (8.19) potential energy of the system = Total work done in assembling the configuration = + + 4 1 2 1 2 0 12 q V r q V r q q r 1 2 fiff fiff fi ff fi ff fflffi (e) Potential energy of a dipole in an external field: Fig. 8.17 : Couple acting on a dipole. Example 8.12: Two charged particles having equal charge of 3 ×10-5 C each are brought from infinity to a separation of 30 cm. Find the increase in electrostatic potential energy during the process. Solution : Taking the potential energy (PE) at ∞ to be zero, Increase in PE = present PE V = q q r 1 2 0 4π = 9 10 3 10 0 3 9 5 2 ( ) . fi fi fi ff = 9910 10 3 10 9 10 1 fl fl fl fl fl fl fi fi fi fi ff ff = 81 3 = 27 J Example 8.13: a) Determine the electrostatic potential energy of a system consisting of two charges -2 µC and +4 µC (with no external field) placed at (-8 cm, 0, 0) and (+8 cm, 0, 0) respectively. b) Suppose the same system of charges is now placed in an external electric field E = A (1/r2 ), where A = 8 × 105 cm-2, what would be the electrostatic potential energy of the configuration Solution: Given : q1 = -2 µC = -2 × 10-6 C, r1 = 0.08 cm q2 = +4 µC = +4 × 10-6 C, r2 = 0.08 cm r = 16 cm = 0.16 m a) Electrostatic potential energy of the system of two charges is V = 1 4 0 1 2 π q q r = 9 10 2 10 4 10 0 16 9 6 6 . fi fiffl ffi ff fi fi fi ff ff = 0.45 J b) In the electric field, total potential energy (PE) = q q r 1 2 0 4π + q1 V (r 1 fiff ) + q2 V (r2 fiff ) E = −dV dr ∴ V = fi ff Edr = fi ffA r dr 2 ffff ,V = A r ∴ Total PE = q q r 1 2 0 4π + Aq r 1 1 + Aq r 2 2 = -0.45+ 8 10 2 10 0 08 5 6 . fi fiffl ffi ff fi ff + 8 10 4 10 0 08 5 6 . fi fiff ffl fi ffi = -0.45 -20 + 40 = 19.55 J Consider a dipole with charges -q and +q separated by a finite distance 2 , placed in a uniform electric field E fiff . It experiences a torqueτ which tends to rotate it. τ fi fffi fffi = ×p E or fi ff ffl pE sin In order to neutralize this torque, let us assume an external torqueτ ext is applied, which rotates it in the plane of the paper from angle θ0 to angle θ , without angular acceleration and at an infinitesimal angular speed. Work done by the external torque W d pE d fi ext ff ffl fi ffi ffi fl 0 0 sin
199 fi pE ff ffl - cos ffi ffi ffi ffi fi ffi ff ffl fl pE - cos - - cos 0 fi ff pE ffl ffi - cosfl fl cos 0 fi pE ff ffl cos - ffi ffi cos 0 This work done is stored as the potential energy of the system in the position when the dipole makes an angle θ with the electric field. The zero potential energy can be chosen as per convenience. We can choose U (θ0 ) = 0, giving fiU U ff ffl ffi ff ffl fl pE ff ffl 0 0 cos - cos a) If initially the dipole is perpendicular to the field E i.e., fi ff 0 2 ffl then U p ffl ffi fi fl E ffl ffi cos - ff cosfi 2 fi - c pE os ff U p ff ffl fi ffi - . E b) If initially the dipole is parallel to the field E then fi0 ff 0 U p ff ffl fi fi ffi E ff ffl cos - cos0 U p ff ffl fi fi ffi E ff ffl 1 - cos 8.7 Conductors and Insulators, Free Charges and Bound Charges Inside a Conductor: a) Conductors and Insulators: When you come in contact with wires in wet condition or while opening the window of your car, you might have experienced a feeling of electric shock. Why don’t you get similar experiences with wooden materials? The reason you get a shock is that there occurs a flow of electrons from one body to another when they come in contact via rubbing or moving against each other. Shock is basically a wild feeling of current passing through your body. Conductors are materials or substances which allow electricity to flow through them. This is because they contain a large number of free charge carriers (free electrons). In a metal the outer (valence) electrons are loosely bound to the nucleus and are thus free for conductivity, when an external electric field is applied. Metals, humans, earth and animal bodies are all conductors. The main reason we get electric shocks is that being a good conductor our human body allows a resistance free path for the current to flow from the wire to our body. Under electrostatic conditions the conductors have following properties. 1. In the interior of a conductor, net electrostatic field is zero. 2. Potential is constant within and on the surface of a conductor. 3. In static situation, the interior of a conductor can have no charge. 4. Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. 5. Surface charge density of a conductor could be different at different points. Example 8.14: An electric dipole consists of two opposite charges each of magnitude 1µC separated by 2 cm. The dipole is placed in an external electric field of 105 N C-1. Find: (i) The maximum torque exerted by the field on the dipole (ii) The work the external agent will have to do in turning the dipole through 180° starting from the position θ = 0° Solution: Given : p = q × 2l = 10-6 × 2 × 10-2 = 2 × 10-8 cm E = 105 NC-1 (i) τmax = p E sin 90°= 2 × 10-8 × 105 × 1 = 2 × 10-3 Nm (ii) W = pE ( cosθ1 − cosθ2 ) = 2 × 10-8 × 105 × (cos 0- cos 180°) = 2 × 10-3 ( 1 + 1 ) = 4 × 10-3 J
200 In insulators, the electrons are tightly bound to the nucleus and are thus not available for conductivity and hence are poor conductors of electricity. There are no free charges since all the charges are bound to the nucleus. An insulating material can be considered as a collection of molecules that are not easily ionized. An insulator can carry any distribution of external electric charges on its surface or in its interior and the electric field in the interior can have non zero values unlike conductors. 8.8 Dielectrics and Electric Polarisation: Dielectrics are insulates which can be used to store electrical energy. This is because when such substances are placed in an external field, their positive and negative charges get displaced in opposite directions and the molecules develop a net dipole moment. This is called polarization of the material and such materials are called dielectrics. In every atom there is a positively charged nucleus and there are negatively charged electrons surrounding it. The negative charges form an electron cloud around the positive charge. These two oppositely charged regions have their own centres of charge (where the effective charge is located). The centre of negative charge is the centre of mass of negatively charged electrons and that of positive charge is the centre of mass of positively charged protons in the nucleus. Thus, dielectrics are insulating materials or non- conducting substances which can be polarised through small localised displacement of charges. e.g. glass, wax, water, wood , mica, rubber, stone, plastic etc. Dielectrics can be classified as polar dielectrics and non polar dielectrics as described below. Polar dielectrics: A molecule in which the centre of mass of positive charges (protons) does not coincide with the centre of mass of negative charges (electrons), because of the asymmetric shape of the molecules is called polar molecule as shown in Fig. 8.18 (a). They have permanent Electrostatic shielding : • To protect a delicate instrument from the disturbing effects of other charged bodies near it, place the instrument inside a hollow conductor where E = 0. This is called electrostatic shielding. • Thin metal foils are used in making the shields. • During lightning and thunder storm it is always advisable to stay inside the car than near a tree in open ground, since the car acts as a shield. Faraday Cages: • It is an enclosure which is used to block the external electric fields in conductive materials. • Electro-magnetic shielding: MRI scanning rooms are built in such a manner that they prevent the mixing of the external radio frequency signals with the MRI machine. b) Free charges and Bound charges inside materials: The electrical behaviour of conductors and insulators can be understood on the basis of free and bound charges. In metallic conductors, the electrons in the outermost shells of the atoms are loosely bound to the nucleus and hence can easily get detached and move freely inside the metal. When an external electric field is applied, they drift in a direction opposite to the direction of the applied electric field. These charges are called free charges. The nucleus, which consist of the positive ions and the electrons of the inner shells, remain held in their fixed positions. These immobile charges are called bound charges. In electrolytic conductors, positive and negative ions act as charge carriers but their movements are restricted by the electrostatic force between them and the external electric field.
201 dipole moments of the order of 10-30 Cm. They act as tiny electric dipoles, as the charges are separated by a small distance. The dielectrics like HCl, water, alcohol, NH3 etc are made of polar molecules and are called polar dielectrics. Water molecule has a bent shape with its two O - H bonds which are inclined at an angle of about 105°. It has a very high dipole moment of 6.1 × 10-30 Cm. Fig. 8.18 (b) and (c) show the structure of HCl and H2 O, respectively. Polarization of a non-polar dielectric in an external electric field: In the presence of an external electric field Eo , the centres of the positive charge in each molecule of a non-polar dielectric is pulled in the direction of Eo , while the centres of the negative charges are displaced in the opposite direction. Therefore, the two centres are separated and the molecule gets distorted. The displacement of the charges stops when the force exerted on them by the external field is balanced by the restoring force between the charges in the molecule. Each molecule becomes a tiny dipole having a dipole moment. The induced dipole moments of different molecules add up giving a net dipole moment to the dielectric in the presence of the external field. Fig. 8.18. Examples of Polar molecules (b) HCI (c) H2 O. Non Polar dielectrics: A molecule in which the centre of mass of the positive charges coincides with the centre of mass of the negative charges is called a non polar molecule as shown in Fig. 8.19 (a). These have symmetrical shapes and have zero dipole moment in the normal state. The dielectrics like hydrogen, nitrogen, oxygen, CO2 , benzene, methane are made up of nonpolar molecules and are called non polar dielectrics. Structures of H2 and CO2 are shown in Fig. 8.19 (b) and (c), respectively. Fig. 8.19. (a) Nonpolar molecule. Examples of Nonpolar molecules (b) H2 (c) CO2 . Fig. 8.20 (a) Shows the non polar dielectric in absence of electric field while. Polarization of a polar dielectric in an external electric field: The molecules of a polar dielectric have tiny permanent dipole moments. Due to thermal agitation in the material in the absence of any external electric field, these dipole moments are randomly oriented as shown in Fig. 8.21 (a) (c) (b) (a) (b) (c) Fig. 8.20 (b) shows it in presence of an external field. Fig. 8.18. (a) A polar molecule.
202 (a). Hence the total dipole moment is zero. When an external electric field is applied the dipole moments of different molecules tend to align with the field. As a result the dielectric develops a net dipole moment in the direction of the external field. Hence the dielectric is polarized. The extent of polarization depends on the relative values of the two opposing energies. Reduction of electric field due to polarization of a dielectric: When a dielectric is placed in an external electric field, the value of the field inside the dielectric is less than the external field as a result of polarization. Consider a rectangular dielectric slab placed in a uniform electric field E fiff acting parallel to two of its faces. Since the electric charges are not free to move about in a dielectric, no current results when it is placed in an electric field. Instead of moving the charges, the electric field produces a slight rearrangement of charges within the atoms, resulting in aligning them with the field. This is shown in Fig. 8.20 and Fig. 8.21. During the process of alignment charges move only over distances that are less than an atomic diameter. As a result of the alignment of the dipole moments there is an apparent sheet of positive charges on the right side and negative charges on the left side of the dielectric. These two sheets of induced surface charges produce an electric field E0 fi fffi called the polarization field in the insulator which opposes the applied electric field E . The net field E fiff ' , inside the dielectric is the vector sum of the applied field E and the polarization field E0 fi fffi ∴ E' = E - E0 (in magnitude) This is shown in Fig. 8.22 (a), (b) and (c). 1. The applied external electric field which tends to align the dipole with the field. 2. Thermal energy tending to randomise the alignment of the dipole. The polarization in presence of a strong external electric field is shown in Fig. 8.21 (b) Thus, both polar and nonpolar dielectric develop net dipole moment in the presence of an electric field. The dipole moment per unit volume is called polarization and is denoted by P fiff . For linear isotropic dielectrics P E fiff fiff = χe . χe is a constant called electric susceptibility of the dielectric medium. It describes the electrical behaviour of a dielectric. It has different values for different dielectrics. For vacuum χe = 0. Fig. 8.21 (a) Shows the polar dielectric in absence of electric field while. Fig. 8.21 (b) shows it in presence of an external field. Fig. 8.22 (b) Induced surface charges on the dielectric establish a polarization field E0 fi fffi in the interior. Fig. 8.22 (a) When a dielectric is placed in an external electric field, the dipoles become aligned.
203 Do you know? The greater the applied field, greater is the degree of alignment of the dipoles and hence greater is the polarization field. The induced dipole moment disappears when the field is removed. The induced dipole moment is often responsible for the attraction of a charged object towards an uncharged insulator such as charged comb and bits of paper. Table 1:Dielectric constants of various materials: Material Min Max Air Ebonite Glass Mica Paper Paraffin Porcelain Quartz Rubber Wood dry Metals 1 2.7 3.8 4 1.5 2 5 5 2 1.4 ∞ 1 2.7 14.5 9 3 3 6.5 5 4 2.9 ∞ 8.9 Capacitors and Capacitance, Combination of Capacitors in Series and Parallel: In XIth Std. you have studied about resistors, resistance and conductance. A resistor is an electrical component which allows current to pass through it and dissipates heat but can’t store electrical energy. So there was a need to develop a device that can store electrical energy. The most common arrangement for this consists of a set of conductors (conducting plates) having charges on them and separated by a dielectric material. The conductors 1 and 2 shown in the Fig. 8.23 have charges +Q and -Q with potential difference, V = V1 - V2 between them. The electric field in the region between them is proportional to the charge Q. Fig. 8.23: A capacitor formed by two conductors. Fig. 8.22 (c) The net field E′ fifffi is a vector sum of E and E0 fi fffi . If we apply a large enough electric field, we can ionize the atoms and create a condition for electric charge to flow like a conductor. The fields required for the breakdown of dielectric is called dielectric strength. The potential difference V is the work done to carry a unit positive test charge from the conductor 2 to conductor 1 against the field. As this work done will be proportional to Q, then V ∝ Q and the ratio Qff Vff is a constant. ∴ C = Q V The constant C is called the capacitance of the capacitor, which depends on the size, shape and separation of the system of two conductors. The SI unit of capacitance is farad (F). Dimensional formula is [M-1 L-2T4 A2 ]. 1 farad = 1 coulomb/1volt A capacitor has a capacitance of one farad, if the potential difference across it rises by 1volt when 1 coulomb of charge is given to it. In practice farad is a big unit, the most commonly used units are its submultiples. 1µF = 10-6F 1nF=10-9F 1pF = 10-12F
204 Fig. 8.24: (a) and (b) Parallel plate capacitor. Uses of Capacitors Principle of a capacitor: To understand the principle of a capacitor let us consider a metal plate P1 having area A. Let some positive charge +Q be given to this plate. Let its potential be V. Its capacity is given byC = Q V 1 Now consider another insulated metal plate P2 held near the plate P1 . By induction a negative charge is produced on the nearer face and an equal positive charge develops on the farther face of P2 (Fig. 8.24 (a)). The induced negative charge lowers the potential of plate P1 , while the induced positive charge raises its potential. based on the shape of the conductors. Combination of Capacitors: When there is a combination of capacitors to be used in a circuit we can sometimes replace it with an equivalent capacitor or a single capacitor that has the same capacitance as the actual combination of capacitors. The effective capacitance depends on the way the individual capacitors are combined. Here we discuss two basic combinations of capacitors which can be replaced by a single equivalent capacitor. (a) Capacitors in series: When a potential difference (V)is applied across several capacitors connected end to end in such a way that sum of the potential difference across all the capacitors is equal to the applied potential difference V, then the capacitors are said to be connected in series. Fig. 8.25: Capacitors in series. Fig. 8.26: Effective capacitance of three capacitors in series. (a) (b) As the induced negative charge is closer to P1 it is more effective, and thus there is a net reduction in potential of plate P1 . If the outer surface of P2 is connected to earth, the induced positive charges on P2 being free, flows to earth. The induced negative charge on P2 stays on it, as it is bound to positive charge of P1 . This greatly reduces the potential of P2 , (Fig 8.24 (b)). If V1 is the potential on plate P2 due to charge (- Q) then the net potential of the system will now be +V-V1 . Hence the capacity C = Q V -V 2 1 ∴C C 2 1 > Thus capacity of metal plate P1 , is increased by placing an identical earth connected metal plate P2 near it. Such an arrangement is called capacitor. It is symbolically shown as ⊥⊥ . If the conductors are plane then it is called parallel plate capacitor. We also have spherical capacitor, cylindrical capacitor etc. In series arrangement as shown in Fig. 8.25, the second plate of first conductor is connected to the first plate of the second conductor and so on. The last plate is connected to earth. In a series combination, charges on the plates (± Q)are the same on each capacitor. Potential difference across the series combination of capacitor is V volt, where V = V1 + V2 + V3 ∴V = Q C + Q C + Q 1 2 C3
205 Remember this Let Cs represent the equivalent capacitance shown in Fig. 8.26, thenV = Q Cs ∴ Q C = Q C + Q C + Q s 1 2 C3 ∴ 1 = 1 + 1 + 1 C C s 1 2 C C3 ( for 3 capacitors in series) This argument can be extended to yield an equivalent capacitance for n capacitors connected in series which is equal to the sum of the reciprocals of individual capacitances of the capacitors. fi ff ffl ffl ffl 1 1 1 1 C C eq 1 2 C Cn ................. If all capacitors are equal then 1 C n C C C n eq = = eq or Fig. 8.27: Parallel combination of capacitors. b) Capacitors in Parallel: The parallel arrangement of capacitors is as shown in Fig. 8.27 below, where the insulated plates are connected to a common terminal A which is joined to the source of potential, while the other plates are connected to another common terminal B which is earthed. Example 8.15 When 108 electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Find the capacitance of the two conductors. Solution : Given : Number of electrons n = 108 V = 10 volt ∴charge transferred Q = ne = 108 × 1.6 × 10-19 (∵ e = 1.6 × 10-19 C) = 1.6 × 10-11 C ∴ Capacitance between two conductors C = Q V = 1 6 10 10 11 . flfi ff = 1 6 10 10 . flfi ff F Example 8.16: From the figure given below find the value of the capacitance C if the equivalent capacitance between A and B is to be 1 µF. All other capacitors are in micro farad. Series combination is used when a high voltage is to be divided on several capacitors. Capacitor with minimum capacitance has the maximum potential difference between the plates. In this combination all the capacitors have the same potential difference but the plate charges (± Q1 ) on capacitor1, (± Q2 ) on the capacitor 2 and (± Q3 ) on capacitor 3 are not necessarily the same. If charge Q is applied at point A then it will be distributed to the capacitors depending on the capacitances. ∴Total charge Q can be written as Q = Q1 + Q2 + Q3 = C1 V + C2 V + C3 V Let C p be the equivalent capacitance of the combination then Q = C p V ∴C Vp =C1 2 V +C V +C3 V ∴C p = C1 + C2 + C3 The general formula for effective capacitance C p for parallel combination of n capacitors follows similarly C p = C1 + C2 + .............. + Cn If all capacitors are equal then C eq = nC Remember this Capacitors are combined in parallel when we require a large capacitance at small potentials.
206 A parallel plate capacitor consists of two thin conducting plates each of area A, held parallel to each other, at a suitable distance d apart. The plates are separated by an insulating medium like paper, air, mica, glass etc. One of the plates is insulated and the other is earthed as shown in Fig. 8.28. Fig. 8.28: Capacitor with dielectric. 8.10 Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between the Plates: In section 8.8 we have studied the behaviour of dielectrics in an external field. Let us now see how the capacitance of a parallel plate capacitor is modified when a dielectric is introduced between its plates. a) Capacitance of a parallel plate capacitor without a dielectric: Solution : Given : C1 = 8 µF , C2 = 4 µF , C3 = 1µF , C4 = 4 µF , C5 = 4 µF The effective capacitance of C4 and C5 in parallel = C4 + C5 = 4 + 4 = 8 µF The effective capacitance of C3 and 8 µF in series = 1 8 1 8 ffl ffl ffl ffl fi ff = 8 9 µF The capacitance 8 µF is in parallel with the series combination of C1 and C2 . Their effective combination is C C C C 1 2 1 2 + + 8 9 ⇒ 8 4 12 flfl flfl × + 8 9 ⇒ 32 9 µF This capacitance of 32 9 µF is in series with C and their effective capacitance is given to be 1µF 32 9 32 9 1 32 9 32 9 fi ff ffl ffi fi ffl ff C C C C = 1.39 µF When a charge +Q is given to the insulated plate, then a charge -Q is induced on the inner face of earthed plate and +Q is induced on its farther face. But as this face is earthed the charge +Q being free, flows to earth. In the outer regions the electric fields due to the two charged plates cancel out. The net field is zero. E = 2 - 2 = 0 0 0 fi ff fi ff In the inner regions between the two capacitor plates the electric fields due to the two charged plates add up. The net field is thus E Q = 2 + 2 = = 0 0 0 0 A fi ff fi ff fi ff ff --- (8.20) The direction of E is from positive to negative plate. Let V be the potential difference between the 2 plates. Then electric field between the plates is given by E V d = or = V Ed --- (8.21) Substituting Eq. (8.20) in Eq. (8.21) we get V Q A = ε 0 d Capacitance of the parallel plate capacitor is given by
207 Remember this (1) If there are n parallel plates then there will be (n-1) capacitors, hence C A d = (n - 1) 0 ε (2) For a spherical capacitor, consisting of two concentric spherical conducting shells with inner and outer radii as a and b respectively, the capacitance C is given by C ab b - a = 4fiff 0 ffl ffi fl (3) For a cylindrical capacitor, consisting of two coaxial cylindrical shells with radii of the inner and outer cylinders as a and b, and length l, the capacitance C is given by C b a e fi 2ffffl 0 log Remember this --- (8.22) C Q V Q Qd A A d = = = fi fi 0 0 ff ffl ffi fl Let E0 be the electric field intensity between the plates before the introduction of the dielectric slab. Then the potential difference between the plates is given by V0 = E0 d, where E Q A o o o fi fi ff ffl ffl , and σ is the surface charge density on the plates. Let a dielectric slab of thickness t (t <d) be introduced between the plates of the capacitor. The field E0 polarizes the dielectric, inducing charge - Qp on the left side and +Qp on the right side of the dielectric as shown in Fig. 8.29. These induced charges set up a field E p inside the dielectric in the opposite direction of E0 . The induced field is given by E Q A p p o p o fi fi ff ffl ffl fi p Qp A ff ffl ffi fl The net field (E) inside the dielectric reduces to E0 - E p . Hence, E = E - E = E k E E - E = k o p o o o p fi ff ffl ffl ffi fl , where k is a constant called the dielectric constant. fi ff E ff Q A K Q AK E ffl ffl 0 0 or --- (8.23) Fig. 8.29: Dielectric slab in the capacitor. The dielectric constant of a conductor is infinite. b) Capacitance of a parallel plate capacitor with a dielectric slab between the plates: Let us now see how Eq. (8.22) gets modified with a dielectric slab in between the plates of the capacitor. Consider a parallel plate capacitor with the two plates each of area A separated by a distance d. The capacitance of the capacitor is given by C A d 0 = ε 0 The field E p exists over a distance t and E0 over the remaining distance (d - t) between the capacitor plates. Hence the potential difference between the capacitor plates is V = E d - t + E t o fi ff fi ff = E d - t + E k t o o fi ff fi ff E E k fi ff ffl ffi fl 0 = E d - t + t k o fi ff ffl ffi fl = o Q A d - t + t fi k ff ffl ffi fl The capacitance of the capacitor on the introduction of dielectric slab becomes
208 Fig. 8.30 : Capacitor filled with n dielectric slabs. 8.11 Displacement Current: C = Q V Q Q A d - t + d k A d - t + t k = = 0 0 fi fi ff ffl ffi fl ff ffl ffi fl Special cases: 1. If the dielectric fills up the entire space then t = d fiC A k d = = k C 0 0 ff ∴ capacitance of a parallel plate capacitor increases k times i.e. k = C C0 2. If the capacitor is filled with n dielectric slabs of thickness t 1 , t 2 ....... t n then this arrangement is equivalent to n capacitors connected in series as shown in Fig. 8.30. fi ff ffl ffi fl C A t k + t k t k n n = + ............. + 0 1 1 2 2 3. If the arrangement consists of n capacitors in parallel with plate areas A1 , A2 , .............. An and plate separation d C d A k + A k +.........+ A k = n n fi 0 1 2 2 1 ff ffl if A A A A n 1 2 = .............. =n then C A dn = + k k + .........+ k 0 1 2 n fi ff ffl 4. If the capacitor is filled with a conducting slab (k = ∞) then C = d d - t C fi ff ffl ffi fl o ∴ C > Co The capacitance thus increases by a factor d d - t fi ff ffl ffi fl We know that electric current in a DC circuit constitutes a flow of free electrons. In a circuit as shown in Fig 8.31, a parallel plate capacitor with a dielectric is connected across a DC source. In the conducting part of the circuit free electrons are responsible for the flow of current. But in the region between the plates of the capacitor, there are no free electrons available for conduction in the dielectric. As the circuit is closed, the current flows through the circuit and grows to its maximum value (i c ) in a finite time (time constant of the circuit). The conduction current, i c is found to be same everywhere in the circuit except inside the capacitor. As the current passes through the leads of the capacitor, the electric field between the plates increases and this in turn causes polarisation of the dielectric. Thus, there is a current in the dielectric due to the movement of the bound charges. The current due to bound charges is called displacement current (i d ) or charge- separation current. We can now derive an expression between i c and i d . From Eq (8.23) we can infer that the charge produced on the plates of a capacitor is due to the electric field E. q = Akε0 E Differentiating the above equation, we get dq dt Ak dE dt fi ff 0 --- (8.24) dq/dt is the conduction current (i c )in the conducting part of the circuit. Fig. 8.31: Displacement current in the space between the plates of the capacitor.
209 Example 8.17 A parallel plate capacitor has an area of 4 cm2 and a plate separation of 2 mm (i) Calculate its capacitance (ii) What is its capacitance if the space between the plates is filled completely with a dielectric having dielectric constant of constant 6.7. Solution : Given A = 4 cm2 = 4 × 10-4 m2 d = 2 mm = 2 × 10-3 m ε0 = 8.85 × 10-12 C2 / Nm2 (i) Capacitance C = A d ε 0 = 8 85 10 4 10 2 10 12 4 3 . fi fi fi fi ff ff ff = 1.77 × 10-12 F i dq dt Ak dE dt c fi fi ff 0 dE dt i Ak dE dt i c fi ff ffl c ffi 0 (for fixed value of A) The rate of change of electric field (dE/dt) across the capacitor is directly proportional to the current (i c ) flowing in the conducting part of the circuit. The quantity on the RHS of Eq (8.24) is having the dimension of electric current and is caused by the displacement of bound charges in the dielectric of the capacitor under the influence of the electric field. This current, called displacement current (i d ), is equivalent to the rate of flow of charge (dq/dt=i c ) in the conducting part of the circuit. In the absence of any dielectric between the plates of the capacitor, k=1 (for air or vacuum), the displacement current i d = Aε0 (dE/dt). As a broad generalization of displacement current in a circuit containing a capacitor, it can be stated that the displacement currents do not remain confined to the space between the plates of a capacitor. A displacement current (i d ) exists at any point in space where, timevarying electric field (E) exists (i.e. dE/dt ≠0). 8.12 Energy Stored in a Capacitor: A capacitor is a device used to store energy. Charging a capacitor means transferring electron from one plate of the capacitor to the other. Hence work will have to be done by the battery in order to remove the electrons against the opposing forces. These opposing forces arise since the electrons are being pushed to the negative plate which repels them and electrons are removed from the positive plate which tends to attract them. In both the cases, the forces oppose the transfer from one plate to another. As the charges on the plate increases, opposition also increases. This work done is stored in the form of electrostatic energy in the electric field between the plates, which can later be recovered by discharging the capacitor. (ii) Capacitance C′ = A k d ε 0ff = 8 85 10 4 10 6 7 2 10 12 4 3 . . fi fi fi fi fi ff ff ff = 7.90 × 10-12 F Example 8.18: In a capacitor of capacitance 20 µF, the distance between the plates is 2 mm. If a dielectric slab of width 1 mm and dielectric constant 2 is inserted between the plates, what is the new capacitance ? Solution: Given C = 20 µF = 20 × 10-6 F d = 2 mm = 2 × 10-3 m t = 1 × 10-3 m k = 2 C = A d ε 0 and C′ = A d t t k fi 0ff ff–ff ff ff ff ⇒ C C ' = d t t k d fi–fi fi fi + ⇒ 20 C ' = 2 10 1 10 1 10 2 2 10 3 3 3 3 flfl flfl fl fl fl fl fi ff fi ffl ffi fi fl fi ff ff ff ff ⇒ C′ = 26.6 µF
210 Observe and discuss Fig. 8.32: Capacitor charged by a DC source. Consider a capacitor of capacitance C being charged by a DC source of V volts as shown in Fig. 8.32. During the process of charging, let q' be the charge on the capacitor and V be the potential difference between the plates. Hence C = q' V A small amount of work is done if a small charge dq is further transferred between the plates. fi ff dW V dq ff q C dq ' Total work done in transferring the charge W dw q C dq C q dq O Q O Q fi fi fi ff ff ff ' ' 1 fi ff ffl ffi fl fi 1 2 1 2 2 0 2 C q Q C Q ' This work done is stored as electrical potential energy U of the capacitor. This work done can be expressed in different forms as follows. fiU ff ffl Q C = CV QV Q =CV 1 2 = 1 2 = 1 2 2 2 The energy supplied to the battery is QV but energy stored in the electric field is 1 2 QV. The rest half 1 2 QV of energy is wasted as heat in the connecting wires and battery itself. Example 8.19: A parallel plate air capacitor has a capacitance of 3 × 10–9 Farad. A slab of dielectric constant 3 and thickness 3 cm completely fills the space between the plates. The potential difference between the plates is maintained constant at 400 volt. What is the change in the energy of capacitor if the slab is removed ? Solution : Energy stored in the capacitor with air Ea = 1 2ffl CV2 = 1 2 fl fl × ×3 10 –9 × (400)2 = 24 × 10–5 J when the slab of dielectric constant 3 is introduced between the plates of the capacitor, the capacitance of the capacitor increases to C′ = kC C′ = 3 × 3 × 10–9 = 9 × 10–9 F Energy stored in the capacitor with the dielectric (Ed ) Ed = 1 2fflC' V2 Ed = 1 2 × 9 × 10-9 × (400)2 = 72 × 10-5 J Change in energy = Ed – Ea = (72 - 24) × 10-5 = 48 ×10–5 J There is, therefore, an increase in the energy on introducing the slab of dielectric material. 8.13 Van de Graaff Generator: Van de Graaff generator is a device used to develop very high potentials of the order of 107 volts. The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies needed for experiments to probe the small scale structure of matter and for various experiments in Nuclear Physics. It was designed by Van de Graaff (1901- 1967) in the year 1931. Principle: This generator is based on (i) the phenomenon of Corona Discharge (action of sharp points), (ii) the property that charge given to a hollow conductor is transferred to its outer surface and is distributed uniformly over it, (iii) if a charge is continuously supplied to an insulated metallic conductor, the potential of the conductor goes on increasing.
211 Construction: Fig. 8.33 shows the schematic diagram of Van de Graaff generator. Fig. 8.33: Schematic diagram of van de Graff generator. P1 P2 = Pulleys BB = Conveyer belt A = Spray brush C = Collector brush D = Dome shaped hollow conductor E = Evacuated accelerating tube I = Ion source P = DC power supply S = Steel vessel filled with nitrogen M = Earthed metal plate filled with nitrogen at high pressure. A small quantity of Freon gas is mixed with nitrogen to ensure better insulation between the vessel S and its contents. A metal plate M held opposite to the brush A on the other side of the belt is connected to the vessel S, which is earthed. Working: The electric motor connected to the pulley P1 is switched on, which begins to rotate setting the conveyor belt into motion. The DC supply is then switched on. From the pointed ends of the spray brush A, positive charge is continuously sprayed on the belt B. The belt carries this charge in the upward direction, which is collected by the collector brush C and sent to the dome shaped conductor. As the dome is hollow, the charge is distributed over the outer surface of the dome. Its potential rises to a very high value due to the continuous accumulation of charges on it. The potential of the electrode I also rises to this high value. The positive ions such as protons or deuterons from a small vessel (not shown in the figure) containing ionised hydrogen or deuterium are then introduced in the upper part of the evacuated accelerator tube. These ions, repelled by the electrode I, are accelerated in the downward direction due to the very high fall of potential along the tube, these ions acquire very high energy. These high energy charged particles are then directed so as to strike a desired target. Uses: The main use of Van de Graff generator is to produce very high energy charged particles having energies of the order of 10 MeV. Such high energy particles are used 1. to carry out the disintegration of nuclei of different elements, 2. to produce radioactive isotopes, 3. to study the nuclear structure, 4. to study different types of nuclear reactions, 5. accelerating electrons to sterilize food and to process materials. An endless conveyor belt BB made of an insulating material such as reinforced rubber or silk, can move over two pulleys P1 and P2 . The belt is kept continuously moving by a motor (not shown in the figure) driving the lower pulley (P1 ). The spray brush A, consisting of a large number of pointed wires, is connected to the positive terminal of a high voltage DC power supply. From this brush positive charge can be sprayed on the belt which can be collected by another similar brush C. This brush is connected to a large, dome-shaped, hollow metallic conductor D, which is mounted on insulating pillars (not shown in the figure). E is an evacuated accelerating tube having an electrode I at its upper end, connected to the dome-shaped conductor. To prevent the leakage of charge from the dome, the pulley and belt arrangement, the dome and a part of the evacuated tube are enclosed inside a large steel vessel S, Internet my friend 1. https://en.m.wikipedia.org 2. hyperphyrics.phy-astr.gsu.edu 3. https://www.britannica.com/science 4. https://www.khanacademy.org>in-i
212 Q1. Choose the correct option i) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are (A) Constant, decreases, decreases (B) Increases, decreases, decreases (C) Constant, decreases, increases (D) Constant, increases, decreases ii) A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is (A) C A d k k fi ffl ff ffi fl 0 3 4 (B) C A d k k fi ff ffl ffi fl 0 2 3 (C) C A d k k fi ffl ff ffi fl 0 3 2 (D) C A d k k fi ff ffl ffi fl 0 4 3 iii) Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio. (A) 1:1 (B) 1:2 (C) 2:1 (D) 1:3 iv) Charge +q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD as shown in the figure below is (C) qQ 6fiff 0L (D) fiqQ 4ffffl 0L v) A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100 V is applied? (A) 1.78 × 10-8 C (B) 1.78 × 10-5 C (C) 4.3 × 104 C (D) 2 × 10-9 C Q2. Answer in brief. i) A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process. Exercises ii) If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease. iii) A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor? iv) The safest way to protect yourself from lightening is to be inside a car. Justify. v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process. 3. A dipole with its charges, -q and +q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the YZ planes. (A) fiqQ 6ffffl 0L (B) qQ 2fiff 0L
213 (a) What is the direction of the electric field E? (b) How much torque would the dipole experience in this field? 4. Three charges -q, +Q and -q are placed at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q:q? 5. A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant. 6. Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same. 7. Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin? 8. A 6 µF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? [Ans: 9 × 10-2 J] 9. One hundred twenty five small liquid drops, each carrying a charge of 0.5 µC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop. [Ans: 2.25 × 106 V] 10. The dipole moment of a water molecule is 6.3 × 10–30 Cm. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 × 105 N /C. Calculate the work to be done to rotate the dipoles from their initial orientation θ1 = 0 to one in which all the dipoles are perpendicular to the field, θ2 = 90°.[Ans: 1.575 × 10-3 J] 11. A charge 6 µC is placed at the origin and another charge –5 µC is placed on the y axis at a position A (0, 6.0) m. a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m b) Calculate the work done to bring a proton from infinity to the point P ? What is the significance of the negative sign ? [Ans: (a) Vp = 2.25 × 103 V (b) W = -5.4 × 10-16 J] 12. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the separation between the plates is 2 mm. a) Calculate the capacitance of the capacitor, b) If this capacitor is connected to 100 V supply, what would be the charge on each plate? c) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ? [Ans: (a) 2.655 × 10-11 F, (b) 2.655 × 10-9 C, (c) 15.93 × 10-9 C] 13. Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d. [Ans: (a) 2A 0 d ε (b) 4A 0 d ε ]
214 Can you recall? 9.1 Introduction: In XIth Std. we have studied the origin of electrical conductivity, in particular for metals. We have also studied how to calculate the effective resistance of two or more resistances in series and in parallel. However, a circuit containing several complex connections of electrical components cannot be easily reduced into a single loop by using the rules of series and parallel combination of resistors. More complex circuits can be analyzed by using Kirchhoff’s laws. Gustav Robert Kirchhoff (1824-1887) formulated two rules for analyzing a complicated circuit. In this chapter we will discuss these laws and their applications. 9.2 Kirchhoff’s Laws of Electrical Network: Before describing these laws we will define some terms used for electrical circuits. Junction: Any point in an electric circuit where two or more conductors are joined together is a junction. Loop: Any closed conducting path in an electric network is called a loop or mesh. Branch: A branch is any part of the network that lies between two junctions. In Fig. 9.1, there are two junctions, labeled a and b. There are three branches: these are the three possible paths 1, 2 and 3 from a to b. For a steady current flowing through an electrical network of resistors, the following Kirchhoff's laws are applicable. 9.2.1 Kirchhoff’s First Law: (Current law/ Junction law) The algebraic sum of the currents at a junction is zero in an electrical network, i.e., Ii i n fi fi ff 0 1 , where Ii is the current in the ith conductor at a junction having n conductors. Fig 9.1: Electric network. Fig. 9.2: Kirchhoff first law. Sign convention: The currents arriving at the junction are considered positive and the currents leaving the junction are considered negative. Consider a junction P in a circuit where six conductors meet (Fig.9.2). Applying the sign convention, we can write I1 - I2 + I 3 +I 4 -I5 -I6 = 0 --- (9.1) Arriving currents I1, I3 and I4 are considered positive and leaving currents I2 , I5 and I6 are considered negative. Equation (9.1) can also be written as I1 + I3 + I 4 = I 2 +I 5 + I 6 Thus the total current flowing towards the junction is equal to the total current flowing away from the junction. 9. Current Electricity • There can be three types of electrical conductors: good conductors (metals), semiconductors and bad conductors (insulators). • Does a semiconductor diode and resistor have similar electrical properties? • Can you explain why two or more resistors connected in series and parallel have different effective resistances? P
215 Remember this 9.2.2 Kirchhoff’s Voltage Law: The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero. IR fi ff fflffl ffi 0 --- (9.2) Sign convention: 1. While tracing a loop through a resistor, if we are travelling along the direction of conventional current, the potential difference across that resistance is considered negative. If the loop is traced against the direction of the conventional current, the potential difference across that resistor is considered positive. 2. The emf of an electrical source is positive while tracing the loop within the source from the negative terminal of the source to its positive terminal. It is taken as negative while tracing within the source from positive terminal to the negative terminal. sense. Applying the sign conventions to Eq. (9.2), we get, -I1 R1 -I3 R5 -I1 R3 +ε1 = 0 ∴ε1 = I 1 R1 + I 3 R5 + I 1 R3 Now consider the loop BFDCB in anticlockwise direction. Applying the sign conventions, we get, fi fi I R2 2 I R3 5 fi ff I R2 4 ffl 2 ffi 0 ∴fi 2 2 ff ffl2 3 5 2 ffl 4 fl fl I R fl fl I R I R Fig. 9.3: Electrical network. Steps usually followed while solving a problem using Kirchhoff’s laws: i) Choose some direction of the currents. ii) Reduce the number of variables using Kirchhoff’s first law. iii) Determine the number of independent loops. iv) Apply voltage law to all the independent loops. v) Solve the equations obtained simultaneously. vi) In case, the answer of a current variable is negative, the conventional current is flowing in the direction opposite to that chosen by us. Example 9.1: Figure shows currents in a part of electrical circuit. Find the current X? Solutions: At junction B, current I1 is split into I2 and I3 therefore I1 = I 2 + I 3 Substituting values we get I 3 = 14 A At C, I 5 = I 3 + I 4 therefore I 5 = 16 A At D, I 5 = I 6 + I 7 therefore I 6 = 7 A Kirchhoff’s first law is consistent with the conservation of electrical charge while the voltage law is consistent with the law of conservation of energy. Some charge is received per unit time due to the currents arriving at a junction. For conservation of charge, same amount of charge must leave the junction per unit time which leads to the law of currents. Algebraic sum of emfs (energy per unit charge) corresponds to the electrical energy supplied by the source. According to the law of conservation of energy, this energy must appear in the form of electrical potential difference across the electrical elements/devices in the loop. This leads to the law of voltages. Consider an electrical network shown in Fig. 9.3. Consider the loop ABFGA in clockwise
216 9.3 Wheatstone Bridge: Resistance of a material changes due to several factors such as temperature, strain, humidity, displacement, liquid level, etc. Therefore, measurement of these properties is possible by measuring the resistance. Measurable values of resistance vary from a few milliohms to hundreds of mega ohms. Depending upon the resistance range (milliohm to tens of ohm, tens of ohm to hundreds of ohms, hundreds of ohm to mega ohm, etc.), various methods are used for resistance measurement. Wheatstone’s bridge is generally used to measure resistances in the range from tens of ohm to hundreds of ohms. Example 9.2: Two batteries of 7 volt and 13 volt and internal resistances 1 ohm and 2 ohm respectively are connected in parallel with a resistance of 12 ohm. Find the current through each branch of the circuit and the potential difference across 12-ohm resistance. Solutions: Let the currents passing through the two batteries be I 1 and I2 . Applying Kirchhoff second law to the loop AEFBA, fi ff 12ffl ffi fi ff 1 7 fl 0 1 2 1 I I I flfl 12 1 7 1 2 1 ff ffl I I fi fi ffi I ffiffi --- (1) For the loop CEFDC fi ff 12ffl ffi I I 1 2 fi ff 2 1 I 2flfl 3 0 fl 12 2 13 1 2 2 ff ffl I I fi fi ffi I ffiffi --- (2) From (1) and (2) 2 13 7 6 2 1 I I fi ff fi ff I I 1 2 2 6 ffiffi ffiffi fi ff Substituting I1ff value in (2) I A 2 85 38 2 237 = = . I I 1 2 2 6 ffiffi ffiffi fi ff I A 1 2 85 38 6 1 526 fi ff . ffl fi ffl I I fi ff1 2 I fi ffl1 526 2 237 0 711 . . AAA ff fi . Potential difference across 12 Ω resistance V I fi fi R V 0 711 12 . . ff fi 8 53 Example 9.3: For the given network, find the current through 4 ohm and 3 ohm. Assume that the cells have negligible internal resistance. Solution: Applying Kirchhoff first law At junction F, I I I 1 3 = 2 – fififiI I I 1 2 fi ff 3 --- (1) Applying Kirchhoff second law, (i) loop EFCDE, 3 4 10 0 2 1 I I fi ff ffl 4 3 10 1 2 I I fi ff --- (2) (ii) loop FABCF fi fi 4 3 ff ffl 5 0 3 2 I I 4 3 5 3 2 I I fi ff --- (3) From Eq. (1) and Eq. (2) 4 3 3 2 2 ff ffl I I fi fi I = 10 fi ff 3 4 fi ffl 4 10 2 3 2 I I I 4 7 10 3 2 I I fi ff --- (4) From Eq. (3) and Eq. (4) 10 5 2 I fi ff I A 2 fi ff . 0 5 Negative sign indicates that I 2 current flows from F to C From Eq. (2) 4 3 0 5 10 1I fi fiff ffl . ffi I A 1 = 2 1. 2 ∴ I I 3 1 I A 2 fi ff fi ffl 2 1. . 2 0 5 1 fi .62
217 The Wheatstone Bridge was originally developed by Charles Wheatstone (1802- 1875) to measure the values of unknown resistances. It is also used for calibrating measuring instruments, voltmeters, ammeters, etc. Four resistances P, Q, R and S are connected to form a quadrilateral ABCD as shown in the Fig. 9.4. A battery of emf ε along with a key is connected between the points A and C such that point A is at higher potential with respect to the point C. A galvanometer of internal resistance G is connected between points B and D. When the key is closed, current I flows through the circuit. It divides into I 1 and I 2 at point A. I1 is the current through P and I2 is the current through S. The current I 1 gets divided at point B. Let I g be the current flowing through the galvanometer. The currents flowing through Q and R are respectively (I 1 – I g ) and (I 1 + I g ), From the Fig. 9.4, I = I 1 + I 2 --- (9.3) Consider the loop ABDA. Applying Kirchhoff’s voltage law in the clockwise sense shown in the loop we get, – I 1 P – I g G + I2 S = 0 --- (9.4) Now consider loop BCDB, applying Kirchhoff’s voltage law in the clockwise sense shown in the loop we get, – (I 1 – I g ) Q + (I2 + I g ) R + I g G = 0 --- (9.5) A special case occurs when the current passing through the galvanometer is zero. In this case, the bridge is said to be balanced. Condition for the balance is I g = 0. This condition can be obtained by adjusting the values of P, Q, R and S. Substituting I g = 0 in Eq. (9.4) and Eq. (9.5) we get, – I 1 P + I2 S = 0 ∴ I1 P = I2 S --- (9.6) – I 1 Q + I2 R = 0 ∴ I1 Q = I2 R --- (9.7) Dividing Eq. (9.6) by Eq. (9.7), we get P Q S R = --- (9.8) This is the condition for balancing the Wheatstone bridge. If any three resistances in the bridge are known, the fourth resistance can be determined by using Eq. (9.8). Fig. 9.4 : Wheatstone bridge. From these three equations (Eq. (9.3), (9.4), (9.5) we can find the current flowing through any branch of the circuit. Example 9.4: At what value should the variable resistor be set such that the bridge is balanced? If the source voltage is 30 V find the value of the output voltage across XY, when the bridge is balanced. When the bridge is balanced PQ RS / / = Q P = S R/ 1 36 10 4 4 10 300 19 94 10 3 3 3 . . . fi fi fi ff fi ffl Total resistance of the arm ADC = 19940 + 4400 = 24340 Ω To find output voltage across XY: Potential difference across AC = I1 fi ff 24340 30 I A 1 30 24340 = Potential difference across X Y
218 Temporary contact with the wire AB can be established with the help of the jockey. A cell of emf ε along with a key and a rheostat are connected between the points A and B. A suitable resistance R is selected from resistance box. The jockey is brought in contact with AB at various points on the wire AB and the balance point (null point), D, is obtained. The galvanometer shows no deflection when the jockey is at the balance point. Let the respective lengths of the wire between A and D, and that between D and C be x and R . Then using the conditions for the balance, we get X R R R AD DB = where RAD and RDB are resistance of the parts AD and DB of the wire resistance of the wire. If l is length of the wire, ρ its specific resistance, and A its area of cross section then R A AD AD fi ff R A DB DB fi ff X R R R AD DC = fi ff ff x R A A / / ∴ X R x R = Therefore, X = x R R --- (9.9) Knowing R, x and R , the value of the unknown resistance can be determined. Application of Wheatstone bridge: Figure 9.4 is a basic circuit diagram of Wheatstone bridge, however, in practice the circuit is used in different manner. In all cases it is used to determine some unknown resistance. Few applications of Wheatstone bridge circuits are discussed in the following article. 9.3.1 Metre Bridge: Fig. 9.5: Metre bridge. Example 9.5: Two resistances 2 ohm and 3 ohm are connected across the two gaps of the metre bridge as shown in figure. Calculate the current through the cell when the bridge is balanced and the specific resistance of the material of the metre bridge wire. Given the resistance of the bridge wire is 1.49 ohm and its diameter is 0.12 cm. Solution: When the bridge is balanced, the resistances 2 and 3 ohm are in series and the total resistance is 5 ohm. Let R1 be the resistance of the wire =1.49 Ω, and R2 be the total resistance (2+3)=5 Ω AD = I1 ×19940 fi ff ffl ffi 30 19940 24340 24 58 / . fi V I 2 30 1360 300 30 1660 fi ff fi A So, Potential difference across AB= I V 2 1360 30 1660 fiff fiff 1360 24 58 . V V V V V V V V V out B D A B A D AB AD fifi fi ff fi ff ffl ffi ff ff ffl ffi fi ff = 24.58-24.58 = 0V Metre bridge (Fig. 9.5) consists of a wire of uniform cross section and one metre in length, stretched on a metre scale which is fixed on a wooden table. The ends of the wire are fixed below two L shaped metallic strips. A single metallic strip separates the two L shaped strips leaving two gaps, left gap and right gap. Usually, an unknown resistance X is connected in the left gap and a resistance box is connected in the other gap. One terminal of a galvanometer is connected to the terminal C on the central strip, while the other terminal of the galvanometer carries the jockey (J).
219 Observe and discuss Do you know? Remember this detect whether there is a current through the central branch. This is possible only by tapping the jokey. 1. Kelvin’s method to determine the resistance of galvanometer (G) by using meter bridge. R R R R R pfifi . . fi . ff fi ffl ff fi 1 2 1 2 1 49 5 1 49 5 1 15ffi The current through the cell = fi R A p ff ff 2 1 15 1 74 . . Specific resistance of the wire fi fi ff R r ffl l 2 l m fi fi 1 r R fi fi 0 12 2 , 0 06 1 49 . c . , m . ff fi ff ffl ffl ffi ffifl ffi ffl ffi R r l 2 2 2 6 1 49 3 14 0 06 10 1 1 68 10 . . . . m Source of errors. 1. The cross section of the wire may not be uniform. 2. The ends of the wire are soldered to the metallic strip where contact resistance is developed, which is not taken into account. 3. The measurements of x and R may not be accurate. To minimize the errors (i) The value of R is so adjusted that the null point is obtained to middle one third of the wire (between 34 cm and 66 cm) so that percentage error in the measurement of x and R are minimum and nearly the same. (ii) The experiment is repeated by interchanging the positions of unknown resistance X and known resistance box R. (iii) The jockey should be tapped on the wire and not slided. We use jockey to Applications: • The Wheatstone bridge is used for measuring the values of very low resistance precisely. • We can also measure the quantities such as galvanometer resistance, capacitance, inductance and impedance using a Wheatstone bridge. The galvanometer whose resistance (G) is to be determined is connected in one gap and a known resistance (R) in the other gap. Working : 1. A suitable resistance is taken in the resistance box. The current is sent round the circuit by closing the key. Without touching the jockey at any point of the wire, the deflection in the galvanometer is observed. 2. The rheostat is adjusted to get a suitable deflection Around (2/3)rd of range. 3. Now, the jockey is tapped at different points of the wire and a point of contact D for which, the galvanometer shows no change in the deflection, is found. 4. As the galvanometer shows the same deflection with or without contact Wheatstone bridge along with operational amplifier is used to measure the physical parameters like temperature, strain, etc.
220 Do you know? between the point B and D, these two points must be equipotential points. 5. The length of the bridge wire between the point D and the left end of the wire is measured. Let l g be the length of the segment of wire opposite to the galvanometer and l r be the length of the segment opposite to the resistance box. Calculation : Let RAD and RDC be the resistance of the two parts of the wire AD and DC respectively. Since bridge is balanced G R = R R AD DC fi ff R R AD DC l l g r fi ff G R l l g r fi ff G R 100 l - l g g {lg+lr= 100 cm} G R fi ff ffl ffi ffi fl l - l g g 100 Using this formula, the unknown resistance of the galvanometer can be calculated. 2. Post Office Box A post office box (PO Box) is a practical form of Wheatstone bridge as shown in the figure. It consists of three arms P, Q and R. The resistances in these three arms are adjustable. The two ratio arms P and Q contain resistances 10 ohm, 100 ohm and 1000 ohm each. The third arm R contains resistances from 1 ohm to 5000 ohm. The unknown resistance X forms the fourth resistance. There are two tap keys K1 and K2 . The resistances in the arms P and Q are fixed to desired ratio. The resistance in the arm R is adjusted so that the galvanometer shows no deflection. Now the bridge is balanced. The unknown resistance X R = Q P/ , where P and Q are the fixed resistances in the ratio arms and R is an adjustable known resistance. If L is the length of the wire and r is its radius then the specific resistance of the material of the wire is given by fi ff ffl X r L 2 Wheatstone Bridge for Strain Measurement: Strain gauges are commonly used for measuring the strain. Their electrical resistance is proportional to the strain in the device. In practice, the range of strain gauge resistance is from 30 ohms to 3000 ohms. For a given strain, the resistance change may be only a fraction of full range. Therefore, to measure small resistance changes with high accuracy, Wheatstone bridge configuration is used. The figure below shows the Wheatstone bridge where the unknown resistor is replaced with a strain gauge as shown in the figure. In these circuit, two resistors R1 and R2 are equal to each other and R3 is the variable resistor. With no force applied to the strain gauge, rheostat is varied and
221 finally positioned such that the voltmeter will indicate zero deflection, i.e., the bridge is balanced. The strain at this condition represents the zero of the gauge. If the strain gauge is either stretched or compressed, then the resistance changes. This causes unbalancing of the bridge. This produces a voltage indication on voltmeter which corresponds to the strain change. If the strain applied on a strain gauge is more, then the voltage difference across the meter terminals is more. If the strain is zero, then the bridge balances and meter shows zero reading. This is the application of precise resistance measurement using a Wheatstone bridge. Fig. 9.6: Potentiometer. Fig. 9.7: Emf comparison by individual method. 9.4 Potentiometer: A voltmeter is a device which is used for measuring potential difference between two points in a circuit. An ideal voltmeter which does not change the potential difference to be measured, should have infinite resistance so that it does not draw any current. Practically, a voltmeter cannot be designed to have an infinite resistance. Potentiometer is one such device which does not draw any current from the circuit. It acts as an ideal voltmeter. It is used for accurate measurement of potential difference. 9.4.1 Potentiometer Principle: A potentiometer consists of a long wire AB of length L and resistance R having uniform cross sectional area A. (Fig. 9.6) A cell of emf ε having internal resistance r is connected across AB as shown in the Fig. 9.6. When the circuit is switched on, current I passes through the wire. Current through AB, I = fi R r ff Potential difference across AB is VAB = I R VAB = fi R ( ) R r ff Therefore, the potential difference per unit length of the wire is, V L AB = fi R L R( ) ff r As long as ε remains constant, V L AB will remain constant. V L AB is known as potential gradient along AB and is denoted by K. Potential gradient can be defined as potential difference per unit length of wire. Consider a point C on the wire at distance from the point A, as shown in the figure. The potential difference between A and C is VAC. Therefore, VAC = K i.e. VAC ∝ Thus, the potential difference between two points on the wire is directly proportional to the length of the wire between them provided the wire is of uniform cross section, the current through the wire is the same and temperature of the wire remains constant. Uses of potentiometer are discussed below. 9.4.2 Use of Potentiometer: A) To Compare emf. of Cells Method I : A potentiometer circuit is set up by connecting a battery of emf ε , with a key K and a rheostat such that point A is at higher
222 potential than point B. The cells whose emfs are to be compared are connected with their positive terminals at point A and negative terminals to the extreme terminals of a twoway key K1 K2 . The central terminal of the two ways key is connected to a galvanometer. The other end of the galvanometer is connected to a jockey (J). (Fig. 9.7) Key K is closed and then, key K1 is closed and key K2 is kept open. Therefore, the cell of emf ε 1 comes into circuit. The null point is obtained by touching the jockey at various points on the potentiometer wire AB. Let 1 be the length of the wire between the null point and the point A. 1 corresponds to emf ε 1 of the cell. Therefore, ε 1 = K1 where K is the potential gradient along the potentiometer wire. Now key K1 is kept open and key K2 is closed. The cell of emf ε 2 now comes in the circuit. Again, the null point is obtained with the help of the Jockey. Let 2 be the length of the wire between the null point and the point A. This length corresponds to the emf ε 2 of the cell. ∴ ε 2 = K 2 From the above two equations we get fi fi 1 2 1 2 ff --- (9.10) Thus, we can compare the emfs of the two cells. If any one of the emfs is known, the other can be determined. Method II: The emfs of cells can be compared also by another method called sum and difference method. When two cells are connected so that the positive terminal of the first cell is connected to the negative terminal of the second cell as shown in Fig 9.8 (a). The emf of the two cells are added up and the effective emf of the combination of two cells is ε 1 + ε 2 . This method of connecting two cells is called the sum method. When two cells are connected so that their negative terminals are together or their positive terminals are connected together as shown in Fig. 9.8 (b). In this case their emf oppose each other and effective emf of the combination of two cells is ε 1 – ε 2 (ε 1 > ε 2 assumed). This method of connecting two cells is called the difference method. Remember that this combination of cells is not a parallel combination of cells. Circuit is connected as shown in Fig.9.9. When keys K1 and K3 are closed the cells ε 1 and ε 2 are in the sum mode. The null point is obtained using the jockey. Let 1 be the length of the wire between the null point and the point A. This corresponds to the emf (ε 1 + ε 2 ). ∴ ε 1 + ε 2 = k 1 Now the key K1 and K3 are kept open and keys K2 and K4 are closed. In this case the two cells are in the difference mode. Again the null point is obtained. Let 2 be the length of the wire between the null point and the point A. This corresponds to ε 1 - ε 2 ∴ ε 1 - ε 2 = 2 Fig. 9.8 (a):Sum method. Fig. 9.9: Emf comparison, sum and difference method. Fig. 9.8 (b): Difference method.
223 From the above two equations, fi fi fi fi 1 2 1 2 1 2 ff ffl ffi By componendo and dividendo method, we get, fi fi 1 2 1 2 1 2 ff ffl ffi --- (9.11) Thus, emf of two cells can be compared. B) To Find Internal Resistance (r) of a Cell: The experimental set up for this method consists of a potentiometer wire AB connected in series with a cell of emf ε , the key K1 , and rheostat as shown in Fig. 9.10. The terminal A is at higher potential than terminal B. A cell of emf ε 1 whose internal resistance r1 is to be determined is connected to the potentiometer wire through a galvanometer G and the jockey J. A resistance box R is connected across the cell ε 1 through the key K2. The length of the wire 2 between the null point and point A is measured. This corresponds to the voltage between the null point and point A. ∴ V k = 2 ∴ fi 1 1 2 1 V 2 k k ff ff Consider the loop PQSTP. ε 1 = IR + Ir and V = IR ∴ fi 1 1 V 2 IR Ir IR R r R ff ffl ff ffl ff ∴ r R fi ff ffl ffi fl 1 2 1 --- (9.12) This equation gives the internal resistance of the cell. C) Application of potentiometer: The applications of potentiometer discussed above are used in laboratory. Some practical applications of potentiometer are given below. 1) Voltage Divider: The potentiometer can be used as a voltage divider to continuously change the output voltage of a voltage supply (Fig. 9.11). As shown in the Fig. 9.11, potential V is set up between points A and B of a potentiometer wire. One end of a device is connected to positive point A and the other end is connected to a slider that can move along wire AB. The voltage V divides in proportion of lengths l 1 and l 2 as shown in the figure 9.11. Fig. 9.10 : Internal resistance of a cell. The key K1 is closed and K2 is open. The circuit now consists of the cell ε , cell ε 1 , and the potentiometer wire. The null point is then obtained. Let 1 be length of the potentiometer wire between the null point and the point A. This length corresponds to emf ε 1 . ∴ ε 1 = k 1 where k is potential gradient of the potentiometer wire which is constant. Now both the keys K1 and K2 are closed so that the circuit consists of the cell ε , the cell ε 1 , the resistance box, the galvanometer and the jockey. Some resistance R is selected from the resistance box and null point is obtained. 2) Audio Control: Sliding potentiometers, are commonly used in modern low-power audio systems as audio control devices. Both sliding Fig. 9.11 : Potentiometer as a voltage divider.
224 difference of the order 10–6 volt can be measured with it. Least count of a potentiometer is much better compared to that of a voltmeter. Demerits: Potentiometer is not portable and direct measurement of potential difference or emf is not possible. 9.5 Galvanometer: A galvanometer is a device used to detect weak electric currents in a circuit. It has a coil pivoted (or suspended) between concave pole faces of a strong laminated horse shoe magnet. When an electric current passes through the coil, it deflects. The deflection is proportional to the current passing through the coil. The deflection of the coil can be read with the help of a pointer attached to it. Position of the pointer on the scale provided indicates the current passing through the galvanometer or the potential difference across it. Thus, a galvanometer can be used as an ammeter or voltmeter with suitable modification. The galvanometer coil has a moderate resistance (about 100 ohms) and the galvanometer itself has a small current carrying capacity (about 1 mA). 9.4.3 Advantages of a Potentiometer Over a Voltmeter: Merits: i) Potentiometer is more sensitive than a voltmeter. ii) A potentiometer can be used to measure a potential difference as well as an emf of a cell. A voltmeter always measures terminal potential difference, and as it draws some current, it cannot be used to measure an emf of a cell. iii) Measurement of potential difference or emf is very accurate in the case of a potentiometer. A very small potential Example 9.7: In an experiment to determine the internal resistance of a cell of emf 1.5 V, the balance point in the open cell condition at is 76.3 cm. When a resistor of 9.5 ohm is used in the external circuit of the cell the balance point shifts to 64.8 cm of the potentiometer wire. Determine the internal resistance of the cell. Solution: Open cell balancing length l 1 = 76.3 cm Closed circuit balancing length l 2 = 64.8 cm External resistance R = 9.5 Ω Internal resistance r l l l fi R ffl ff ffi fl fi ffl ff ffi fl fi 1 2 2 76 3 64 8 64 8 9 5 1 686 . . . . . Fig. 9.12 Internal structure of galvanometer. 9.5.1 Galvanometer as an Ammeter: Let the full scale deflection current and the resistance of the coil G of moving coil galvanometer (MCG ) be I s and G. It can be converted into an ammeter, which is a current measuring instrument. It is always connected in series with a resistance R through which the current is to be measured. (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals. 3) Potentiometer as a senor: If the slider of a potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes changes in potential which is further amplified using an amplifier circuit. The potential difference is calibrated in terms of the displacement of the moving part.
225 To convert a moving coil galvanometer (MCG ) into an ammeter To convert an MCG into an ammeter, the modifications necessary are 1. Its effective current capacity must be increased to the desired higher value. 2. Its effective resistance must be decreased. The finite resistance G of the galvanometer when connected in series, decreases the current through the resistance R which is actually to be measured. In ideal case, an ammeter should have zero resistance. 3. It must be protected from the possible damages, which are likely due to the passage of an excess electric current to be passed. In practice this is achieved by connecting a low resistance in parallel with the galvanometer, which effectively reduces the resistance of the galvanometer. This low resistance connected in parallel is called shunt (S). This arrangement is shown in Fig. 9.13. Uses of the shunt: a. It is used to divert a large part of total current by providing an alternate path and thus it protects the instrument from damage. b. It increases the range of an ammeter. c. It decreases the resistance between the points to which it is connected. The shunt resistance is calculated as follows. In the arrangement shown in the figure, I g is the current through the galvanometer. Therefore, the current through S is (I – I g ) = I s ∴ GI g = S (I – I g ) fi ff ffl ffi fl S I I I G g g --- (9.13) Equation 9.13 is useful to calculate the range of current that the galvanometer can measure. (i) If the current I is n times current I g , then I = n I g . Using this in the above expression we get S GI nI I g g g fi ff OR S G n fi ff1 This is the required shunt to increase the range n times. (ii) Also if I s is the current through the shunt resistance, then the remaining current (I – I s ) will flow through galvanometer. Hence G (I – Is ) = S Is i.e. G I – G Is = S Is i.e. S Is + G Is = G I fi ff ffl ffi fl I I G S G s This equation gives the fraction of the total current through the shunt resistance. Example 9.8: A galvanometer has a resistance of 100 Ω and its full scale deflection current is 100 µ A. What shunt resistance should be added so that the ammeter can have a range of 0 to 10 mA ? Solution: Given I G = 100 µ A = 0.1 mA The upper limit gives the maximum current to be measured, which is I = 10 mA . The galvanometer resistance is G = 100 Ω. Now n s G n fi fi ff fi ffl fi ffl fi 10 0 1 100 1 100 100 1 10 0 . 99 ffi Example 9.9: What is the value of the shunt resistance that allows 20% of the main current through a galvanometer of 99 Ω? Solution: Given G = 99 Ω and I G =(20/100)I = 0.2 I Now S I G I I I I I G G fi ff fi ffl ffi fl ff fi ffl fi . fi . . . . 0 2 99 0 2 0 2 99 0 8 24 75 Fig. 9.13 Ammeter. Since S and G are parallel, ∴ GI g = S Is
226 9.5.2 Galvanometer as a Voltmeter: A voltmeter is an instrument used to measure potential difference between two points in an electrical circuit. It is always connected in parallel with the component across which voltage drop is to be measured. A galvanometer can be used for this purpose. To Convert a Moving Coil Galvanometer into a Voltmeter. To convert an MCG into a Voltmeter the modifications necessary are: 1. Its voltage measuring capacity must be increased to the desired higher value. 2. Its effective resistance must be increased, and 3. It must be protected from the possible damages, which are likely due to excess applied potential difference. All these requirements can be fulfilled, if we connect a resistance of suitable high value (X) in series with the given MCG. A voltmeter is connected across the points where potential difference is to be measured. If a galvanometer is used to measure voltage, it draws some current (due to its low resistance), therefore, actual potential difference to be measured decreases. To avoid this, a voltmeter should have very high resistance. Ideally, it should have infinite resistance. Fig. 9.14 : Voltmeter. Example 9.10: A Galvanometer has a resistance of 25 Ω and its full scale deflection current is 25 µA. What resistance should be added to it to have a range of 0 -10 V? Solution: Given G = 25 µA. Maximum voltage to be measured is V =10 V. The Galvanometer resistance G = 25 Ω. The resistance to be added in series, X V I G G fi ff fi ffl ff fi ffl ff 10 25 10 25 399 975 10 6 3 . ffi Example 9.11: A Galvanometer has a resistance of 40 Ω and a current of 4 mA is needed for a full scale deflection . What is the resistance and how is it to be connected to convert the galvanometer (a) into an ammeter of 0.4 A range and (b) into a voltmeter of 0.5 V range? Solution: Given G = 40 Ω and I G = 4 mA (a) To convert the galvanometer into an ammeter of range 0.4 A, ff ffl I I fi G G S I ffi G ff ffl 0 4. . fi 0 004 0 004 40 S ffi fl . S fi ff fi fi 0 004 40 0 396 0 16 0 396 0 4040 . . . . . fflfflffi (b)To convert the galvanometer into a voltmeter of range of 0.5 V V I fi ff G ffl ffi G X 0 5. . fi ff 0 004 40 ffl ffi X X fi ff fi 0 5 0 004 40 85 . . ffl A very high resistance X is connected in series with the galvanometer for this purpose as shown in Fig. 9.14. The value of the resistance X can be calculated as follows. If V is the voltage to be measured, then V = I g X + I g G. ∴ I g X = V – I g G fi ff X ffl V I G g , --- (9.14) where I g is the current flowing through the galvanometer. Eq. (9.14) gives the value of resistance X. If n V V V g g I G V fi fi ( ) ff is the factor by which the voltage range is increased, it can be shown that X = G (nv -1)
227 Comparison of an ammeter and a voltmeter: AMMETER VOLTMETER 1. It measures current. 2. It is connected in series. 3. It is an MCG with low resistance. (Ideally zero) 4. Smaller the shunt, greater will be the current measured. 5. Resistance of ammeter is R S G S G G n A fi ff ffl fi 1. It measures potential difference 2. It is connected in parallel. 3. It is an MCG with high resistance. (Ideally infinite) 4. Larger its r e s i s t a n c e , greater will be the potential difference measured. 5. Resistance of voltmeter is R G X G n V fi ff fi ffl V THERMOELECTRICITY When electric current is passed through a resistor, electric energy is converted into thermal energy. The reverse process, viz., conversion of thermal energy directly into electric energy was discovered by Seebeck and the effect is called thermoelectric effect. Seebeck Effect If two different metals are joined to form a closed circuit (loop) and these junctions are kept at different temperatures, a small emf is produced and a current flows through the metals. This emf is called thermo emf this effect is called the Seebeck effect and the pair of dissimilar metals forming the junction is called a thermocouple. An antimony-bismuth thermo-couple is shown in a diagram. For this thermo couple the current flows from antimony to bismuth at the cold junction. (ABC rule). For a copper-iron couple (see diagram) the current flows from copper to iron at the hot junction, This effect is reversible. The direction of the current will be reversed if the hot and cold junctions are interchanged. The thermo emf developed in a thermocouple when the cold junction is at 00 C and the hot junction is at T°C is given by fi ffffi fl T Tffl 1 2 2 Here α and b are called the thermoelectric constants. This equation tells that a graph showing the variation of ε with temperature is a parabola. Do you know? Accelerator in India: Cyclotron for medical applications. Picture credit: Director, VECC, Kolkata, Department of Atomic Energy, Govt. of India
228 1. Choose the correct option. i) Kirchhoff’s first law, i.e., ΣI = 0 at a junction, deals with the conservation of (A) charge (B) energy (C) momentum (D) mass ii) When the balance point is obtained in the potentiometer, a current is drawn from (A) both the cells and auxiliary battery (B) cell only (C) auxiliary battery only (D) neither cell nor auxiliary battery iii) In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and B is (A) 2 Ω (B) 4 Ω (C) 8 Ω (D) 16 Ω 2. Answer in brief. i) Define or describe a Potentiometer. ii) Define Potential Gradient. iii) Why should not the jockey be slided along the potentiometer wire? iv) Are Kirchhoff’s laws applicable for both AC and DC currents? v) In a Wheatstone’s meter-bridge experiment, the null point is obtained in middle one third portion of wire. Why is it recommended? vi) State any two sources of errors in meterbridge experiment. Explain how they can be minimized. vii) What is potential gradient? How is it measured? Explain. viii) On what factors does the potential gradient of the wire depend? ix) Why is potentiometer preferred over a voltmeter for measuring emf? x) State the uses of a potentiometer. xi) What are the disadvantages of a potentiometer? xii) Distinguish between a potentiometer and a voltmeter. xiii) What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is (i) increased (ii) decreased. 3. Obtain the balancing condition in case of a Wheatstone’s network. 4. Explain with neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge. 5. Describe Kelvin’s method to determine the resistance of a galvanometer by using a meter bridge. 6. Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually. (A) infinite (B) zero (C) 2 Ω (D) 1.5 Ω iv) Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge (A) 10 Ω (B) 15 Ω (C) 20 Ω (D) 30 Ω v) A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery. (A) 0.5 A (B) 1A (C) 2A (D) 3A vi) To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of 4 Ω is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end? Exercises
229 7. Describe how a potentiometer is used to compare the emfs of two cells by combination method. 8. Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula. 9. On what factors does the internal resistance of a cell depend? 10. A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance. [Ans: 1 A] 11. Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances of 1 Ω and 2 Ω are connected in parallel so as to send current in same direction through an external resistance of 5 Ω. Find the current through the external resistance. [Ans: 5/17 A] 12. A voltmeter has a resistance 30 Ω. What will be its reading, when it is connected across a cell of emf 2 V having internal resistance 10 Ω? [Ans: 1.5 V] 13. A set of three coils having resistances 10 Ω, 12 Ω and 15 Ω are connected in parallel. This combination is connected in series with series combination of three coils of the same resistances. Calculate the total resistance and current through the circuit, if a battery of emf 4.1 Volt is used for drawing current. [Ans: 0.1 A] 14. A potentiometer wire has a length of 1.5 m and resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire. [Ans: 0.0178 V/cm] 15. When two cells of emfs. ε 1 and ε 2 are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells. [Ans: 1.25] 16. The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell. [Ans: r = 2 Ohm] 17. A potential drop per unit length along a wire is 5 x 10-3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell. [Ans: 0.01080 V] 18. The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1µV/mm? [Ans: 15992 ohm] 19. Find the equivalent resistance between the terminals of A and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm. I2 I-I2 I-I2 E I 1 -I2 I1 [Ans: 14 Ohm] 20. A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Ω? [Ans: 1.66 V]
230 Try this Do you know? 10.1 Introduction: In this Chapter you will be studying how magnetic fields are produced by an electric current. Important foundation for further developments will also be laid down. Hans Christian Oersted first discovered that magnetic field is produced by an electric current passing through a wire. Later, Gauss, Henry, Faraday and others showed that magnetic field is an important partner of electric field. Maxwell’s theoretical work highlighted the close relationship of electric and magnetic fields. This resulted into several practical applications in day today life, for example electrical motors, generators, communication systems and computers. In electrostatics, we have considered static charges and the force exerted by them on other charge or test charge. We now consider forces between charges in motion. You have learnt in Xth Std. that if a magnetic needle is held in close proximity of a current carrying wire, it shows the direction of magnetic field circling around the wire. Imagine that a current carrying wire is grabbed with your right hand with the thumb pointing in the direction of the current, then your fingers curl around in the direction of the magnetic field (Fig. 10.2). 10. Magnetic Fields due to Electric Current You must have noticed high tension power transmission lines, the power lines on the big tall steel towers. Strong magnetic fields are created by these lines. Care has to be taken to reduce the exposure levels to less than 0.5 milligauss (mG). You can show that wires having currents passing through them, (a) in opposite directions repel and (b) in the same direction attract. Hang two conducting wires from an insulating support. Connect them to a cell first as shown in Fig. 10.1 (a) and later as shown Fig. 10.1 (b), with the help of binding posts. You will notice that the wires in (a) repel each other and those in (b) come closer, i.e., they attract each other as soon as the current starts. The force in this experiment is certainly not of electrostatic origin, even through the current is due to the electrons flowing in the wires. The overall charge neutrality is maintained throughout the wire, hence the electrostatic forces are ruled out. Fig. 10.1 (a) Fig. 10.1 (b) Can you recall? • Do you know that a magnetic field is produced around a current carrying wire? • What is right hand rule? • Can you suggest an experiment to draw magnetic field lines of the magnetic field around the current carrying wire? • Do you know solenoid? Can you compare the magnetic field due to a current carrying solenoid with that due to a bar magnet?
231 How can one account for the force on the neighbouring current carrying wire? The magnetic field due to current in the wire 1 at any point on wire 2 is directed into the plane of the paper. The electrons flow in a direction opposite to the conventional current. Then the wire 2 experiences a force F fiff towards wire 1. 10.2 Magnetic Force: From the above discussion and Fig. 10.3, you must have realized that the directions of v , B fiff and F fiff follow a vector cross product relationship. Actually the magnetic force Fm on an electron with a charge -e, moving with velocity v in a magnetic field B fiff is Fm fiff = -e( v × B fiff ) --- (10.1) In general for a charge q, the magnetic force will be Fm fiff = q( v × B fiff ) --- (10.2) If both electric field E fiff and the magnetic field B fiff are present, the net force on charge q moving with the velocity v in F fiff = q[ E fiff +( v × B fiff )] --- (10.3) = q E fiff +q( v × B fiff ) = Fe fiff + Fm fiff --- (10.4) Justification for this law can be found in experiments such as the one described in Fig. 10.1 (a) and (b). The force described in Fig. (10.4) is known as Lorentz force. Here Fe fiff is the force due to electric field and Fm fiff is the force due to magnetic field. There are interesting consequences of the Lorentz force law. (i) If the velocity v of a charged particle is parallel to the magnetic field B fiff , the magnetic force is zero. Fig. 10.4: The cross product is in the direction of the unit vector perpendicular to both v and B fiff . Thus the vectors v and F fiff are always perpendicular to each other. Hence. F fiff . v = 0, for any magnetic field B fiff . Magnetic force Fm fiff is thus perpendicular to the displacement and hence the magnetic force never does any work on moving charges. The magnetic forces may change the direction of motion of a charged particle but they can never affect the speed. Interestingly, Eq. (10.2) leads to the definition of units of B fiff . From Eq. (10.2), F fiff = q | v × B fiff | = qvB sin θ , --- (10.5) where θ is the angle between v and B fiff and is unit vector in the direction of force . If the force F is 1 N acting on the charge of 1 C moving with a speed of 1m s-1 perpendicular to B fiff , then we can define the unit of B. . . . B F q = v ∴unit of B is N s C m . . . Dimensionally, [B] = [F/qv] (ii) If the charge is stationary, v =0, the force = 0, even if B fiff ≠ 0. From Eq. (10.4) it may be observed that the force on the charge due to electric field depends on the strength of the electric field and the magnitude of the charge. However, the magnetic force depends on the velocity of the charge and the cross product of the velocity vector v the magnetic field vector B fiff , and the charge q. Consider the vectors v and B fiff with certain angle between them. Then v × B fiff will be a vector perpendicular to the plane containing the vectors v and B fiff (Fig. 10.4). Fig. 10.2: Right hand thumb rule. Fig. 10.3: Force on wire 2 due to current in wire 1. I
232 Do you know? Do you know? Magnetic Resonance Imaging (MRI) technique used for medical imaging requires a magnetic field with a strength of 1.5 T and even upto 7 T. Nuclear Magnetic Resonance experiments require a magnetic field upto 14 T. Such high magnetic fields can be produced using superconducting coil electromagnet. On the other hand, Earth’s magnetic field on the surface of the Earth is about 3.6 ×10-5 T = 0.36 gauss. 10.3 Cyclotron Motion: In a magnetic field, a charged particle typically undergoes circular motion. Figure 10.5 shows a uniform magnetic field directed perpendicularly into the plane of the paper (parallel to the -ve z axis). Fig. (a) Fig. (b) Fig. (c) Can you recall? Electromagnetic crane: How does it work? This SI unit is called tesla (T) 1 T = 104 gauss. Gauss is not an SI unit, but is used as a convenient unit. Example 10.1: A charged particle travels with a velocity v through a uniform magnetic field B fiff as shown in the following figure, in three different situations. What is the direction of the magnetic force Fm fiff due to the magnetic field, on the particle? B B B Solution: In Fig. (a), the direction of the vector v × B fiff will be in the positive y direction. Hence Fm fiff will be in the positive y direction. In Fig. (b) v × B fiff will be in the positive x direction. Hence the force Fm will be in the same direction. In Fig. (c) v and B fiff are antiparallel, the angle between them is 180° and because sin 180° = 0, Fm fiff will be equal to zero. B F R x y Fig. 10.5: Charged particle moving in a magnetic field. Figure 10.5 shows a particle with charge q moving with a speed v, and a uniform magnetic field B fiff is directed into the plane of the paper. According to the Lorentz force law, the magnetic force on the particle will act towards the centre of a circle of radius R, and this force will provide centripetal force to sustain a uniform circular motion. Thus q B m R v v = 2 --- (10.6) ∴mv = p = qBR --- (10.7) Equation (10.7) represents what is known as cyclotron formula. It describes the circular motion of a charged particle in a particle accelerator, the cyclotron. Let us look at a charged particle which is moving in a circle with a constant speed. This is uniform circular motion that you have studied earlier. Thus, there must be a net force acting on the particle, directed towards the centre of the circle. As the speed is constant, the force also must be constant, always perpendicular to the velocity of the particle at any given instant of time. Such a force is provided by the uniform magnetic field B fiff perpendicular to the plane of the circle along which the charged particle moves.
233 Remember this 10.3.1 Cyclotron Accelerator: Particle accelerators have played a key role in providing high energy (MeV to GeV) particle beams useful in studying particlematter interactions and some of these are also useful in medical treatment of certain tumors/ diseases. The Cyclotron is a charged particle accelerator, accelerating charged particles to high energies. It was invented by Lawrence and Livingston in the year 1934 for the purpose of studying nuclear structure. Both electric as well as magnetic fields are used in a Cyclotron, in combination. These are applied in directions perpendicular to each other and hence they are called crossed fields. The magnetic field puts the particle (ion) into circular path and a high frequency electric field accelerates it. Frequency of revolution of a charged particle is independent of its energy, in a magnetic field. This fact is used in this machine. Cyclotron consists of two semicircular disc-like metal chambers, D1 and D2 , called the dees (Ds). Figure 10.6 shows a schematic diagram of a cyclotron. A uniform magnetic field B is applied perpendicular to plane of the Ds. This magnetic field is produced using an electromagnet producing a field upto 1.5 T. An alternating voltage upto 10000 V at high frequency, 10 MHz (f a ), is applied between the two Ds. Positive ions are produced by a gas ionizing source kept at the point O in between the two Ds. The electric field provides acceleration to the charged particle (ion). Once the ion in emitted, it accelerates due to the negative voltage of a D and performs a semi circular motion within the D. Whenever the ion moves from one D to the other D, it accelerates due to the potential difference between the two Ds and again performs semicircular motion in the other D. Thus the ion is acted upon by the electric field every time it moves from one D to the other D. As the electric field is alternating, its sign is changed in accordance with the circular motion of the ion. Hence the ion is always accelerated, its energy increases and the radius of its circular path also increases, making the entire path a spiral (See Fig. 10.6). Consider an ion source placed at P. An ion moves in a semi circular path in one of the Ds and reaches the gap between the two Ds in a time interval T/2, T being the period of a full revolution. Using the Cyclotron formula Eq. (10.7), mv = qBR, where q is the change on the ion. fi ff ff ff T R m R qBR m qB 2 2 2 ffl ffl ffl v --- (10.8) The frequency of revolution (Cyclotron frequency) is Fig 10.6: Schematic diagram of a Cyclotron with the two Ds. A uniform magnetic field B fiff is perpendicular to the plane of the paper, coming out. The ions are injected into the D at point P. An alternating voltage is supplied to the Ds. The entire assembly is placed in a vacuum chamber. Field penetrating into the paper is represented as ⊗, while that coming out of the paper is shown by .
234 Internet my friend Do you know? f T qB m c fi fi 1 2ff --- (10.9) The frequency of the applied voltage (f a ) between the two Ds is adjusted so that polarity of the two Ds is reversed as the ion arrives at the gap after completing one semi circle. This condition f a = f c is the resonance condition. The ions do not experience any electric field while they travel within the D. Their kinetic energy increases by eV every time they cross over from one D to the other. Here V is the voltage difference across the gap. The ions move in circular path with successively larger and larger radius to a maximum radius at which they are deflected by a magnetic field so that they can be extracted through an exit slit. From Eq. (10.7), v exit = qB m R , where Rexit is the radius of the path at the exit. The kinetic energy of the ions/ protons will be K.E. = 1 2 mv2 = q B R m exit 2 2 2 2 --- (10.10) Thus the final energy is proportional to the square of the radius of the outermost circular path (Rexit). 10.4 Helical Motion: So far it has been assumed that the charged particle moves in a plane perpendicular to magnetic field B fiff . If such a particle has some component of velocity parallel to B fiff , ( v // ) then it leads to helical motion. Since a component v // is parallel to B fiff , the magnetic force Fm fiff will be: Particle accelerators are important for a variety of research purposes. Large accelerators are used in particle research. There have been several accelerators in India since 1953. The Department of Atomic Energy (DAE), Govt. of India, had taken initiative in setting up accelerators for research. Apart from ion accelerators, the DAE has developed and commissioned a 2 GeV electron accelerator which is a radiation source for research in science. This accelerator, 'Synchrotron', is fully functional at Raja Ramanna Centre for Advanced Technology, Indore. An electron accelerator, Microtron with electron energy 8-10 MeV is functioning at Physics Department, Savitribai Phule Pune University, Pune. (i) Existing and upcoming particle accelerators in India http://www. researchgate.net (ii) Search the internet for particle accelerators and get more information. 10.5 Magnetic Force on a Wire Carrying a Current: We have seen earlier the Lorentz force law (Eq. (10.4)). From this equation, we can obtain the force on a current carrying wire. (i) Straight wire: Consider a straight wire of length L as shown in Fig. 10.8. An external magnetic field B fiff is applied perpendicular to the wire, coming Fig. 10.7: Helical Motion of a charged particle in a magnetic field B fiff . Fm fiff = v // × B fiff =v.B sin (0°) = 0 --- (10.11) Thus, v // will not be affected and the particle will move along the direction of B fiff . At the same time the perpendicular component of the velocity ( v ⊥) leads to circular motion as stated above. As a result, the particle moves parallel to the field B fiff while moving along a circular path perpendicular to B fiff . Thus the path becomes a helix (Fig. 10.7). z x y B
235 Fig. 10.9: Wire with arbitrary shape. extended to a wire of arbitrary shape as shown in Fig. 10.9. Fig. 10.8 Electrons in the wire having drift velocity v d experience a magnetic force Fm fiff upwards as the applied magnetic field lines come out of the plane of the paper. out of the plane of the paper. Let a current I flow through the wire under an applied potential difference. If v d is the drift velocity of conduction electrons in the part of length L of the wire, the charge q flowing across the plane pp in time t will be q = I t q = IL d v --- (10.12) The magnetic force Fm fiff on this charge, according to Eq. (10.2), due to the applied magnetic field B fiff is given by F q B IL B n IL B n m d d d fi fffi ff fiff ffl ffl fi ff fi ffl fi ffi ( ) sin . .sin v v v fl 90 , where is a unit vector perpendicular to both B fiff and v d , in the direction of Fm fiff Fm fiff = ILB --- (10.13) This is, therefore, the magnetic force acting on the portion of the straight wire having length L. If B fiff is not perpendicular to the wire, then the above Eq. (10.13) takes the form F fiff m = I L fiff × B fiff , --- (10.14) where L fiff is the length vector directed along the portion of the wire of length L. (ii) Arbitrarily shaped wire: In the previous section we considered a straight wire. Equation (10.14) can be Consider a segment of infinitesimal length dl along the wire. If I in the current flowing, using Eq. (10.14), the magnetic force due to perpendicular magnetic field B fiff (coming out of the plane of the paper) is given by d Fm fiff = I d l × B fiff --- (10.15) The force on the total length of wire is thus F d m m F I dl B fiff fiff ff fiff fi fi ff ffl ffl --- (10.16) If B fiff is uniform over the whole wire, F I m dl B fiff ff fiff fi ff ffl ffi fl --- (10.17) Fm I Example 10.2: A particle of charge q follows a trajectory as shown in the figure. Obtain the type of the charge (positive or negatively charged). Obtain the momentum p of the particle in terms of B, L, s, q, s being the distance travelled by the particle. Particle trajectory: A uniform magnetic field B fiff is applied in the region pp, perpendicular to the plane of the paper, coming out of the plane of the paper. Solution: B fiff is coming out of the paper. Since the particle moves upwards, there must be a force in that direction. The velocity is in the positive x direction. Fm L I e Vd
236 10.7 Torque on a Current Loop: It will be very interesting to apply the results of the above sections to a current carrying loop of a wire. You have learnt about an electric motor in Xth Std. An electric motor works on the principle you have studied in the preceding sections, i.e., the magnetic force on a current carrying wire due to a magnetic field. Figure 10.10 shows a current carrying loop (abcd) in a uniform magnetic field. There will, therefore, be the magnetic forces Fm fiff acting in opposite directions on the segments of the loop ab and cd. This results into rotation of the loop about its central axis. Without going into the details of contact carbon brushes and external circuit, we can visualize the rotating action of a motor. 10.6 Force on a Closed Circuit in a Magnetic Field B fiff : Equation (10.17) can be extended to a closed wire circuit C F I m dl B C fiff ff fiff ffl fi ff ffl --- (10.18) Here, the integral is over the closed circuit C. For uniform B fiff , F I m dl B C fiff ff fiff ffl fi ff ffl ffi fl --- (10.19) The term in the bracket in Eq. (10.19) is the sum of vectors along a closed circuit. Hence it must be zero. ∴ Fm fiff = 0 ( B fiff uniform) --- (10.20) ∴ v × B fiff is in -ve y direction. As the force is in +y direction, i.e., opposite, the charge must be negative. According to Eq (10.5), Force = Bqv in the y direction. ∴ acceleration = Bq m v , where m is the mass of the particle. Using Newton’s equation of motion, the distance travelled in the y direction is given by s = ut + 1 2 a t2 = 0 + 1 2 Bq m v t 2 as the initial velocity in the y direction is zero.But in the same time t, the particle travels the distance L along the x direction, with uniform velocity v . ∴ L = v.t ∴ s = 1 2 2 BqL mv ∴ momentum p = mv = 1 2 Bq 2 s L Example 10.3: Consider a square loop of wire loaded with a glass bulb of mass m hanging vertically, suspended in air with its one part in a uniform magnetic field B fiff with its direction coming out of the plane of the paper (). Due to the current I flowing through the loop, there is a magnetic force in upward direction. Calculate the current I in the loop for which the magnetic force would be exactly balanced by the force on mass m due to gravity. Solution: The current I in the loop with its part in the magnetic field B causes an upward force Fm in the horizontal part of the loop, given by Fm = IBa, where a is the length of one arm of the loop. This force is balanced by the force due to gravity. ∴ Fm = I Ba = mg ∴ I mg Ba = For this current, the wire loop will hang in air. a
237 Fig. 10.10: A current loop in a magnetic field: principle of a motor. Now we can calculate the net force and the net torque on the loop in a situation depicted in Fig. 10.12 (a) and (b). Let us obtain the forces on all sides of the loop. The force F fiff 4 on side 4 (bc) will be F fiff 4 = Il2 B sin (90-θ) --- (10.21) The force F fiff 2 on side 2 (da) will be equal and opposite to F fiff 4 and both act along the same line. Thus, F fiff 2 and F fiff 4 will cancel out each other. The magnitudes of forces F fiff 1 and F fiff 3 on sides 3 (ab) and 1 (cd) will be Il 1 B sin 90° i.e., Il1 B. These two forces do not act along the same line and hence they produce a net torque. This torque results into rotation of the loop so that the loop is perpendicular to the direction of B fiff , the magnetic field. The moment arm is 1 2 (l 2 sin θ ) about the central axis of the loop. The torque τ due to forces F fiff 1 and F fiff 3 will then be τ = (Il1 B 1 2 l 2 sin θ ) + (Il1 B 1 2 l 2 sin θ ) = Il1 l 2 B sin θ --- (10.22) If the current carrying loop is made up of multiple turns N, in the form of a flat coil, the total torque will be τ' = Nτ = N I l 1 l 2 B sin θ τ' = (NIA)B sin θ --- (10.23) A = l 1 l 2 Here A is the area enclosed by the coil. The above equation holds good for all flat (planar) coils irrespective of their shape, in a uniform magnetic field. Fig. 10.12 (a): Loop abcd placed in a uniform magnetic field emerging out of the paper. Electric connections are not shown. Fig. 10.11: The right hand rule. The current carrying wire loop is of rectangular shape and is placed in the uniform magnetic field in such a way that the segments ab and cd of the loop are perpendicular to the field B fiff . We can use the right hand rule (Fig. 10.11) to find out the direction of the magnetic force Fm fiff . Let the pointing finger of the right hand show the direction of the current, let the middle finger show the direction of the magnetic field B fiff , then the stretched thumb shows the direction of the force. Let us now look at the action of rotation in detail. For this purpose, consider Fig. 10.12 a, showing the rectangular loop abcd placed in a uniform magnetic field B fiff such that the sides ab and cd are perpendicular to the magnetic field B fiff but the sides bc and da are not. Fm Fm Fig. 10.12 (b): Side view of the loop abcd at an angle θ. F3 B Fm (90- ) l 2
238 Can you recall? Larger the current is, larger is the deflection and larger is the torque due to the spring. If the deflection is φ, the restoring torque due to the spring is equal to K φ where K is the torsional constant of the spring. Thus, K φ = NIAB, and the deflection φ = NAB K fi ff ffl ffi fl I --- (10.24) Thus the deflection φ is proportional to the current I. Modern instruments use digital ammeters and voltmeters and do not use such a moving coil galvanometer. 10.8 Magnetic Dipole Moment: In the preceding section, we have dealt with a current carrying coil. This current carrying coil can be described with a vector µ fiff , its magnetic dipole moment. If n is a unit vector normal to the plane of the coil, the direction of µ fiff is the direction of n shown in Fig. 10.12 (b). We can then define the magnitude of µ fiff as µ = NIA, --- (10.25) where N is the number of turns of the coil, I the current passing through the coil, A the area enclosed by each turn of the coil. If held in uniform magnetic field B fiff , the torque responsible for the rotation of the coil, according to Eq. (10.23) will be τ = µB sinθ , How does the coil in a motor rotate by a full rotation? In a motor, we require continuous rotation of the current carrying coil. As the plane of the coil tends to become parallel to the magnetic field B fiff , the current in the coil is reversed externally. Referring to Fig. 10.10, the segment ab occupies the position cd. At this position of rotation, the current is reversed. Instead of from b to a, it flows from a to b, force Fm fiff continues to act in the same direction so that the torque continues to rotate the coil. The reversal of the current is achieved by using a commutator which connects the wires of the power supply to the coil via carbon brush contacts. 10.7.1 Moving Coil Galvanometer: A current in a circuit or a voltage of a battery can be measured in terms of a torque exerted by a magnetic field on a current carrying coil. Analog voltmeters and ammeters work on this principle. Figure 10.13 shows a cross sectional diagram of a galvanometer. It consists of a coil of several turns mounted (suspended or pivoted) in such a way that it can freely rotate about a fixed axis, in a radial uniform magnetic field. A soft iron cylindrical core makes the field radial and strong. The coil rotates due to a torque acting on it as the current flows through it. This torque is given by (Eq. 10.23) τ = N I A.B, where A is the area of the coil, B the strength of the magnetic field, N the number of turns of the coil and I the current in the coil. Here, sin θ = 1 as the field is radial (plane of the coil will always be parallel to the field). However, this torque is counter balanced by a torque due to a spring fitted as shown in the Fig. 10.13. This counter torque balances the magnetic torque, so that a fixed steady current I in the coil produces a steady angular deflection φ . Fig. 10.13: Moving coil galvanometer. d
239 θ being an angle between µ fiff (i.e., ) and B fiff . ∴ τ = µ fiff × B fiff --- (10.26) You have learnt in XIth Std. about the torque on an electric dipole exerted by an electric field, E fiff . ∴ τ = P fiff × E fiff --- (10.27) Here P fiff is the electric dipole moment. The two expression Eq. (10.26) and Eq. (10.27) are analogous to each other. 10.10 Magnetic Field due to a Current : Biot-Savart Law: In sections 10.1 and 10.2, we have seen that magnetic field is produced by a current carrying wire. Can we calculate this magnetic field? Fig. 10.14: Minimum and maximum magnetic potential energy of a magnetic dipole µ in a magnetic field B fiff . Figure 10.15: A current carrying wire of arbitrary shape, carrying a current I. The current in the differential length element dl produces differential magnetic field d B fiff at a point P at a distance r from the element dl. The indicates that d B fiff is directed into the plane of the paper. 10.9 Magnetic Potential Energy of a Dipole: A magnetic dipole freely suspended in a magnetic field possesses magnetic potential energy because of its orientation in the field. You have learnt about an electric dipole in Chapter 8. Electrical Potential energy is associated with an electric dipole on account of its orientation in an electric field. It has been shown that the potential energy U of an electric dipole P fiff in an electric field E fiff is given by U = - P fiff . E fiff --- (10.28) Analogously, the magnetic potential energy of a magnetic dipole µ fiff in a magnetic field B fiff is given by U = - µ fiff . B fiff --- (10.29) = - µB.cos θ , --- (10.30) where θ is the angle between µ fiff and B fiff . Case (i) : If θ = 0, U = - µ B.cos(0°) = - µB This is the minimum potential energy of a magnetic dipole in a magnetic field i.e., when and are parallel to each other. Case (ii) : If = 180°, U = - µ.B.cos (180°) = µB. This is the maximum potential energy of a magnetic dipole in a magnetic field, i.e., when and are antiparallel to each other. Figure 10.15 shows an arbitrarily shaped wire carrying a current I. dl is a length element along the wire. The current in this element is in the direction of the length vector dl fifffi . Let us calculate the differential field d B fiff at the point P, produced by the current I through the length element dl. Net magnetic field at the point P can be obtained by superimposition of magnetic fields d B fiff at that point due to different length elements along the wire. This can be done by integrating i.e., summing up of magnetic fields d B fiff from these length elements. Experimentally, the magnetic fields d B fiff produced by current I in the length element d l is B µ B µ Case (i) Case (ii) Example 10.4: A circular coil of conducting wire has 500 turns and an area 1.26×10-4 m2 is enclosed by the coil. A current 100 µA is passed through the coil. Calculate the magnetic moment of the coil. Solution: µ = NIA = 500 × 100 × 10-6 × 1.26 × 10-4 Am2 = 630 × 10-8 = 6.3 × 10-6 Am2 or J/T. dB I dl I r
240 dB Idl r fi ff ffl 0 ffi 2 4 sin --- (10.31) Here, θ is the angle between the directions of dl fifffi and r . µ0 is called permeability constant given by µ0= 4π × 10-7 T. m/A --- (10.32) ≈ 1.26 × 10-6 T. m/A --- (10.33) The direction of d B fiff is dictated by the cross product dl fifffi × r . Vectorially, d B Idl r r fiff ff ff fi ffl ff ffi 0 3 4 --- (10.24) Equation (10.31) and Eq. (10.34) are known as the Biot and Savart law. This inverse square law is experimentally deduced. It may be noted that this is still inverse square law as r appears in the numerator and r3 in the denominator. Using the Biot-Savart law, we can calculate the magnetic field produced by various distributions of currents as discussed below: (i) Current in a straight, long wire: You are aware of the right hand thumb rule which gives the direction of the magnetic field produced by a current flowing in a wire. Figure 10.16 shows a long wire of length l. We want to calculate magnetic field B fiff at a point P which is at a perpendicular distance R from the wire. Let us consider a current length element (the infinitesimal length d l of the wire, multiplied by the current I passing through it) I. dl fifffi situated at a distance r from the point P. Using Eq. (10.31), the magnetic field d B fiff produced at P due to the current length element I. dl fifffi becomes d B Idl r fiff ff fi ff ffl 0 ffi 2 4 sin --- (10.35) Here, the direction of d B fiff is given by the cross product d l × r (see Eq. (10.34)), hence into the plane of the paper. (a) We now calculate the magnitude of the magnetic field produced at P by all current length elements in the upper half of the infinitely long wire. This we do by integrating Eq. (10.35) from o to ∞. (b) Let us now calculate the magnitude of the magnetic field produced at P by a current length element in the lower half of the wire. By symmetry, this magnitude is the same as that from the upper half of the wire. The direction of this field is also the same as from the upper half of the wire, going into the plane of the paper. Adding both the contributions (a) and (b), the total magnetic field B at point P is B dB Idl r fi fi ff ff ffl ffl 2 2 4 0 0 2 0 ffi fl sin --- (10.36) But r l fi ff R 2 2 and sinθ = sin (π-θ )= R r R l R fi ff2 2 --(10.37) fi ff ffi ffl flffi fl ffl ff ffi fl ffl B I Rdl l R l R I R dl l R 0 2 2 2 2 0 0 2 2 3 2 0 2 2 / fi ff B ff I R R I R ffl ffi ffl ffi 0 2 0 2 1 2 --- (10.38) From Eq. (10.36), this is the magnetic field at a point P at a perpendicular distance R from the infinitely straight wire. This is due to both the upper semi-infinite part and the lower semi-infinite part of the wire. Thus, the magnetic field B due to semi-infinite straight wire is fi ff B I R ffl ffi 0 4 --- (10.39) In Eq. (10.38) and Eq. (10.39), the field is inversely proportional to the distance from the wire. Fig. 10.16: The magnetic field d B fiff at P going into the plane of the paper, due to current I through the wire. R I r θ
241 To solve I dl l R fi ffl ffi ff fl 2 2 3 2 0 / , we substitute l = R tan θ ; dl = r sec2 θ dθ Now the limits of the integral also change. l = 0, tan θ = 0 ∴ θ = 0 l = ∞, tan θ = ∞ ∴ θ = π/2 fi ff ffl ff ffi I R d R R d sec (tan ) (cos ) sec ( / / / 2 3 2 3 2 0 2 2 2 3 2 2 1 1 fl fl fl fl fl fl sin cos ) cos sin / / / 2 2 3 2 0 2 2 0 2 2 1 1 fl fl fl fl fl ffl ff ff ffi ffi R d R 0 2 2 2 1 1 0 / 1 ff [ ] ff R R Fig. 10.17 : Two long parallel wires, distance d apart. For infinitely long wires, this force will be infinite! Force per unit length of the wire will be Ffi ff I I d ffl ffi 0 1 2 2 --- (10.42) If the currents I 1 and I 2 are antiparallel, the force will be repulsive. Let us consider a section of length L of the wire 2. The force on this section due to the current in wire 1 is given by F = I2 B.L --- (10.43) fi fi ff ffl 0 1 2 21 2 I I d L F --- (10.44) We will denote this force by F21 i.e., the force on a section of length L of wire 2 due to the current in wire 1. Similarly, the force on a section of the same length L of wire 1 will experience a force due to the current in wire 2. This force we denote as F12, which is equal and opposite to F21 ∴ F21 = - F12 --- (10.44 A) The force of attraction per unit length is then, from Eq. (10.44), F L I I d fi ff ffl 0 1 2 2 --- (10.45) If the currents I1 and I2 are flowing in opposite directions, then there is a force of repulsion on the sector of length L of each of the wires. The magnitude of the repulsive force per unit length of the wire is also given by F L I I d fi ff ffl 0 1 2 2 --- (10.46) We can summarize these result as: Parallel currents attract, antiparallel currents repel. 10.11 Force of Attraction between two Long Parallel Wires: As an application of the result obtained in the last section, let us obtain the force of attraction between two long, parallel wires separated by a distance d (Fig. 10.17). Let the currents in the two wires be I 1 and I2 . The magnetic field at the second wire due to the current I 1 in the first one, according to Eq. (10.38), B I d fi ff ffl 0 1 2 --- (10.40) By the right hand rule, the direction of this field is into the plane of the paper. We now apply the Lorentz Force law. Accordingly, the force on the wire 2, because of the current I2 and the magnetic field B due to current in wire 1, is given by (Eq. 10.13). F I I d fi dl ff ffl ffi fl 2 0 1 2 --- (10.41) The direction of this force is towards wire 1, i.e., it will be attractive force. The ampere: Definition of the unit of electrical current ampere, was adopted a few decades ago. Consider two parallel conducting wires having infinite length, have a separation of 1 m, and are placed in vacuum. The constant current through these wires producing a force on each other of magnitude 2×10-7 N per meter of their length, is 1 ampere (A). I1 I 2