e.g. Suppose in the following network of 12 identical resistances, equivalent resistance between point A and C
is to be calculated. B
RR
E F
R RR
AR C
R
R R R
H G
RR
D
According to the above guidelines we can solve this problem as follows
Step (1) B Step (2) B
R
R R R R R R R
A E Ei i i 2i
F Ri i F C
R R R A 2i 2i R R
Ri R R
R C Hi i 2i
4i R D G 4i
R R
H G E R
RR
D
E
An imaginary battery of emf E is assumed The current in each branch is distributed by
across the terminals A and C assuming 4i current coming out of the battery.
Step (3) Applying KVL along the loop including the nodes A, B, C and the battery E. Voltage equation is
− 2iR − iR − iR − 2iR + E = 0
Step (4) After solving the above equation, we get 6iR = E ⇒ equivalent resistance between A and C is
R = E = 6iR = 3 R
4i 4i 2
Concepts
Using Kirchoff’s law while dividing the current having a junction through different arms of a network, it will be same through
different arms of same resistance if the end points of these arms are equilocated w.r.t. exit point for current in network and
will be different through different arms if the end point of these arms are not equilocated w.r.t. exit point for current of the
network.
e.g. In the following figure the current going in arms AB, AD and AL will be same because the location of end points B, D and L
of these arms are symmetrically located w.r.t. exit point N of the network.
B i Di C
6i 2i i
i
Ai i N
2i 2i
M
i 2i 6i
Li K
Example
Example: 67 In the following circuit E1 = 4V, R1 = 2Ω E1 R1 [MP PET 2003]
E2 = 6V, R2 = 2Ω and R3 = 4Ω. The current i1 is R2
Solution : (b) (a) 1.6 A i1 i1
Example: 68 i2
Solution : (a) (b) 1.8 A i2 R3
Example: 69 (c) 2.25 A E2
(d) 1 A
For loop (1) −2i1 − 2(i1 − i2 ) + 4 = 0 ⇒ 2i1 − i2 = 2 …… (i) 4V 2Ω
For loop (2) −4i2 + 2(i1 − i2 ) + 6 = 0 ⇒ 3i2 − i1 = 3 …… (ii) i1 1 2Ω (i1 – i2)
After solving equation (i) and (ii) we get i1 = 1.8 A and i2 = 1.6 A i2 2
6V
4Ω
Determine the current in the following circuit 10 V 2Ω
(a) 1 A
(b) 2.5 A
(c) 0.4 A 5 V 3Ω
(d) 3 A
Applying KVL in the given circuit we get −2i + 10 − 5 − 3i = 0 ⇒ i = 1A
Second method : Similar plates of the two batteries are connected together, so the net emf = 10 – 5 = 5V
Total resistance in the circuit = 2 + 3 = 5Ω
∴ i = ΣV = 5 = 1A
ΣR 5
In the circuit shown in figure, find the current through the branch BD
(a) 5 A A 6Ω B 3Ω C
(b) 0 A 15 V
(c) 3 A 3Ω
30 V
(d) 4 A D
Solution : (a) The current in the circuit are assumed as shown in the fig. A 6Ω i1 B 3Ω i1 – i2 C
Example: 70 Applying KVL along the loop ABDA, we get
Solution : (c)
– 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 …… (i) 3Ω 30 V
Applying KVL along the loop BCDB, we get
15 V
– 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 …… (ii) i2
Solving equation (i) and (ii) for i2, we get i2 = 5 A i1 D
The figure shows a network of currents. The magnitude of current is shown here. The current i will be [MP PMT 1995]
(a) 3 A 15 A 3A
(b) 13 A
(c) 23 A 5A i
(d) – 3 A
i = 15 + 3 + 5 = 23A
Example: 71 Consider the circuit shown in the figure. The current i3 is equal to [AMU 1995]
Solution : (d) (a) 5 amp
Example: 72 28Ω 54Ω
(b) 3 amp 6V
(c) – 3 amp i3
(d) – 5/6 amp 8V 12 V
Suppose current through different paths of the circuit is as follows. 28Ω 54Ω
After applying KVL for loop (1) and loop (2)
We get 28i1 = −6 − 8 ⇒ i1 = − 1 A 1 6V 2
2 i3
and 54i2 = −6 − 12 ⇒ i2 = − 1 A 8 V 12 V
3
Hence i3 = i1 + i2 = − 5 A
6
A part of a circuit in steady state along with the current flowing in the branches, with value of each resistance is
shown in figure. What will be the energy stored in the capacitor C0
[IIT-JEE 1986]
(a) 6 × 10–4 J 1A
(b) 8 × 10–4 J 3Ω A 3Ω 5Ω D
(c) 16 × 10–4 J 2A i1
(d) Zero 4µF i2 1Ω i3
2A 1Ω B 2Ω C 4Ω
3Ω
1A
Solution : (b) Applying Kirchhoff’s first law at junctions A and B respectively we have 2 + 1 – i1 = 0 i.e., i1 = 3A
Example: 73 and i2 + 1 – 2 – 0 = 0 i.e., i2 = 1A
Solution : (c) Now applying Kirchhoff’s second law to the mesh ADCBA treating capacitor as a seat of emf V in open circuit
– 3 × 5 – 3 × 1 – 1 × 2 + V = 0 i.e. V(= VA – VB) = 20 V
So, energy stored in the capacitor U = 1 CV 2 = 1 (4 × 10 −6 ) × (20) 2 = 8 × 10 −4 J
2 2
In the following circuit the potential difference between P and Q is
(a) 15 V P RQ
(b) 10 V 15V 2A
(c) 5 V
5Ω
(d) 2.5 V E1 E2
By using KVL −5 × 2 − VPQ + 15 = 0 ⇒ VPQ = 5V
Tricky Example: 9
As the switch S is closed in the circuit shown in figure, current passed through it is
20 V 2Ω 4Ω 5 V
AB
2Ω
S
(a) 4.5 A (b) 6.0 A (c) 3.0 A (d) Zero
Solution : (a) Let V be the potential of the junction as shown in figure. Applying junction law, we have
or 20 − V + 5−V = V −0 or 40 – 2V + 5 – V = 2V 20 V 2Ω 4Ω 5V
2 4 2 A i1 i2 B
2Ω
or 5V = 45 ⇒ V = 9V
i3
∴ i3 = V = 4.5 A 0V
2
Different Measuring Instruments.
(1) Galvanometer : It is an instrument used to detect small current passing through it by showing deflection.
Galvanometers are of different types e.g. moving coil galvanometer, moving magnet galvanometer, hot wire
galvanometer. In dc circuit usually moving coil galvanometer are used.
(i) It’s symbol : G ; where G is the total internal resistance of the galvanometer.
(ii) Principle : In case of moving coil galvanometer deflection is directly proportional to the current that
passes through it i.e. i ∝ θ .
(iii) Full scale deflection current : The current required for full scale deflection in a galvanometer is called
full scale deflection current and is represented by ig.
(iv) Shunt : The small resistance connected in parallel to galvanometer coil, in order to control current flowing
through the galvanometer is known as shunt.
Merits of shunt Demerits of shunt the G
SK
(a) To protect the galvanometer coil Shunt resistance decreases
from burning sensitivity of galvanometer.
(b) It can be used to convert any
galvanometer into ammeter of
desired range.
(2) Ammeter : It is a device used to measure current and is always connected in series with the ‘element’
through which current is to be measured. R
(i) The reading of an ammeter is always lesser than actual current in the i A
circuit.
(ii) Smaller the resistance of an ammeter more accurate will be its reading. +V–
An ammeter is said to be ideal if its resistance r is zero.
(iii) Conversion of galvanometer into ammeter : A galvanometer may be converted into an ammeter by
connecting a low resistance (called shunt S) in parallel to the galvanometer G as shown in figure.
(a) Equivalent resistance of the combination = GS S
G+S
i – ig
(b) G and S are parallel to each other hence both will have equal potential i ig G
difference i.e. igG = (i − ig )S ; which gives
Required shunt S ig G Ammeter
− ig
= (i )
(c) To pass nth part of main current (i.e. ig = i ) through the galvanometer, required shunt S = (n G .
n − 1)
(3) Voltmeter : It is a device used to measure potential difference and is always put in parallel with the
‘circuit element’ across which potential difference is to be measured.
(i) The reading of a voltmeter is always lesser than true value. V
(ii) Greater the resistance of voltmeter, more accurate will be its reading. i R
A voltmeter is said to be ideal if its resistance is infinite, i.e., it draws no current
from the circuit element for its operation. + V–
(iii) Conversion of galvanometer into voltmeter : A galvanometer
may be converted into a voltmeter by connecting a large resistance R in series with the galvanometer as shown in
the figure. G R
(a) Equivalent resistance of the combination = G + R V – Vg
(b) According to ohm’s law V = ig (G + R); which gives Vg = igG
ig
Required series resistance R V −G V − 1 G V
ig Vg
= =
(c) If nth part of applied voltage appeared across galvanometer (i.e. Vg V ) then required series resistance
=n
R = (n − 1) G .
(4) Wheatstone bridge : Wheatstone bridge is an arrangement of four resistance which can be used to
measure one of them in terms of rest. Here arms AB and BC are called ratio P B Q
arm and arms AC and BD are called conjugate arms A K1 C
(i) Balanced bridge : The bridge is said to be balanced when deflection G
in galvanometer is zero i.e. no current flows through the galvanometer or in
other words VB = VD. In the balanced condition P = R , on mutually changing R S
Q S +– D K2
the position of cell and galvanometer this condition will not change.
(ii) Unbalanced bridge : If the bridge is not balanced current will flow from D to B if VD > VB i.e.
(VA − VD ) < (VA − VB ) which gives PS > RQ.
(iii) Applications of wheatstone bridge : Meter bridge, post office box and Carey Foster bridge are
instruments based on the principle of wheatstone bridge and are used to measure unknown resistance.
(5) Meter bridge : In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of
tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC (100 – l) so that Q = (100 − l) .
P l
Also P = R ⇒ S = (100 − l) R
Q S l
S
R R.B.
G
A PB Q C
l cm (100 – l) cm
EK
Concepts
Wheatstone bridge is most sensitive if all the arms of bridge have equal resistances i.e. P = Q = R = S
If the temperature of the conductor placed in the right gap of metre bridge is increased, then the balancing length decreases and
the jockey moves towards left.
In Wheatstone bridge to avoid inductive effects the battery key should be pressed first and the galvanometer key afterwards.
The measurement of resistance by Wheatstone bridge is not affected by the internal resistance of the cell.
Example
Example: 74 The scale of a galvanometer of resistance 100 Ω contains 25 divisions. It gives a deflection of one division on
Solution : (b) passing a current of 4 × 10–4 A. The resistance in ohms to be added to it, so that it may become a voltmeter of
Example: 75
Solution : (c) range 2.5 volt is [EAMCET 2003]
Example: 76
(a) 100 (b) 150 (c) 250 (d) 300
Current sensitivity of galvanometer = 4 × 10–4 Amp/div.
So full scale deflection current (ig) = Current sensitivity × Total number of division = 4 × 10–4 × 25 = 10–2 A
To convert galvanometer in to voltmeter, resistance to be put in series is R = V −G = 2.5 − 100 = 150 Ω
ig 10 −2
A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. the length in
meter of a resistance wire of area of cross-section 2.97 × 10–2 cm2 that can be used to convert the
galvanometer into an ammeter which can read a maximum of 5A current is : (Specific resistance of the wire =
5 × 10–7 Ωm)
[EAMCET 2003]
(a) 9 (b) 6 (c) 3 (d) 1.5
Given G = 50 Ω, ig = 0.05 Amp., i = 5A, A = 2.97 × 10–2 cm2 and ρ = 5 × 10–7Ω-m
By using i = 1 + G ⇒ S = G.ig ⇒ ρl = Gig ⇒ l = Gig A on putting values l = 3 m.
ig S (i − ig ) A (i − ig ) (i − ig )ρ
100 mA current gives a full scale deflection in a galvanometer of resistance 2 Ω. The resistance connected with
the galvanometer to convert it into a voltmeter of 5 V range is
[KCET 2002; UPSEAT 1998; MNR 1994 Similar to MP PMT 1999]
(a) 98 Ω (b) 52 Ω (c) 80 Ω (d) 48 Ω
Solution : (d) R = V −G = 5 − 2 = 50 − 2 = 48 Ω .
Example: 77 Ig 100 × 10 −3
Solution : (a)
Example: 78 A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as voltmeter of range 10 volt, the
Solution : (c) resistance that must be connected in series with it will be [KCET (Engg./Med.) 2001]
(a) 999 Ω (b) 99 Ω (c) 1000 Ω (d) None of these
By using R = V − G ⇒ R = 10 −1 = 999Ω
ig 10 × 10 −3
In the following figure ammeter and voltmeter reads 2 amp and 120 volt respectively. Resistance of voltmeter is
(a) 100 Ω AX 75 Ω Y
(b) 200 Ω
(c) 300 Ω V
(d) 400 Ω
Let resistance of voltmeter be RV. Equivalent resistance between X and Y is R XY = 75RV
75 + RV
Reading of voltmeter = potential difference across X and Y = 120 = i × RXY = 2 × 75RV ⇒ RV = 300Ω
75 + RV
Example: 79 In the circuit shown in figure, the voltmeter reading would be 1Ω 2Ω
(a) Zero
Solution : (a)
Example: 80 (b) 0.5 volt AV
Solution : (d)
(c) 1 volt + 3V–
Example: 81 (d) 2 volt
Ammeter has no resistance so there will be no potential difference across it, hence reading of voltmeter is zero.
Voltmeters V1 and V2 are connected in series across a d.c. line. V1reads 80 V and has a per volt resistance of
200 Ω, V2 has a total resistance of 32 kΩ. The line voltage is [MNR 1992]
(a) 120 V (b) 160 V (c) 220 V (d) 240 V
Resistance of voltmeter V1 is R1 = 200 × 80 = 16000 Ω and resistance of voltmeter V2 is R2 = 32000 Ω
By using relation V'= R' V ; where V′ = potential difference across any resistance R′ in a series grouping.
Req
R1 R2
So for voltmeter V1 potential difference across it is V1 V2
80 = R1 .V ⇒ V = 240 V 80 V
R1 + R2 +–
V
The resistance of 1 A ammeter is 0.018 Ω. To convert it into 10 A ammeter, the shunt resistance required will
be [MP PET 1982]
(a) 0.18 Ω (b) 0.0018 Ω (c) 0.002 Ω (d) 0.12 Ω
Solution : (c) By using i = 1 + 4 ⇒ 10 = 1+ 0.018 ⇒ S = 0.002 Ω
Example: 82 ig S 1 S
Solution: (a) In meter bridge the balancing length from left and when standard resistance of 1 Ω is in right gas is found to be
20 cm. The value of unknown resistance is
[CBSE PMT 1999]
(a) 0.25 Ω (b) 0.4 Ω (c) 0.5 Ω (d) 4 Ω
The condition of wheatstone bridge gives X = 20 r , r- resistance of wire per cm, X- unknown resistance
R 80 r
X R = 1Ω
∴ X = 20 × R = 1 ×1 = 0.25 Ω
80 4
P = 20 r Q = 80 r
20 cm 80 cm
Example: 83 A galvanometer having a resistance of 8 Ω is shunted by a wire of resistance 2 Ω. If the total current is 1 amp,
the part of it passing through the shunt will be [CBSE PMT 1998]
(a) 0.25 amp (b) 0.8 amp (c) 0.2 amp (d) 0.5 amp
Solution: (b) Fraction of current passing through the galvanometer
ig S or ig = 2 = 0.2
i = S+G i 2+8
So fraction of current passing through the shunt
is = 1 − ig = 1 − 0.2 = 0.8 amp
i i
Example: 84 A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of
Solution: (b) resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r connected
across it. What is the maximum current which can be through this galvanometer if no shunt is used
[MP PMT 1996]
(a) 0.01 A (b) 0.02 A (c) 0.03 A (d) 0.04 A
For ammeter, S = (i ig ) G ⇒ ig G = (i − ig )S
−i
g
So ig G = (0.03 − ig )4r …… (i) and ig G = (0.06 − ig )r …… (ii)
Dividing equation (i) by (ii) 1= (0.03 − ig ) 4 ⇒ 0.06 − ig = 0.12 − 4ig
0.06 − ig
⇒ 3ig = 0.06 ⇒ ig = 0.02 A
Tricky Example: 10
The ammeter A reads 2 A and the voltmeter V reads 20 V. The value of resistance R is [JIPMER 1999]
R
A
(a) Exactly 10 ohm V
(c) More than 10 ohm
(b) Less than 10 ohm
(d) We cannot definitely say
Solution: (c) If current goes through the resistance R is i then iR = 20 volt ⇒ R = 20 . Since i < 2A so R > 10Ω.
i
Potentiometer.
Potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. It is also
used to measure internal resistance of a given cell.
(1) Superiority of potentiometer over voltmeter : An ordinary voltmeter cannot measure the emf
accurately because it does draw some current to show the deflection. As per definition of emf, it is the potential
difference when a cell is in open circuit or no current through the cell. Therefore voltmeter can only measure
terminal voltage of a give n cell.
Potentiometer is based on no deflection method. When the potentiometer gives zero deflection, it does not
draw any current from the cell or the circuit i.e. potentiometer is effectively an ideal instrument of infinite resistance
for measuring the potential difference.
(2) Circuit diagram : Potentiometer consists of a long resistive wire AB of length L (about 6m to 10 m long)
made up of mangnine or constantan. A battery of known voltage e and internal resistance r called supplier battery
or driver cell. Connection of these two forms primary circuit.
One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and
the other terminal at any point on the resistive wire through a galvanometer G. This forms the secondary circuit.
Other details are as follows e, r K Rh
J = Jockey J B
K = Key Primary A
R = Resistance of potentiometer wire, circuit G
ρ = Specific resistance of potentiometer wire. E
Secondary
circuit
Rh = Variable resistance which controls the current through the wire AB
(3) Points to be remember
(i) The specific resistance (ρ) of potentiometer wire must be high but its temperature coefficient of resistance
(α) must be low.
(ii) All higher potential points (terminals) of primary and secondary circuits must be connected together at
point A and all lower potential points must be connected to point B or jockey.
(iii) The value of known potential difference must be greater than the value of unknown potential difference to
be measured.
(iv) The potential gradient must remain constant. For this the current in the primary circuit must remain
constant and the jockey must not be slided in contact with the wire.
(v) The diameter of potentiometer wire must be uniform everywhere.
(4) Potential gradient (x) : Potential difference (or fall in potential) per unit length of wire is called potential
gradient i.e. x = V volt where V = iR = R + e + r .R . So =x V iR iAθ= (R + e + r) . R
L m Rh L= L= Rh L
(i) Potential gradient directly depends upon
(a) The resistance per unit length (R/L) of potentiometer wire.
(b) The radius of potentiometer wire (i.e. Area of cross-section)
(c) The specific resistance of the material of potentiometer wire (i.e. ρ)
(d) The current flowing through potentiometer wire (i)
(ii) x indirectly depends upon
(a) The emf of battery in the primary circuit (i.e. e)
(b) The resistance of rheostat in the primary circuit (i.e. Rh)
Note : ≅When potential difference V is constant then x1 = L2
x2 L1
≅ Two different wire are connected in series to form a potentiometer wire then x1 = R1 . L2
x2 R2 L1
≅ If the length of a potentiometer wire and potential difference across it’s ends are kept constant and
if it’s diameter is changed from d1 → d2 then potential gradient remains unchanged.
≅ The value of x does not change with any change effected in the secondary circuit.
(5) Working : Suppose jocky is made to touch a point J on wire then potential difference between A and J
will be V = xl e, r K Rh
At this length (l) two potential difference are obtained l B
(i) V due to battery e and A J1 J J2
(ii) E due to unknown cell
If V > E then current will flow in galvanometer circuit in one direction GGG
E
If V < E then current will flow in galvanometer circuit in opposite direction
If V = E then no current will flow in galvanometer circuit this condition to known as null deflection position,
length l is known as balancing length.
In balanced condition E = xl or E = xl = V l = iR l = R + e + r . R × l
L L Rh L
Note : ≅ If V is constant then L ∝ l ⇒ L1 = l1
L2 l2
(6) Standardization of potentiometer : The process of determining potential gradient experimentally is
known as standardization of potentiometer.
Let the balancing length for the standard emf E0 is l0 then by the principle of e K R
B
potentiometer E0 = xl0 ⇒ x = E0 A J
l0 E G
E
(7) Sensitivity of potentiometer : A potentiometer is said to be more sensitive,
if it measures a small potential difference more accurately.
(i) The sensitivity of potentiometer is assessed by its potential gradient. The sensitivity is inversely proportional
to the potential gradient.
(ii) In order to increase the sensitivity of potentiometer
(a) The resistance in primary circuit will have to be decreased.
(b) The length of potentiometer wire will have to be increased so that the length may be measured more accuracy.
(8) Difference between voltmeter and potentiometer
Voltmeter Potentiometer
(i) It’s resistance is high but finite Its resistance is high but infinite
(ii) It draws some current from source of emf It does not draw any current from the source of known emf
(iii) The potential difference measured by it is lesser than The potential difference measured by it is equal to
the actual potential difference actual potential difference
(iv) Its sensitivity is low Its sensitivity is high
(v) It is a versatile instrument It measures only emf or potential difference
(vi) It is based on deflection method It is based on zero deflection method
Application of Potentiometer.
(1) To determine the internal resistance of a primary cell
(i) Initially in secondary circuit key K' remains open and balancing length (l1) is obtained. Since cell E is in
open circuit so it’s emf balances on length l1 i.e. E = xl1 ……. (i) e, r K Rh
(ii) Now key K′ is closed so cell E comes in closed circuit. If the
process is repeated again then potential difference V balances on J B
G
length l2 i.e. V = xl2 ……. (ii) A
K′
(iii) By using formula internal resistance r = E − 1 . R' E
V R′
r = l1 − l2 . R'
l2
(2) Comparison of emf’s of two cell : Let l1 and l2 be the balancing lengths with the cells E1 and E2
respectively then E1 = xl1 and E2 = xl2 ⇒ E1 = l1 e, r K Rh
E2 l2 J B
A
Note : ≅ Let E1 > E2 and both are connected in series. If balancing length E1 1G
E2 2
is l1 when cell assist each other and it is l2 when they oppose each
other as shown then :
+ E1– + E2– + E1– – E2+
(E1 + E2 ) = xl1 (E1 − E2 ) = xl 2
⇒ E1 + E2 = l1 or E1 = l1 + l2
E1 − E2 l2 E2 l1 − l2
(3) Comparison of resistances : Let the balancing length for resistance R1 (when XY is connected) is l1 and
let balancing length for resistance R1 + R2 (when YZ is connected) is l2. K Rh
Then iR1 = xl1 and i(R1 + R2) = xl2 J
⇒ A
B
R2 = l2 − l1 XY G Z
R1 l1
i R1 R2
(4) To determine thermo emf K1 Rh1
(i) The value of thermo-emf in a thermocouple for ordinary temperature difference is very low (10–6 volt). For
this the potential gradient x must be also very low (10–4 V/m). Hence a high resistance (R) is connected in series with
the potentiometer wire in order to reduce current.
(ii) The potential difference across R must be equal to the emf of +– K Rh
standard cell i.e. iR = E0 ∴ i = E0
R
R A
A HRB
B
(iii) The small thermo emf produced in the thermocouple e = xl
G
(iv) x = iρ = iR ∴ e = iRl where L = length of potentiometer +– G
L L E0 1 2 3
wire, ρ = resistance per unit length, l = balancing length for e Cold ice Hot sand
(5) To calibrate ammeter and voltmeter
Calibration of ammeter
(i) If p.d. across 1Ω resistance is measured by potentiometer, then current through this (indirectly measured) is thus
known or if R is known then i = V/R can be found.
(ii) Circuit and method
(a) Standardisation is required and per formed as already described earlier. (x = E0/l0)
(b) The current through R or 1Ω coil is measured by the connected e K1
ammeter and same is calculated by potentiometer by finding a balancing +–
length as described blow.
A + E– 1 B
+
1
Let i ' current flows through 1Ω resistance giving p.d. as V' = i' (1) = xl1
where l1 is the balancing length. So error can be found as [i (measured by 2G
1Ω 3
+A –
E0
ammeter) ∆i' = i − i' = xl1 = l0 l1 +– K2
Calibration of voltmeter
(i) Practical voltmeters are not ideal, because these do not have infinite resistance. The error of such practical
voltmeter can be found by comparing the voltmeter reading with calculated value of p.d. by potentiometer.
(ii) Circuit and procedure e K1 Rh
+–
(a) Standardisation : If l0 is balancing length for E0 the emf of standard
cell by connecting 1 and 2 of bi-directional key, then x = E0/l0. A + E–0 1 C B
+ G
(b) The balancing length l1 for unknown potential difference V′ is given by
(by closing 2 and 3) V ' = xl1 = (E0 / l0 )l1 . +V – 2
3
RB
If the voltmeter reading is V then the error will be (V – V′) which may +– K2
be +ve, – ve or zero.
Rh
Concepts
In case of zero deflection in the galvanometer current flows in the primary circuit of the potentiometer, not in the
galvanometer circuit.
A potentiometer can act as an ideal voltmeter.
Example
Example: 85 A battery with negligible internal resistance is connected with 10m long wire. A standard cell gets balanced on
600 cm length of this wire. On increasing the length of potentiometer wire by 2m then the null point will be
Solution : (b) displaced by
Example: 86
(a) 200 cm (b) 120 cm (c) 720 cm (d) 600 cm
By using L1 = l1 ⇒ 10 = 600 ⇒ l2 = 720 cm .
L2 l2 12 l2
Hence displacement = 720 – 600 = 120 cm
In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R1 and an
accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection
in galvanometer is zero. The current supplied by the accumulator will be
+– R1
i B
5m G
A
Hot Cold
Junction Junction
(a) 4 × 10–4 A (b) 8 × 10–4 A (c) 4 × 10–3 A (d) 8 × 10–3 A
E = x l = iρ l
Solution : (a) ∴ i= E = E = 2.4 × 10 −3 = 4 × 10 −4 A .
Example: 87 ρl ρl 1.2 × 5
Solution : (a) The resistivity of a potentiometer wire is 40 × 10–8 Ωm and its area of cross section is 8 × 10–6 m2. If 0.2
Example: 88
Solution: (b) amp. Current is flowing through the wire, the potential gradient will be [MP PET/PMT 1998]
Example: 89
(a) 10–2 volt/m (b) 10–1 volt/m (c) 3.2 × 10–2 volt/m (d) 1 volt/m
Solution : (a)
Potential gradient = V = iR = iρL = iρ = 0.2 × 40 × 10−8 = 10−2 V/ m
Example: 90 L L AL A 8 × 10−6
Solution : (a)
Example: 91 A deniel cell is balanced on 125 cm length of a potentiometer wire. When the cell is short circuited with a
2 Ω resistance the balancing length obtained is 100 cm. Internal resistance of the cell will be [RPMT 1998]
Solution : (b)
(a) 1.5 Ω (b) 0.5 Ω (c) 1.25 Ω (d) 4/5 Ω
By using r = l1 − l2 × R' ⇒ r = 125 − 100 × 2 = 1 = 0.5 Ω
l2 100 2
A potentiometer wire of length 10 m and a resistance 30 Ω is connected in series with a battery of emf
2.5 V and internal resistance 5 Ω and an external resistance R. If the fall of potential along the
potentiometer wire is 50 µV/mm, the value of R is (in Ω) [KCET 1998]
(a) 115 (b) 80 (c) 50 (d) 100
By using x = e + r) . R
(R + Rh L
⇒ 50 × 10 −6 2.5 30 ⇒ R = 115
10 −3 = (30 + R + 5) × 10
A 2 volt battery, a 15 Ω resistor and a potentiometer of 100 cm length, all are connected in series. If the
resistance of potentiometer wire is 5 Ω, then the potential gradient of the potentiometer wire is [AIIMS 1982]
(a) 0.005 V/cm (b) 0.05 V/cm (c) 0.02 V/cm (d) 0.2 V/cm
By using x = e + r) . R ⇒ x = (5 2 0) × 5 = 0.5 V/m = 0.005 V/cm
(R + Rh L + 15 + 1
In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance
point is at a length of 2 m when the cell is shunted by a 5 Ω resistance; and is at a length of 3 m when the cell
is shunted by a 10 Ω resistance. The internal resistance of the cell is, then [Haryana CEE 1996]
(a) 1.5 Ω (b) 10 Ω (c) 15 Ω (d) 1 Ω
By using r = l1 − l 2 R' ⇒ r = l1 − 2 × 5 …… (i)
l2 2
and r = l1 − 3 × 10 ……. (ii)
3
On solving (i) and (ii) r = 10 Ω
Example: 92 A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a
cell of emf 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across
300 cm of the wire. The emf E of each cell is [RPET 2001; MP PMT 1997]
4Ω
10 V, 1Ω
3m 5Ω5m
E
G
E
(a) 1.5 V (b) 3.0 V (c) 0.67 V (d) 1.33 V
Solution : (b) By using Eeq = (R + e + r) . R × l ⇒ E = (5 10 1) × 5 × 3 ⇒ E = 3 volt
Example: 93 Rh L +4+ 5
Solution : (d)
A potentiometer has uniform potential gradient across it. Two cells connected in series (i) to support each
Example: 94
other and (ii) to oppose each other are balanced over 6 m and 2 m respectively on the potentiometer wire.
The emf’s of the cells are in the ratio of [MP PMT 2002; RPMT 2000]
(a) 1 : 2 (b) 1 : 1 (c) 3 : 1 (d) 2 : 1
If suppose emf’s of the cells are E1 and E2 respectively then
E1 + E2 = x × 6 ……. (i) [x = potential gradient) ]
and E1 – E2 = x × 2 …….(ii)
⇒ E1 + E2 = 3 ⇒ E1 = 2
E1 − E2 1 E2 1
In the following circuit the potential difference between the points B and C is balanced against 40 cm length of
potentiometer wire. In order to balance the potential difference between the points C and D, where should
jockey be pressed Rh
+–
40 cm Y
X
10 Ω G D
B 10 Ω C 4Ω F
A +–
r = 1Ω K
6V
(a) 32 cm (b) 16 cm (c) 8 cm (d) 4 cm
Solution : (a) 1 = 1 + 1 = 2 = 1 or R1 = 5 Ω
R 10 10 10 5
R2 = 4Ω, l1 = 40 cm, l2 = ?
l2 = l1 R2 or l2 = 40 × 4 = 32 cm
R1 5
Example: 95 In the following circuit diagram fig. the lengths of the wires AB and BC are same but the radius of AB is three
times that of BC. The ratio of potential gradients at AB and BC will be
(a) 1 : 9
(b) 9 : 1 +–
(c) 3 : 1
(d) 1 : 3
r22 2 AB C
r12
Solution : (a) x ∝ Rp ∝ 1 ⇒ x1 = = r = 1
r2 x2 3r 9
Example: 96 With a certain cell the balance point is obtained at 0.60 m from one end of the potentiometer. With another
cell whose emf differs from that of the first by 0.1 V, the balance point is obtained at 0.55 m. Then, the two
emf’s are
(a) 1.2 V, 1.1 V (b) 1.2 V, 1.3 V (c) – 1.1 V, – 1.0 V (d) None of the above
Solution : (a) E1 = x (0.6) and E2 = E1 – 0.1 = x (0.55) ⇒ E1 = 0.6
E1 − 0.1 0.55
or 55 E1 = 60 E1 – 6 ⇒ E1 = 6 = 1.2 V thus E2 = 1.1 V
5
Tricky Example: 11
A cell of internal resistance 1.5Ω and of emf 1.5 volt balances 500 cm on a potentiometer wire. If a
wire of 15Ω is connected between the balance point and the cell, then the balance point will shift
[MP PMT 1985]
(a) To zero (b) By 500 cm
(c) By 750 cm (d) None of the above
Solution : (d) In balance condition no current flows in the galvanometer circuit. Hence there will be no shift in
balance point after connecting a resistance between balance point and cell.
Magnetic effects of current
Oersted found that a magnetic field is established around a current carrying Magnetic lines
conductor. i of forces
Magnetic field exists as long as there is current in the wire. AB
The direction of magnetic field was found to be changed when direction of i
current was reversed.
dl
Note : ≅ A moving charge produces magnetic as well as electric field,
unlike a stationary charge which only produces electric field.
Biot Savart's Law.
Biot-Savart’s law is used to determine the magnetic field at any point due to a current carrying conductors.
This law is although for infinitesimally small conductors yet it can be used for long conductors. In order to
understand the Biot-Savart’s law, we need to understand the term current-element.
Current element
It is the product of current and length of infinitesimal segment of current carrying wire.
The current element is taken as a vector quantity. Its direction is same as the direction of current.
Current element AB = i dl
In the figure shown below, there is a segment of current carrying wire and P is a point where magnetic field is
to be calculated. i dl is a current element and r is the distance of the point ‘P’ with respect to the current element
i dl . According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element i dl is given by the
i dlsinθ µ0i dl sinθ
r2 4π r2
expression, ∫ ∫dB= k also B= dB = .
In C.G.S. : k = 1 ⇒ dB = idl sinθ Gauss P
r2 dl r
In S.I. : k = µ0 ⇒ dB = µ0 ⋅ idl sinθ Tesla i
4π 4π r2
where µ0 = Absolute permeability of air or vacuum = 4π × 10−7 Wb . It's other units are Henry
Amp − metre metre
or N or Tesla − metre
Amp 2 Ampere
(1) Different forms of Biot-Savarts law
Vector form Biot-Savarts law in terms of Biot-savarts law in terms of
current density charge and it's velocity
Vectorially, In terms Jo×f current density In terms of charge and it’s
r3 r dV (v × r)
dB µ0 i(dl × rˆ) µ0 i(dl × r ) dB µ0 dB = µ0 r3
= 4π ⋅ = 4π ⋅ r ⇒ = 4π velocity, 4π q
r2 3 is
Direction of dB
perpendicular to both dl and where j = i = idl = idl = = q = q dl = qv
ˆr . This is given by right hand A Adl dV idl dt dl dt
screw rule.
current density at any point of
the element, dV = volume of
element
(2) Similarities and differences between Biot-Savart law and Coulomb’s Law
(i) The current element produces a magnetic field, whereas a point charge produces an electric field.
(ii) The magnitude of magnetic field varies as the inverse square of the distance from the current element, as
does the electric field due to a point charge.
dB = µ0 idl × rˆ Biot-Savart Law F = 1 q1 q 2 rˆ Coulomb’s Law
4π r2 4πε 0 r2
(iii) The electric field created by a point charge is radial, but the magnetic field created by a current element is
perpendicular to both the length element dl and the unit vector rˆ.
dl E
i
+q ˆr
rˆ B
Direction of Magnetic Field.
The direction of magnetic field is determined with the help of the following simple laws :
(1) Maxwell’s cork screw rule
According to this rule, if we imagine a right handed screw placed along the
current carrying linear conductor, be rotated such that the screw moves in the direction
of flow of current, then the direction of rotation of the thumb gives the direction of
magnetic lines of force.
(2) Right hand thumb rule
According to this rule if a current carrying conductor is held in the right hand B
such that the thumb of the hand represents the direction of current flow, then the
direction of folding fingers will represent the direction of magnetic lines of force.
B
(3) Right hand thumb rule of circular currents
According to this rule if the direction of current in circular conducting coil is in i
the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched
thumb.
(4) Right hand palm rule
If we stretch our right hand such that fingers point towards the point. At which
magnetic field is required while thumb is in the direction of current then normal to the
palm will show the direction of magnetic field.
B
Note : ≅ If magnetic field is directed perpendicular and into the plane of the paper it is represented by ⊗
(cross) while if magnetic field is directed perpendicular and out of the plane of the paper it is
represented by (dot)
ii i i
B BB B CW ACW
Out In In Out In Out
In : Magnetic field is away from the observer or perpendicular inwards.
Out : Magnetic field is towards the observer or perpendicular outwards.
Application of Biot-Savarts Law.
(1) Magnetic field due to a circular current
If a coil of radius r, carrying current i then magnetic field on it's axis at a distance x from its centre given by
Baxis = μ0 . ( 2πNir 2 ; where N = number of turns in coil.
4π x 2 + r 2 )3/2
r
Different cases O x PB
i
Case 1 : Magnetic field at the centre of the coil
(i) At centre x = 0 ⇒ Bcentre = µ0 . 2πNi = µ0 Ni = Bmax
4π r 2r
(ii) For single turn coil N = 1 ⇒ Bcentre = µ0 . 2πi = µ0i (iii) In C.G.S. µ0 =1 ⇒ Bcentre = 2πi
4π r 2r 4π r
Note : ≅ Bcentre ∝ N (i, r constant), Bcentre ∝ i (N, r constant), Bcentre ∝ 1 (N, i constant)
r
Case 2 : Ratio of Bcentre and Baxis
The ratio of magnetic field at the centre of circular coil and on it's axis is given by Bcentre = 1 + x2 3/2
Baxis r2
(i) If x = ±a, Bc = 2 2 Ba x = ± a , Bc = 55 Ba x=± a, Bc = 3 3 / 2 Ba
2 8 2 2
(ii) If Ba = Bc then x == ±r (n2 / 3 − 1) and if Ba = Bc then x == ±r (n1/ 3 − 1)
n n
Case 3 : Magnetic field at very large/very small distance from the centre
(i) If x >> r (very large distance) ⇒ Baxis µ0 . 2π Nir 2 µ0 . 2NiA where A = πr2 = Area of each turn of the coil.
4π x3 4π x3
= =
(ii) If x << r (very small distance) ⇒ Baxis ≠ Bcentre , but by using binomial theorem and neglecting higher
power of x2 ; Baxis = Bcentre 1 − 3 x2
r2 2 r2
Case 4 : B-x curve
The variation of magnetic field due to a circular coil as the distance x varies as shown in the figure.
B varies non-linearly with distance x as shown in figure and is maximum when x 2 = min = 0 , i.e., the point is
at the centre of the coil and it is zero at x = ± ∞.
Point of inflection (A and A′) : Also known as points of curvature change or pints of zero curvature.
(i) At these points B varies linearly with x ⇒ dB = constant ⇒ d2B =0.
dx dx 2
(ii) They locates at x = ± r from the centre of the coil. A′ B0 A
2
(iii) Separation between point of inflextion is equal to radius of coil (r)
(iv) Application of points of inflextion is "Hamholtz coils" arrangement. x = – r/2 x = 0 x = r/2
Note : ≅The magnetic field at x= r is B= 4µ0 Ni
2 5 5r
(2) Helmholtz coils
(i) This is the set-up of two coaxial coils of same radius such that distance between their centres is equal to their radius.
(ii) These coils are used to obtain uniform magnetic field of short range which is obtained between the coils.
(iii) At axial mid point O, magnetic field is given by B = 8µ0 Ni = 0.716 µ 0 Ni = 1.432 B , where B = µ0 Ni
5 5R R 2R
(iv) Current direction is same in both coils otherwise this arrangement is not called Helmholtz's coil arrangement.
(v) Number of points of inflextion ⇒ Three (A, A′, A′′)
Resultant field (Uniform)
a O a A′ A′′ A
O2
O1 + O1 O O2
– a a
+ x
– x=− x=
2 2
Note : ≅The device whose working principle based on this arrangement and in which uniform magnetic field
is used called as "Halmholtz galvanometer".
(3) Magnetic field due to current carrying circular arc : Magnetic field at centre O
ii
i
Or θr O r
O θ
B = µ0 . πi = µ0i B = µ0 . θr i B = µ0 . (2π −θ)i
4π r 4r 4π 4π r
Special results
If magnetic field at the centre of circular coil Angle at centre Magnetic field at
centre in term of B0
is denoted by B0 = µ0 . 2πi 360o (2π) B0
4π r 180o (π) B0 / 2
120o (2π/3) B0 / 3
Magnetic field at the centre of arc which is 90o (π/2) B0 / 4
60o (π/3) B0 / 6
making an angle θ at the centre is 30o (π/6) B0 / 12
Barc = B0 .θ
2π
(4) Concentric circular loops (N = 1)
(i) Coplanar and concentric : It means both coils are in same plane with common centre
(a) Current in same direction (b) Current in opposite direction
i B1 = µ0 2πi 1 + 1 i B2 = µ0 2πi 1 − 1
i 4π r1 r2 i 4π r1
r2 r2 r2
r1 r1
Note : ≅ B1 = r2 + r1
B2 r2 − r1
(ii) Non-coplanar and concentric : Plane of both coils are perpendicular to each other
Magnetic field at common centre B2
i1
B12 2 µ0 i12 + i22
B= + B 2 = 2r
i2 B1
(5) Magnetic field due to a straight current carrying wire
Magnetic field due to a current carrying wire at a point P which lies at a perpendicular distance r from the wire
as shown is given as
B = µ0 . i (sin φ1 + sinφ2 ) Yβ
4π r
φ2
From figure α = (90o − φ1) and β = (90o + φ2 ) i r φ1 P
Hence B= µo . i (cosα − cos β ) α
4π r
X
Different cases
Case 1 : When the linear conductor Case 2 : When the linear conductor Case 3 : When the linear conductor
XY is of finite length and the point P XY is of infinite length and the point P is of infinite length and the point P
lies on it's perpendicular bisector as lies near the centre of the conductor lies near the end Y or X
shown
Y
φ P P
i
i φ i
r
φ1 =X φ2 = φ φ1 = φ2 = 90o. P
φ1 = 90o and φ 2 = 0 o .
So B= µ0 . i (2 sin φ) So, B = µ0 i [sin 90 o + sin 90o ] = µ0 2i So, B= µ0 i [sin 90 o + sin 0o ] = µ0 i
4π r 4π r 4π r 4π r 4π r
Note : ≅ When point P lies on axial position of current carrying conductor then magnetic field at P
B=0 i
P
≅ The value of magnetic field induction at a point, on the centre of separation of two linear
parallel conductors carrying equal currents in the same direction is zero.
(6) Zero magnetic field : If in a symmetrical geometry, current enters from one end and exists from the
other, then magnetic field at the centre is zero.
ii
O
OO
OO O
In all cases at centre B = 0
Concepts
If a current carrying circular loop (n = 1) is turned into a coil having n identical turns then magnetic field at the centre of the coil
becomes n2 times the previous field i.e. B (n turn) = n2 B(single turn)
When a current carrying coil is suspended freely in earth's magnetic field, it's plane stays in East-West direction.
Magnetic field ( B ) produced by a moving charge q is given by B= µ0 q(v × r ) = µ0 q(v × rˆ) ; where v = velocity of charge and
4π r3 4π r2
v << c (speed of light). B
r v
q
If an electron is revolving in a circular path of radius r with speed v then magnetic field produced at the centre of circular path
B = µ0 . ev .
4π r2
Examples
Example: 1 Current flows due north in a horizontal transmission line. Magnetic field at a point P vertically above it directed
(a) North wards N P
(b) South wards WE i
(c) Toward east
(d) Towards west S
Solution : (c) By using right hand thumb rule or any other rule which helps to determine the direction of magnetic field.
Example: 2
Magnetic field due to a current carrying loop or a coil at a distant axial point P is B1 and at an equal distance
Solution : (a)
in it's plane is B2 then B1 is
B2
(a) 2 (b) 1 (c) 1 (d) None of these
2
Current carrying coil behaves as a bar magnet as shown in figure.
We also knows for a bar magnet, if axial and equatorial distance are Q B2
same then Ba = 2Be x
Hence, in this equation B1 = 2 S N P
B2 1 B1
x
Example: 3 Find the position of point from wire 'B' where net magnetic field is zero due to following current distribution
Solution : (c)
(a) 4 cm
(b) 30 cm A B
7 5i 2i
(c) 12 cm 6cm
7
(d) 2 cm
Suppose P is the point between the conductors where net magnetic field is zero.
So at P |Magnetic field due to conductor 1| = |Magnetic field due to conductor 2|
i.e. µ0 . 2(5i) = µ0 . 2(2i) ⇒ 5 9 ⇒ x = 30 cm 5i 2i
4π i 4π (6 − x) x = 6−x 7 B1
Hence position from B 6 30 12 cm B2 P
7 7
= − = 6 cm
x cm (6 – x)
1 2
cm
Example: 4 Find out the magnitude of the magnetic field at point P due to following current distribution
Solution : (a) (a) µ 0 ia
Example: 5 πr 2
i B
µ 0ia 2 r
πr
(b) a
µ 0 ia M θ P
2πr 2 θ
(c) a
r
i B
(d) 2µ 0 ia
πr 2
Net magnetic field at P, Bnet = 2B sinθ ; where B = magnetic field due to one wire at P = µ0 . 2i
4π r
and sin θ = a ∴ Bnet = 2× µ0 . 2i × a = µ 0 ia .
r 4π r r πr 2
What will be the resultant magnetic field at origin due to four infinite length wires. If each wire produces
magnetic field 'B' at origin
(a) 4 B Y
(b) 2 B i4 1 ix 2
(c) 2 2 B x iX
(d) Zero 3i
Solution : (c) Direction of magnetic field (B1, B2, B3 and B4) at origin due to wires 1, 2, 3 and 4 are shown in the following figure.
Example: 6 B1 = B2 = B3 = B4 = µ0 . 2i = B . So net magnetic field at origin O i 1
Solution : (c) 4π x i2
i
Bnet = (B1 + B2 )2 + (B2 + B4 )2 B2B4
O B3 B1
= (2B)2 + (2B)2 = 2 2B 4i
3
Two parallel, long wires carry currents i1 and i2 with i1 > i2 . When the currents are in the same direction, the
magnetic field at a point midway between the wires is 10 µT. If the direction of i2 is reversed, the field
becomes 30 µT. The ratio i1 / i2 is
(a) 4 (b) 3 (c) 2 i1 (d) 1 i2
Initially when wires carry currents in the same direction as shown. O
Magnetic field at mid point O due to wires 1 and 2 are respectively xx
B1 = µ0 . 2i1 ⊗ and B2 = µ0 . 2i2
4π x 4π x
12
Hence net magnetic field at O Bnet = µ0 × 2 (i1 − i2)
4π x
⇒ 10 × 10 −6 = µ0 . 2 (i1 − i2) .....(i) i1 i2
4π x O
If the direction of i2 is reversed then xx
B1 = µ0 . 2i1 ⊗ and B2 = µ0 . 2i2 ⊗ 12
4π x 4π x
So Bnet = µ0 . 2 (i1 + i2) ⇒ 30 × 10 −6 = µ0 . 2 (i1 + i2) ......(ii)
4π x 4π x
Dividing equation (ii) by (i) i1 + i2 = 3 ⇒ i1 = 2
i1 − i2 1 i2 1
Example: 7 A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of three turns. If in
Solution : (a)
both cases same amount of current is passed, then the ratio of the intensities of magnetic field produced at the
Example: 8 centre of a coil will be
[MP PMT 2002]
Solution : (c)
Example: 9 (a) 9 times of first case (b) 1 times of first case (c) 3 times of first case (d) 1 times of first case
Solution : (b) 9 3
Magnetic field at the centre of n turn coil carrying current i B = µ0 . 2πni ......(i)
4π r
For single turn n =1 B = µ0 . 2πi .....(ii)
4π r
If the same wire is turn again to form a coil of three turns i.e. n = 3 and radius of each turn r' = r
3
So new magnetic field at centre B' = µ0 . 2π (3) ⇒ B' = 9 × µ0 . 2πi ......(iii)
4π r' 4π r
Comparing equation (ii) and (iii) gives B' = 9B .
A wire in the form of a square of side a carries a current i. Then the magnetic induction at the centre of the
square wire is (Magnetic permeability of free space = µ0) [EAMCET 2001]
(a) µ0i i
2πa O
(b) µ0i 2
πa
(c) 2 2µ0i a
πa
(d) µ0i i
2πa
Magnetic field due to one side of the square at centre O i
B1 µ0 . 2i sin 45 o B1 µ0 . 2 2i 45o O
4π a/ 2 4π a 45o
= ⇒ =
Hence magnetic field at centre due to all side Bnet = 4B1 = µ0 (2 2 i) a/2
πa a
The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the
centre of a square coil made from the same length of wire will be
(a) π 2 (b) π 2 (c) π (d) π
42 82 22 42
Circular coil Square coil i
i i 45o
45o
r O
i a/2
a
Length L = 2π r
Length L = 4a
Magnetic field B= µ0 . 2πi µ0 . 4π 2 i B= µ0 . 2 2 i B= µ0 . 8 2 i
4π r 4π r 4π a 4π a
=
Hence Bcircular = π2
Bsquare 82
Example: 10 Find magnetic field at centre O in each of the following figure
(i) (ii) (iii)
ir r1 r2
O O
r1
r2 O
(a) µ0i ⊗ (a) µ0 i 1 − 1 ⊗ (a) µ0 i 1 − 1 ⊗
r 4 r1 r2 4 r1 r2
(b) µ0i (b) µ0 i 1 + 1 ⊗ (b) µ0 i 1 + 1 ⊗
2r 4 r1 r2 4 r1 r2
(c) µ0i ⊗ (c) µ0 i 1 − 1 (c) µ0 i 1 − 1
4r 4 r1 r2 4 r1 r2
(d) µ0i (d) Zero (d) Zero
4r
Solution : (i) (c) Magnetic field at O due to parts 1 and 3, B1 = B3 = 0 (2)
(ii)
(iii) While due to part (2) B2 = µ0 . πi ⊗ r
4π r
Example: 11 (1) O (3)
∴ Net magnetic field at centre O, 2
Bnet = B1 + B2 + B3 = µ0 . π i ⊗ ⇒ Bnet = µ0i ⊗
4π r 4r
1 r1 3
O
(b) B1 = B3 = 0
B2 = µ0 . πi ⊗ r2
4π r1
B4 = µ0 . πi ⊗ 4
4π r2
So Bnet = B2 + B4 = µ0 .π i 1 + 1 ⊗
4π r1 r2
(a) B1 = B3 = 0
B2 = µ0 . πi ⊗ 4
4π r1
B4 = µ0 . πi ⊗ As | B2 | > | B4 | r2 2
4π r2
r1
So Bnet = B2 − B4 ⇒ Bnet = µ0 i 1 − 1 ⊗ 1O 3
4 r1 r2
Find magnetic field at centre O in each of each of the following figure
(i) (ii) (iii)
i i i
r 90o r r
O O
O
(a) µ0i (a) µ0 i (π − 2) ⊗ (a) µ0 2i (π + 1) ⊗
2r 2π r 2r r
(b) µ0i ⊗ (b) µ0i . i (π + 2) (b) µ0i . 2i (π − 1) ⊗
2r 4π r 4r r
(c) 3µ0i ⊗ (c) µ0i ⊗ (c) Zero
8r 4r
(d) 3µ0 i (d) µ0i (d) Infinite
8r 4r
Solution : (i) (d) By using B = µ0 . (2π − θ )i ⇒ B= µ0 . (2π −π / 2)i = 3µ0i
(ii) 4π r 4π r 8r
(iii) (b) Magnetic field at centre O due to section 1, 2 and 3 are respectively
Example: 12 B1 = µ0 . i 1i
4π r i
B2 = µ0 . πi r
4π r 2O
B3 = µ0 . i 3
4π r
⇒ Bnet = B1 + B2 + B3 = µ0 i (π + 2)
4π r
(b) The given figure is equivalent to following figure, magnetic field at O due to long wire (part 1)
B1 = µ0 . 2i
4π r
i
Due to circular coil B2 = µ0 . 2π i ⊗ r 2
4π r O
Hence net magnetic field at O
Bnet = B2 − B1 = µ0 . 2i (π − 1) ⊗
4π r
times that due1 to a long straight wire at a distance r
The field B at the centre of a circular coil of radius r is π
from it, for equal currents here shows three cases; in all cases the circular part has radius r and straight ones
are infinitely long. For same current the field B is the centre P in cases 1, 2, 3 has the ratio
[CPMT 1989]
ir ir i rO
O O 90o
(1) (2) (3)
(a) − π : π : 3π − 1 (b) − π + 1 : π + 1 : 3π + 1
2 2 4 2 2 2 4 2
(c) − π : π : 3π (d) − π − 1 : π − 1 : 3π + 1
2 2 4 2 2 2 4 2
Solution : (a) Case 1 : BA = µ0 . i ⊗ (A)
Example: 13 4π r i
(B)
BB = µ0 . i r
4π r O
(C)
BC = µ0 . i
4π r (B)
So net magnetic field at the centre of case 1
B1 = BB − (BA + BC ) ⇒ B1 = µ0 . πi ..... (i)
4π r
Case 2 : As we discussed before magnetic field at the centre O in this case
B2 = µ0 . πi ⊗ .....(ii) i r (C)
4π r
Case 3 : BA = 0 (A) O
BB = µ0 . (2π −π / 2) = µ0 . 3πi ⊗ (B)
4π r ⊗ 4π 2r
i rO
BC = µ0 . i (A) 90o
4π r
So net magnetic field at the centre of case 3
B3 = µ0 . i 3π − 1 ⊗ ....(iii) (C)
4π r 2
From equation (i), (ii) and (iii) B1 : B2 : B3 =π :π: 3π − 1 ⊗ = − π : π : 3π − 1
2 2 2 4 2
Two infinite length wires carries currents 8A and 6A respectively and placed along X and Y-axis. Magnetic field
at a point P (0, 0, d)m will be
(a) 7µ0 (b) 10µ 0 (c) 14 µ 0 (d) 5µ0
πd πd πd πd
Solution : (d) Magnetic field at P
Example: 14
Due to wire 1, B1 = µ0 . 2 (8) P (0, 0, d) Y
4π d B2
and due to wire 2, B2 = µ0 . 2 (16) B1
4π d
6A
B12 + B22 = µ0 16 2 µ0 12 2 µ0 2 5µ0 X
4π d 4π d 4π d πd
∴ Bnet = . + . = × × 10 = 8A
An equilateral triangle of side 'a' carries a current i then find out the magnetic field at point P which is vertex of triangle
(a) µ0i ⊗ P
2 3πa ii
(b) µ0i
2 3πa
(c) 2 3µ 0i a
πa
(d) Zero
Solution : (b) As shown in the following figure magnetic field at P due to side 1 and side 2 is zero.
Example: 15 Magnetic field at P is only due to side 3, P
Solution : (d) 12
which is B1 = µ0 . 2i sin 30o i 30o 30o i 3a
Example: 16 4π 3a 2
Solution : (b)
Example: 17 2
Solution : (b)
Example: 18 = µ0 . 2i = µ0i 3
4π 3a 2 3πa a
A battery is connected between two points A and B on the circumference of a uniform conducting ring of
radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. The value of, the
magnetic induction at the centre due to the current in the ring is [IIT-JEE 1995]
(a) Proportional to 2(180o − θ ) (b) Inversely proportional to r
(c) Zero, only if θ = 180o (d) Zero for all values of θ
Directions of currents in two parts are different, so directions of magnetic fields due to these currents are
different. i1 R1 = i2 R2 ⇒ i1l1 = i2l 2 …..(i) i2 A
Also applying Ohm's law across AB r l1
Also B1 = µ0 × i1l1 and B2 = µ0 × i2l2 ; ∴ B2 = i1l1 =1 [Using (i)] θ i1
4π r2 4π r2 B1 i2l2 l2 r
B
Hence, two field are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and
is independent of θ .
The earth’s magnetic induction at a certain point is 7 ×10−5 Wb / m2 . This is to be annulled by the magnetic
induction at the centre of a circular conducting loop of radius 5 cm. The required current in the loop is
(a) 0.56 A (b) 5.6 A (c) 0.28 A [MP PET 1999; AIIMS 2000]
(d) 2.8 A
According to the question, at centre of coil B = BH ⇒ µ0 . 2π i = BH
4π r
⇒ 10 −7 × 2π i = 7 × 10 −5 ⇒ i = 5.6 amp.
(5 × 1−2 )
A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular
path of radius 0.8 metre. The value of the magnetic field produced at the centre will be (µ0 − permeability for
vacuum) [CPMT 1986]
(a) 10−7 (b) 10−17 µ0 (c) 10−6 µ0 (d) 10−7 µ0
µ0
Magnetic field at the centre of orbit due to revolution of charge.
B = µ0 . 2π (qν ) ; where ν = frequency of revolution of charge
4π r
So, B= µ0 × 2π × (100e × 1) ⇒ B = 10 −17 µ 0 .
4π 0.8
Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of 3R on its axis is
(a) 10 10 (b) 20 10 (c) 2 10 (d) 10
Bcentre = 1 + x2 3 / 2 Bcentre
Baxis r2 Baxis
Solution : (a) By using ; where x = 3R and r = R ⇒ = (10)3 / 2 = 10 10 .
Example: 19
Solution : (b) A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the
Example: 20
magnetic induction will be 1 th to its value at the centre of the coil, is [MP PMT 1997]
Solution : (c) 8
Example: 21
(a) R (b) R 3 (c) 2 3R (d) 2 R
Solution : (b)
3 3
Bcentre = 1 + x2 3 / 2 1
Baxis r2 8
By using , given r = R and Baxis = Bcentre
x2 3 / 2 1 + x2 1 / 2 3 x2 1 / 2 x2
R2 R2 R2 R2
⇒ 8 = 1 + ⇒ (2)2 = ⇒ 2 = 1 + ⇒ 4 =1+ ⇒ x= 3R
An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The
magnetic field due to this current at the point M is H1. Now, another infinitely long straight conductor QS is
connected at Q so that the current is 1 in QR as well as in QS, the current in PQ remaining unchanged. The
2
magnetic field at M is now H2. The ratio H1 / H2 is given by
[IIT-JEE (Screening) 1999]
(a) 1 90o M
2 i Q
–∞ +∞
(b) 1 P S
(c) 2 90o
3
R
(d) 2
Magnetic field at any point lying on the current carrying conductor is zero.
Here H1 = magnetic field at M due to current in PQ
H2 = magnetic field at M due to R + due to QS + due to PQ = 0+ H1 + H1 = 3 H1
2 2
∴ H1 = 2
H2 3
Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is
2r. The value of magnetic field at the centre of the loop assuming uniform wire is
(a) 2 µ0i B
3π a
(b) 2 µ0i ⊗ AC
3π a iO
(c) 2 µ0i D
π a
(d) 2 µ0i ⊗
π a
According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is
half that of ABC i.e. i2 = 1 . Also i1 + i2 =i ⇒ i1 = 2i and i2 = i
i1 2 3 3
Magnetic field at centre O due to wire AB and BC (part 1 and 2) B1 = µ0 . 2i1 sin 45o ⊗ = µ0 . 2 2 i1 ⊗
4π a/2 4π a
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) B3 = B4 = µ0 2 2 i2
4π a
Also i1 = 2i2. So (B1 = B2) > (B3 = B4) B (2)
Hence net magnetic field at centre O C
Bnet = (B1 + B2 ) − (B3 + B4 ) (1)
A i1
2 2 × 2 i 2 2 i × 2 i i2 O
3 3 (3)
= 2 × µ0 . − µ0 . = µ0 . 42 i (2 − 1) ⊗ = 2 µ0i ⊗ (4)
4π a 4π 4π 3a 3π a D
a
Tricky example: 1
Figure shows a straight wire of length l current i. The magnitude of magnetic field produced by the
current at point P is lP
i
l
(a) 2µ0i (b) µ0i (c) 2µ0i (d) µ0i
πl 4πl 8πl 2 2πl
Solution: (c) The given situation can be redrawn as follow.
As we know the general formula for finding the magnetic field due to a finite length wire
B = µ0 . i (sin φ1 + sin φ 2 ) l
4π r
P
Here φ1 = 0o, φ = 45o i 45o
∴ B = µ0 . i (sin 0 o + sin 45o ) = µ0 . i ⇒ B= 2µ0i l
4π r 4π 2l 8πl
Tricky example: 2
A cell is connected between the points A and C of a circular conductor ABCD of centre 'O' with angle
AOC = 60°, If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC
and ADC respectively, the ratio B1 is i1 B [KCET (Engg./ Med.) 1999]
B2 O
300o
(a) 0.2 60o
AC
(b) 6
i2 D
(c) 1
(d) 5 1A
Solution: (c) B = µ0 . θ i
4π r
⇒ B∝θi
⇒ B1 = θ1 × i1
B2 θ2 i2
Also i1 = l2 = θ2 Hence B1 1
i2 l1 θ1 B2 =1
Magnetic effects of current
Amperes Law.
Amperes law gives another method to calculate the magnetic field due to a given current distribution.
Line integral of the magnetic field B around any closed curve is equal to µ0 times the net current i threading
through the area enclosed by the curve
B
Bdl = µ0 θ
∫ ∑i.e. i = µ0 (i1 + i3 − i2 )
i1 i3
i2 i5
Also using B = µ0 H (where H = magnetising field)
∫ ∫µ0 H.dl = µ0Σi ⇒ H.dl = Σi i4
Note : ≅ Total current crossing the above area is (i1 + i3 − i2 ) . Any current outside the area is not
included in net current. (Outward → +ve, Inward ⊗ → – ve)
≅ When the direction of current is away from the observer then the direction of closed path is
clockwise and when the direction of current is towards the observer then the direction of closed path
is anticlockwise.
OO
Application of Amperes law. P P
(1) Magnetic field due to a cylindrical wire r
(i) Outside the cylinder i
P
R R R1 r
i R2 i
Solid cylinder Thin hollow cylinder Thick hollow cylinder
B.dl = µ0i µ0i
∫ ∫In all above cases magnetic field outside the wire at P ⇒ B dl = µ0i ⇒ B × 2πr = µ0i ⇒ Bout = 2πr
In all the above cases Bsurface = µ0i
2πR
(ii) Inside the cylinder : Magnetic field inside the hollow cylinder is zero.
B≠0 B=0 B≠0
B=0
Cross sectional view ⇒ Solid cylinder Thin hollow cylinder Thick hollow cylinder
Solid cylinder Inside the thick portion of hollow cylinder
Rr r Loop Q Loop
Loop R1
i R2
Current enclosed by loop (i′) is lesser then the total current (i) Current enclosed by loop (i′) is lesser then the total current (i)
Current density is uniform i.e. J = J′ ⇒ i = i' Also i' = i× A' = i× (r 2 − R12 )
A A' A (R22 − R12 )
⇒ i' = i× A' = i r2 ∫Hence at point Q B. dl = µ 0 i' ⇒ B × 2πr = µ0i × (r 2 − R12)
A R2 (R22 − R12)
∫Hence at point Q B. dl = µ0i' ⇒ B × 2πr = µ0ir 2 ⇒ B= µ0i . (r 2 − R12) . If r = R1 (inner surface) B=0
R2 2πr (R22 − R12)
⇒ B = µ0 . ir If r = R2 (outer surface) B = µ0i (max.)
2π R2 2π R2
Note : ≅For all cylindrical current distributions
Baxis = 0 (min.), Bsurface = max (distance r always from axis of cylinder), Bout ∝ 1/r.
(2) Magnetic field due to an infinite sheet carrying current : The figure shows an infinite sheet of
current with linear current density j (A/m). Due to symmetry the field line pattern above and below the sheet is
uniform. Consider a square loop of side l as shown in the figure.
PB
i dc
l
al b
bB.dl + cB.dl + dB.dl + aB.dl
a b c d
∫ ∫ ∫ ∫According to Ampere’s law, = µ0i .
Since B ⊥ dl along the path b → c and d → a, therefore, cB.dl = 0 ; aB.dl = 0
∫ ∫b d
Also, B || dl along the path a → b and c → d, thus bB.dl + aB.dl = 2Bl
∫ ∫a d
The current enclosed by the loop is i = jl
Therefore, according to Ampere’s law 2Bl = µ0 ( jl) or B = µ0 j
2
(3) Solenoid
A cylinderical coil of many tightly wound turns of insulated wire with generally diameter of the coil smaller
than its length is called a solenoid.
One end of the solenoid behaves like the north pole and opposite end behaves like the south pole. As the
length of the solenoid increases, the interior field becomes more uniform and the external field becomes weaker.
B=0
SN
B
i Solenoid i
A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform
and parallel to the axis of solenoid.
(i) Finite length solenoid : If N = total number of turns,
l = length of the solenoid r αβ
P
N
n = number of turns per unit length = l
Magnetic field inside the solenoid at point P is given by B = µ0 (2π ni)[sinα + sin β]
4π
(ii) Infinite length solenoid : If the solenoid is of infinite length and the point is well inside the solenoid i.e.
α = β = (π / 2) .
So Bin = μ0ni
(ii) If the solenoid is of infinite length and the point is near one end i.e. α = 0 and β = (π / 2)
So Bend = 1 (μ 0ni)
2
Note : ≅Magnetic field outside the solenoid is zero.
≅ Bend = 1 Bin
2
(4) Toroid : A toroid can be considered as a ring shaped closed solenoid. Hence it is like an endless
cylindrical solenoid.
Winding Core rP
i O
r d→l
→B
Consider a toroid having n turns per unit length
Let i be the current flowing through the toroid (figure). The magnetic lines of force mainly remain in the core
of toroid and are in the form of concentric circles. Consider such a circle of mean radius r. The circular closed path
∫surrounds N loops of wire, each of which carries a current i therefore from B.dl = µ0inet
⇒ B × (2πr) = µ0 Ni ⇒ B = µ0 Ni = µ o ni where n = N
2πr 2πr
For any point inside the empty space surrounded by toroid and outside the toroid, magnetic field B is zero
because the net current enclosed in these spaces is zero.
Concepts
The line integral of magnetising field (H) for any closed path called magnetomotive force (MMF). It's S.I. unit is amp.
Ratio of dimension of e.m.f. to MMF is equal to the dimension of resistance.
Biot-Savart law is valid for asymmetrical current distributions while Ampere's law is valid for symmetrical current distributions.
Biot-Savart law is based only on the principle of magnetism while Ampere's laws is based on the principle of electromagnetism.
Example
Example: 22 A long solenoid has 200 turns per cm and carries a current of 2.5 A. The magnetic field at its centre is
Solution : (b) [µ0 = 4π × 10–7 Wb/m2] (b) 6.28 × 10–2 Wb/m2 [MP PET 2000]
Example: 23 (a) 3.14 × 10–2 Wb/m2
(c) 9.42 × 10–2 Wb/m2 (d) 12.56 × 10–2 Wb/m2
Solution : (a)
Example: 24 B = µ 0 ni = 4π × 10 −7 × 200 × 2.5 = 6.28 × 10 −2 Wb / m2 .
10 −2
Solution : (a)
Example: 25 A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of 20
Solution : (d) millitesla inside the solenoid will be approximately µ0 = 10 −7 Tesla - metre / ampere [MP PMT 1994]
Example: 26 4π
(a) 8.0 A (b) 4.0 A (c) 2.0 A (d) 1.0 A
B = µ0ni ; where n = 20 turn = 2000 turn . So, 20 × 10 −5 = 4π × 2000 × i ⇒ i = 8 A.
10 cm m
Two solenoids having lengths L and 2L and the number of loops N and 4N, both have the same current, then
the ratio of the magnetic field will be [CPMT 1994]
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
B = µ0 N i ⇒ B ∝ N ⇒ B1 = N1 × L2 = N 2l = 1 .
L L B2 N2 L1 4N × L 2
The average radius of a toroid made on a ring of non-magnetic material is 0.1 m and it has 500 turns. If it
carries 0.5 ampere current, then the magnetic field produced along its circular axis inside the toroid will be
(a) 25 × 10−2 Tesla (b) 5 × 10−2 Tesla (c) 25 × 10−4 Tesla (d) 5 × 10−4 Tesla
B = µ0ni ; where n = N ∴ B = 4π × 10 −7 × 500 × 0.5 = 5 × 10 −4 T.
2πR 2π × 0.1
For the solenoid shown in figure. The magnetic field at point P is
(a) µ 0 ni ( 3 + 1)
4
n turn
(b) 3 µ 0 ni
4 30o 60o
P
µ 0 ni
(c) 2 ( 3 + 1)
(d) µ 0 ni ( 3 − 1)
4
Solution : (a) B = µ0 .2π ni (sinα + sin β ) . From figure α = (90o – 30o) = 60o and β = (90o – 60o) = 30o
Example: 27 4π
Solution : (d) ∴ B = µ 0 ni (sin 60 o + sin 30o ) = µ 0 ni ( 3 + 1) .
2 4
Figure shows the cress sectional view of the hollow cylindrical conductor with inner radius 'R' and outer radius
'2R', cylinder carrying uniformly distributed current along it's axis. The magnetic induction at point 'P' at a
distance 3R from the axis of the cylinder will be
2
(a) Zero
(b) 5µ0i
72π R
R
(c) 7µ0i 2R 3R/2
18π R
(d) 5µ0i
36π R
r2 a2 3R − R 2
b2 a2 2
By using B= µ0i − here r = 3R , a = R, ab = 2R ⇒ B = µ0i × = 5.µ o i .
2π r − 2 36π r
2π 3R (R 2 ) − R 2
2
Tricky example: 3
A winding wire which is used to frame a solenoid can bear a maximum 10 A current. If length of solenoid
is 80cm and it's cross sectional radius is 3 cm then required length of winding wire is (B = 0.2 T)
(a) 1.2 × 10 2 m (b) 4.8 × 10 2 m (c) 2.4 × 10 3 m (d) 6 × 10 3 m
Solution : (c) B= µ0 Ni where N = Total number of turns, l = length of the solenoid
l
⇒ 0.2 = 4π × 10 −7 × N × 10 ⇒ N = 4 × 10 4
0.8 π
Since N turns are made from the winding wire so length of the wire (L) = 2π r × N [2π r = length of each turns]
⇒ L = 2π × 3 × 10 −2 × 4 × 10 4 = 2.4 × 10 3 m.
π
Magnetic effects of current
Motion of Charged Particle in a Magnetic Field.
If a particle carrying a positive charge q and moving with velocity v enters a magnetic field B then it
experiences a force F which is given by the expression × ×× × v × B→ ×
F = q(v × B) ⇒ F = qvB sinθ × ×
× q, m× ××
Here v = velocity of the particle, B = magnetic field
××××××
××××××
(1) Zero force
Force on charged particle will be zero (i.e. F = 0) if
(i) No field i.e. B = 0 ⇒ F = 0 θ = 0o
(ii) Neutral particle i.e. q = 0 ⇒ F = 0 qv
(iii) Rest charge i.e. v = 0 ⇒ F = 0
(iv) Moving charge i.e. θ = 0o or θ = 180o ⇒ F = 0 B
q
θ = 180o
(2) Direction of force
ScrewTRhuelef,otrhcreouFghisvalawnadysBpethrpeemnsdeilcvuelsarmtaoyboorthmtahyenvoetlobceitpyervpeannddictuhlearfiteoldeaBchionthaecrc.ordance with Right Hand
→ →
Fm Fm
→
θ →v →
→v 90° B
B
Direction of force on charged particle in magnetic field can also be find by Flemings Left Hand Rule (FLHR).
Here, First finger (indicates) → Direction of magnetic field F
Middle finger → Direction of motion of positive charge or direction, B
opposite to the motion of negative charge.
Thumb → Direction of force v
(3) Circular motion of charge in magnetic field
Consider a charged particle of charge q and mass m enters in a uniform magnetic field B with an initial
velocity v perpendicular to the field.
× × v × + × v× ×
×××××××
× ×+ × × × +× ×
F
×××××××
×××××××
×v × + v
× ×
× ××
θ = 90o, hence from F = qvB sinθ particle will experience a maximum magnetic force Fmax = qvB which act's
in a direction perpendicular to the motion of charged particle. (By Flemings left hand rule).
(i) Radius of the path : In this case path of charged particle is circular and magnetic force provides the
necessary centripetal force i.e. qvB = mv 2 ⇒ radius of path r = mv
r qB
If p = momentum of charged particle and K = kinetic energy of charged particle (gained by charged particle
after accelerating through potential difference V) then p = mv = 2mK = 2mqV
So r = mv = p = 2mK 1 2mV
qB qB qB = B q
r ∝ v ∝ p ∝ K i.e. with increase in speed or kinetic energy, the radius of the orbit increases.
Note : ≅ Less radius (r) means more curvature (c) i.e. c ∝ 1
r
Less : r More : r r=∞
More : c Less : c c=0
(ii) Direction of path : If a charge particle enters perpendicularly in a magnetic field, then direction of path
described by it will be
Type of charge Direction of magnetic field Direction of it’s circular motion
Negative Outwards B
Anticlockwise
Negative Inward ⊗ –q
⊗ B
Clockwise
Positive Inward ⊗ –q
⊗ B
Anticlockwise
Positive Outward +q
B Clockwise
+q
(iii) Time period : As in uniform circular motion v = rω, so the angular frequency of circular motion, called
cyclotron or gyro-frequency, will be given by ω = v = qB and hence the time period, T = 2π = 2π m
r m ω qB
i.e., time period (or frequency) is independent of speed of particle and radius of the orbit and depends only on
the field B and the nature, i.e., specific charge q , of the particle.
m
(4) Motion of charge on helical path
When the charged particle is moving at an angle to the field (other than 0o, 90o, or 180o).
In this situation resolving the velocity of the particle along and perpendicular to the field, we find that the
particle moves with constant velocity v cosθ along the field (as no force acts on a charged particle when it moves
parallel to the field) and at the same time it is also moving with velocity v sinθ perpendicular to the field due to
which it will describe a circle (in a plane perpendicular to the field) of radius. r = m(vsinθ)
qB
→v → Yp →B
q, m θ r
B v sinθ v
X
θ
v cosθ
Z
Time period and frequency do not depend on velocity and so they are given by T= 2π m and ν = qB
qB 2π m
So the resultant path will be a helix with its axis parallel to the field B as shown in figure in this situation.
The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by
p = T(v cosθ ) = 2π m (v cosθ )
qB
Note : ≅ 1 rotation ≡ 2π ≡ T and 1 pitch ≡ 1 T
≅ Number of pitches ≡ Number of rotations ≡ Number of repetition = Number of helical turns
≅ If pitch value is p, then number of pitches obtained in length l given as
Number of pitches = l and time reqd. t= l
p v cosθ
Some standard results
Ratio of radii of path described by proton and α-particle in a magnetic field (particle enters perpendicular to
the field)
Constant quantity Formula Ratio of radii Ratio of curvature (c)
v - same rp : rα = 1 : 2 cp : cR = 2 :1
r = mv ⇒ r ∝ m rp : rα = 2 : 1 cp : cR = 1: 2
p - same qB q rp : rα = 1 : 1 cp : cR = 1:1
k - same
r = p ⇒r ∝ 1 rp : rα = 1 : 2 cp : cR = 2 :1
V - same qB q
r= 2mk ⇒r ∝ m
qB q
r∝ m
q
Particle motion between two parallel plates (v ⊥ B)
(i) To strike the opposite plate it is essential that d < r . ×××××××
(ii) Does not strike the opposite plate d > r. × d<r × × d=r
(iii) To touch the opposite plate d = r .
(iv) To just not strike the opposite plate d ≥ r . ×××××××
(v) To just strike the opposite plate d ≤ r . × × × × × d>r
×××××××
q, m
(5) Lorentz force EFm a=ndq(vm ×agBn)e; tiscofitehled B ,
net
the When the moving charged particle is subjected sfoimrcueltaFneeo=uqsEly to both electric field
moving charged particle will experience electric and magnetic force
force on it will be F = q[E + (v × B)] . Which is the famous ‘Lorentz-force equation’.
Depending on the directions of v, E and B following situations are possible
(i) When v, E and B all the three are collinear : In this situation as
antiparallel to the field, the magnetic force on it will be zero and only electric force the particle is ma o=vmiFng=pqmaErallel or
will act and so
The particle will pass through the field following a straight line path (parallel field) with change in its speed. So
in this situation speed, velocity, momentum kinetic energy all will change without change in direction of motion as
E
shown v
B
q
(ii) When E is parallel to B and both these fields are perpendicular to v then : Fe is
perpendicular to Fm and they cannot cancel each other. The path of charged particle is curved in both these fields.
E B
v
q
(iii) v , E and B are mutually perpendicular : In this situation if E and B are such that y E
Fe
F = + = 0 i.e., a = (F / m) = 0 +q v
Fe Fm Fm
as shown in figure, the particle will pass through the field with same velocity. +q x
And in this situation, as Fe = Fm i.e., qE = qvB v = E / B B
z
This principle is used in ‘velocity-selector’ to get a charged beam having a specific velocity.
Note : ≅ From the above discussion, conclusion is as follows
≅ If E = 0, B = 0, so F = 0.
≅ If E ≠≠=000,,,BBB=≠≠000,,,sssoooFFF=≠ma00y((bibfee|cFzaeeur|soe=|(ivfFm≠θ = 0o or 180o ). are opposite)
≅ If E | and their directions
≅ If E constant ).
Cyclotron.
Cyclotron is a device used to accelerated positively charged particles (like, α-particles, deutrons etc.) to acquire
enough energy to carry out nuclear disintegration etc. t is based on the fact
that the electric field accelerates a charged particle and the magnetic field
keeps it revolving in circular orbits of constant frequency. Thus a small High frequency N D2
potential difference would impart if enormously large velocities if the oscillator W Energetic
particle is made to traverse the potential difference a number of times.
D1 proton beam
It consists of two hollow D-shaped metallic chambers D1 and D2
called dees. The two dees are placed horizontally with a small gap Target
separating them. The dees are connected to the source of high frequency S
electric field. The dees are enclosed in a metal box containing a gas at a
low pressure of the order of 10–3 mm mercury. The whole apparatus is placed between the two poles of a strong
electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.
Note : ≅ The positive ions are produced in the gap between the two dees by the ionisation of the gas.
To produce proton, hydrogen gas is used; while for producing alpha-particles, helium gas is used.
(1) Cyclotron frequency : Time taken by ion to describe q semicircular path is given by t = πr = πm
v qB
If T = time period of oscillating electric field then T = 2t = 2π m the cyclotron frequency ν = 1 = Bq
qB T 2πm
(2) Maximum energy of position : Maximum energy gained by the charged particle Emax = q2B2 r 2
2m
where r0 = maximum radius of the circular path followed by the positive ion.
Note : ≅Cyclotron frequency is also known as magnetic resonance frequency.
≅ Cyclotron can not accelerate electrons because they have very small mass.
Hall effect : The Phenomenon of producing a transverse emf in a current carrying conductor on applying a
magnetic field perpendicular to the direction of the current is called Hall effect.
Hall effect helps us to know the nature and number of charge carriers in a conductor.
Negatively charged particles Positively charged particles
Consider a conductor having electrons as current carriers. Let the current carriers be positively charged holes. The
hole move in the direction of current
The electrons move with drift velocity v opposite to the
direction of flow of current
→B z →B
x
z +++ + ++ ––– – ––
x
y →v →F – y →F + →v
+ + + + ++ +
– – – – –– – VH VH
Force acting on the hole due to magnetic field
force acting on electron Fm = −e(v × B). This force acts Fm = +e(v × B) force acts along x-axis and hence holes
along x-axis and hence electrons will move towards face move towards face (2) and it becomes positively charged.
(2) and it becomes negatively charged.
Concepts
The energy of a charged particle moving in a uniform magnetic field does not change because it experiences a force in a
direction, perpendicular to it's direction of motion. Due to which the speed of charged particle remains unchanged and hence
it's K.E. remains same.
Magnetic force does no work when the charged particle is displaced while electric force does work in displacing the charged
particle.
Magnetic force is velocity dependent, while electric force is independent of the state of rest or motion of the charged particle.
If a particle enters a magnetic field normally to the magnetic field, then it starts moving in a circular orbit. The point at which it
enters the magnetic field lies on the circumference. (Most of us confuse it with the centre of the orbit)
Deviation of charged particle in magnetic field : If a charged particle (q, m) enters a uniform magnetic field B (extends
upto a length x) at right angles with speed v as shown in figure.
The speed of the particle in magnetic field does not change. But it gets deviated in the magnetic field. v
××××
Bq × θ× × θ×
m
Deviation in terms of time t; θ = ω t = t ××× ×
Deviation in terms of length of the magnetic field ; θ = sin −1 x . This relation can be used only when x ≤ r . q, m× ×× × B→
r
× v ×
×x×
×
×××
For x > r, the deviation will be 180o as shown in the following figure
v ×××
×× × ×
r× ×× ×
×
× v×
x
Example
Example: 28 Electrons move at right angles to a magnetic field of 1.5 × 10 −2 Tesla with a speed of 6 × 10 27 m / s. If the
Solution : (c) specific charge of the electron is 1.7 × 1011 Coul/kg. The radius of the circular path will be [BHU 2003]
Example: 29
(a) 2.9 cm (b) 3.9 cm (c) 2.35 cm (d) 3 cm
Solution : (d)
r = mv ⇒ v = 6 × 10 27 = 2.35 × 10 −2 m = 2.35 cm.
qB (q / m). B 17 × 1011 × 1.5 × 10 −2
An electron (mass = 9 × 10 −31 kg. charge = 1.6 × 10 −19 coul. ) whose kinetic energy is 7.2 × 10 −18 joule is
moving in a circular orbit in a magnetic field of 9 × 10−5 weber / m2. The radius of the orbit is [MP PMT 2002]
(a) 1.25 cm (b) 2.5 cm (c) 12.5 cm (d) 25.0 cm
r= 2mK 2 × q × 10 −31 × 7.2 × 10 −8 = 0.25 cm = 25 cm .
qB = 1.6 × 10 −19 × q × 10 −5
Example: 30 An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic energy. Which of the
Solution : (b)
following is true [MP PET 1999]
Example: 31
Solution : (b) (a) Trajectory of electron is less curved (b) Trajectory of proton is less curved
Example: 32
Solution : (a) (c) Both trajectories are equally curved (d) Both move on straight line path
Example: 33
Solution : (c) By using r = 2mk ; For both particles q → same, B → same, k → same
Example: 34 qB
Solution : (d)
Hence r ∝ m ⇒ re = me mp > me so rp > re
rp mp
Since radius of the path of proton is more, hence it's trajectory is less curved.
A proton and an α − particles enters in a uniform magnetic field with same velocity, then ratio of the radii of
path describe by them
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) None of these
By using r= mv ; v → same, B → same ⇒ r ∝ m ⇒ rp = mp × qα = mp × 2q p = 1
qB 2 rα mα qp 4mp qp 2
A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What
should be the energy of α-particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same
radius [BHU 1997]
(a) 1 MeV (b) 4 MeV (c) 2 MeV (d) 0.5 MeV
By using r = 2mK ; r → same, B → same ⇒ K ∝ q2
qB m
Hence Kα = qα 2 × mp = 2q p 2 × mp 1 ⇒ Kα = K p = 1meV.
Kp qp mα qp 4mp
A proton and an α − particle enter a uniform magnetic field perpendicularly with the same speed. If proton
takes 25µ sec to make 5 revolutions, then the periodic time for the α − particle would be [MP PET 1993]
(a) 50µ sec (b) 25µ sec (c) 10µ sec (d) 5µ sec
Time period of proton Tp = 25 = 5µ sec
5
By using T = 2π m ⇒ Tα = mα × qp = 4mp × qp ⇒ Tα = 2Tp = 10µ sec .
qB Tp mp qα mp 2q p
A particle with 10–11 coulomb of charge and 10–7 kg mass is moving with a velocity of 108 m/s along the y-axis.
A uniform static magnetic field B = 0.5 Tesla is acting along the x-direction. The force on the particle is
[MP PMT 1997]
(a) 5 × 10–11 N along ˆi (b) 5 × 103 N along kˆ (c) 5 × 10–11 N along − ˆj (d) 5 × 10–4 N along − kˆ
By using F = q(v × B) ; where v = 10ˆj and B = 0.5ˆi
⇒ F = 10 −11(108 ˆj × 0.5ˆi ) = 5 × 10 −4 (ˆj × ˆi ) = 5 × 10 −4 (−kˆ) i.e., 5 × 10 −4 N along − kˆ.
Example: 35 An electron is moving along positive x-axis. To get it moving on an anticlockwise circular path in x-y plane, a
Solution : (a)
magnetic filed is applied [MP PMT 1999]
Example: 36
Solution : (b) (a) Along positive y-axis (b) Along positive z-axis
Example: 37
Solution : (b) (c) Along negative y-axis (d) Along negative z-axis
Example: 38 The given situation can be drawn as follows
Solution : (c)
Example: 39 According to figure, for deflecting electron in x-y plane, force must be acting an it towards y-axis.
Hence according to Flemings left hand rule, magnetic field directed along positive y − axis.
y x-y plane
e– e–
z x
A particle of charge − 16 ×10−18 coulomb moving with velocity 10 m/s along the x-axis enters a region where
a magnetic field of induction B is along the y-axis, and an electric field of magnitude 104 V/m is along the
negative z-axis. If the charged particle continuous moving along the x-axis, the magnitude of B is [AIEEE 2003]
(a) 10−3 Wb / m2 (b) 103 Wb / m2 (c) 105 Wb / m2 (d) 1016 Wb / m2
Particles is moving undeflected in the presence of both electric field as well as magnetic field so it's speed
v= E ⇒ B= E 10 4 = 103 Wb / m2.
B v = 10
A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a
region containing a uniform magnetic field B directed along the negative z direction extending from x = a
to x = b. The minimum value of v required so that the particle can just enter the region x > b is
(a) qbB/m (b) q(b – a)B/m (c) qaB/m [IIT-JEE (Screening) 2002]
(d) q(b+a)B/2m
As shown in the following figure, the z − axis points out of the paper and the magnetic fields is directed into
the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the
magnetic field. It can just enter the region x > b for r ≥ (b − q) Y QS
⊗B
Now r = mv ≥ (b − a)
qb
⇒ v ≥ q(b − a)B ⇒ vmin = q(b − a)B . Ov x=a x=b X
m m
PR
At a certain place magnetic field vertically downwards. An electron approaches horizontally towards you and
enters in this magnetic fields. It's trajectory, when seen from above will be a circle which is
(a) Vertical clockwise (b) Vertical anticlockwise
(c) Horizontal clockwise (d) Horizontal anticlockwise
By using Flemings left hand rule.
When a charged particle circulates in a normal magnetic field, then the area of it's circulation is proportional to
(a) It's kinetic energy (b) It's momentum
(c) It's charge (d) Magnetic fields intensity
Solution : (a) r= 2mK and A = Aq 2 ⇒ A = π (2mK) ⇒ A ∝ K.
Example: 40 qB q 2b 2
Solution : (b) An electron moves straight inside a charged parallel plate capacitor at uniform charge density σ . The space
Example: 41
between the plates is filled with constant magnetic field of induction B. Time of straight line motion of the
Solution : (b) electron in the capacitor is
Example: 42
Solution : (d) (a) eσ
ε 0lB
×××××××
(b) ε 0lB ×××××××
σ
e–
×××××××
(c) eσ ×××××××
ε0B l
(d) ε0B
eσ
The net force acting on the electron is zero because it moves with constant velocity, due to it's motion on
straight line.
⇒ Fnet = Fe + Fm = 0 ⇒ | F e |=| F m | ⇒ e E = evB ⇒ ve = E = σ E = σ
B ε0B εo
∴ The time of motion inside the capacitor t = l = ε 0lB .
v σ
A proton of mass 1.67 × 10−27 kg and charge 1.6 × 10−19 C is projected with a speed of 2 × 106 m / s at an
angle of 600 to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of proton
is [IIT-JEE 1995]
(a) A circle of radius = 0.2 m and time period π × 10−7 s
(b) A circle of radius = 0.1 m and time period 2π × 10−7 s
(c) A helix of radius = 0.1 m and time period 2π × 10−7 s
(d) A helix of radius = 0.2 m and time period 4π × 10−7 s Y
→
By using r = mv sinθ ⇒ r = 1.67 × 1527 × 2 × 106 × sin 30° = 0.1m B →v
qB 1.6 × 10−19 × 0.104 X
30o
2πm 2 × π × 9.1× 10−31 60o
qB 1.6 × 10−19 × 0.104
and it's time period T= = = 2π × 10−7 sec .
A charge particle, having charge q accelerated through a potential difference V enter a perpendicular magnetic
field in which it experiences a force F. If V is increased to 5V, the particle will experience a force
(a) F (b) 5F (c) F (d) 5F
5
1 mv 2 = qV ⇒ v= 2qV . Also F = qvB
2 m
⇒ F = qB 2qV hence F ∝ V which gives F' = 5F.
m
Example: 43 The magnetic field is downward perpendicular to the plane of the paper and a few charged particles are
projected in it. Which of the following is true [CPMT 1997]
(a) A represents proton and B and electron × × × ×A × × ×
(b) Both A and B represent protons but velocity of A is more than that of B ×××××××
(c) Both A and B represents protons but velocity of B is more than that of A ×××××××
(d) Both A and B represent electrons, but velocity of B is more than that of A × × × × × ×B ×
×××××××
×××××××
Solution : (c) Both particles are deflecting in same direction so they must be of same sign.(i.e., both A and B represents protons)
Example: 44
Solution : (d) By using r = mv ⇒ r ∝v
qB
From given figure radius of the path described by particle B is more than that of A. Hence vB > vA.
Two very long straight, particle wires carry steady currents i and – i respectively. The distance between the
wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the
plane of the wires. It's instantaneous velocity v is perpendicular to this plane. The magnitude of the force due
to the magnetic field acting on the charge at this instant is [IIT-JEE 1998]
(a) µ 0 iqv (b) µ 0 iqv (c) 2µ 0 iqv (d) Zero
2πd πd πd
According to gives information following figure can be drawn, which shows that direction of magnetic field is
along the direction of motion of charge so net on it is zero.
v
q
d
d/2 d/2
Example: 45 A metallic block carrying current i is subjected to a uniform magnetic induction B as shown in the figure. The
Solution : (c) moving charges experience a force F given by ……. which results in the lowering of the potential of the face
……. Assume the speed of the carriers to be v →B Y [IIT-JEE 1996]
(a) eVBkˆ, ABCD E G
(b) eVBkˆ, ABCD AF B HX
(c) − eVBkˆ, ABCD C I
(d) − eVBkˆ, EFGH D
As the block is of metal, the charge carriers are electrons; so for current along positive x-axis, the electrons are
moving along negative x-axis, i.e. v = − vi E B e– G y
and as the magnetic field is along the y-axis, i.e. B = Bˆj A vF zx
so F = q(v × B) for this case yield F = (−e)[−vˆi × Bˆj] di F B
H
C
D
i.e., F = evBkˆ [As ˆi × ˆj = kˆ ]
As force on electrons is towards the face ABCD, the electrons will accumulate on it an hence it will acquire
lower potential.
Tricky example: 4
Solution : (c) An ionised gas contains both positive and negative ions. If it is subjected simultaneously to an electric
field along the +ve x-axis and a magnetic field along the +z direction then [IIT-JEE (Screening) 2000]
(a) Positive ions deflect towards +y direction and negative ions towards – y direction
(b) All ions deflect towards + y direction
(c) All ions deflect towards – y direction
(d) Positive ions deflect towards – y direction and negative ions towards + y direction.
As the electric field is switched on, positive ion will start to move along positive x-direction and
negative ion along negative x-direction. Current associated with motion of both types of ions is along
positive x-direction. According to Flemings left hand rule force on both types of ions will be along
negative y-direction.
Force on a Current Carrying Conductor in Magnetic Field.
In case of current carrying conductor in a magnetic field force experienced by its small length element is
dF = idl × B ; idl = current element dF = l(dl × B) ×××××××
∫ ∫Total magnetic force F = dF = i(dl × B) ×××××××
If magnetic field is uniform i.e., B = constant × × × dF × × ×
∫F = i dl × B = i(L′ × B) × ×× × × ×B ×
× ×
×i d×l ××
×××××××
∫ dl = L = vector sum of all the length elements from initial to final point. Which is in accordance with the
law of vector addition is equal to length vector L′ joining initial to final point.
(1) Direction of force : The odfirtewcotiovnecotof rfso(rcAe×isB )alwwiathys perpendicular to the plane containing idl and B
is same as that of cross-product
and A = i dl .
→ →
dF i dl
P→ P θ→
B B
→
i dl θ
→
dF
The direction of force when current element i dl and B are perpendicular to each other can also be
determined by applying either of the following rules
Fleming’s left-hand rule Right-hand palm rule
Stretch the fore-finger, central finger and thumb left hand Stretch the fingers and thumb of right hand at right angles to
mutually poefrpfeienlddicBulara.nTdhetnheif the fore-finger points in the eBachanodthtehru.mTbheinn if the fingers point in the direction of field
direction central in the direction of the direction of current i, then normal to
current i, the thumb will point in the direction of force the palm will point in the direction of force
Force Current
Magnetic Magnetic
field field
Current Force
(2) Force on a straight wire : If a current carrying straight conductor (length l) is placed in an uniform
magnetic field (B) such that it makes an angle θ with the direction of field then force experienced by it is
F = Bil sinθ F
If θ = 0o , F = 0 B
If θ = 90o , Fmax = Bil il
(3) Force on a curved wire
The force acting on a curved wire joining points a and b as shown in the figure is the same as that on a straight
wire joining these points. It is given by the expression F = i L × B
×× ×× ××× × × × × × b× → ×
×× ×× b → × × ×F × × × B ×
×× ×× × ×
×× × →× × B
××
× L× ××× ×××××××
a
×× ××× × × i× × → × × ×
××
L
× a× × × × × ×
×××
××× ×××××××
Specific Example
The force experienced by a semicircular wire of radius R when it is carrying a current i and is placed in a
uniform magnetic field of induction B as shown.
Y → Y → ×× ×× ×××
i i ×× ×××
B B ×× ×Y × ×××
×× ×××
P O QX P O QX ×× ×× ×××
××
× P× ×× ×Q × X×
× O×
= 2Rˆi and = Bˆi = 2Rˆi and = Bˆj 2Rˆi and B (−kˆ)
L′ B L′ B L′ = B=
So by using F = i(L' × B) force on F = i× 2BR(ˆi × ˆj) ∴ F = i× 2BR(+ˆj)
the wire = 2BiR kˆ F = 2BiR (along Y-axis)
F F i.e. F = 2BiR
= i(2R)(B)(ˆi × ˆi) ⇒ F =0
(perpendicular to paper outward)
Force Between Two Parallel Current Carrying Conductors.
When two long straight conductors carrying currents i1 and i2 placed parallel to each other at a distance ‘a’
from each other. A mutual force act between them when is given as
F1 = F2 = F = µ0 ⋅ 2i1i2 ×l i1 i2
4π a
where l is the length of that portion of the conductor on which force is to be calculated. a
Hence force per unit length F = µ0 ⋅ 2i1i2 N or F = 2i1i2 dyne
l 4π a m l a cm
Direction of force : If conductors carries current in same direction, then force between them will be
attractive. If conductor carries current in opposite direction, then force between them will be repulsive.
1 × 2× × 1× × 2×
× i2 × × i1 × × i2 ×
••
F1 × ×
i1 ×
× ×
•• × ×
×
• • F1 F2 × × × × F2
• × × × ×
•
•• ×× ××
•• ×× ××
Note : ≅ If a = 1m and in free space F = 2 × 10−7 N / m then i1 = i2 = 1Amp in each identical wire. By this
l
concept S.I. unit of Ampere is defined. This is known as Ampere’s law.
Force Between Two Moving Charges.
If two charges q1 and q2 are moving with velocities v1 and v2 respectively and at any instant the distance
between them is r, then
q1 q2 Fe v1 Fm Fm v2
Fe r Fe q1 q2 Fe
r
Stationary charges Moving charges
Magnetic force between them is Fm = µ0 . q1 q 2 v1v 2 ..... (i)
4π r
2
and Electric force between them is Fe = 1 . q1q 2 ..... (ii)
4πε 0 r2
Fm 1 Fm v 2
Fe = c2 Fe c
From equation (i) and (ii) = µ0ε0v2 but µ0ε 0 ; where c is the velocity light in vacuum. So =
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