If v << c then Fm << Fe
Standard Cases for Force on Current Carrying Conductors.
Case 1 : When an arbitrary current carrying loop placed in a magnetic field (⊥ to the plane of loop), each
element of loop experiences a magnetic force due to which loop stretches and open into circular loop and tension
developed in it’s each part.
TA B T
⊗B
dθ
⊗B R
O
Specific example
In the above circular loop tension in part A and B.
In balanced condition of small part AB of the loop is shown below dF
2T sin dθ = dF = B i dl ⇒ 2T sin dθ = BiRdθ i B
2 2 A
If dθ is small so, sin dθ ≈ dθ ⇒ 2T. dθ = BiRdθ T dθ/2 dθ/2 dθ/2 dθ/2 T
2 2 2
T sin dθ
T = BiR , if 2πR = L so T = BiL 2 O T sin dθ
2π 2
Note : ≅If no magnetic field is present, the loop will still open into a circle as in it’s adjacent parts current will
be in opposite direction and opposite currents repel each other.
i
i
Case 2 : Equilibrium of a current carrying conductor : When a finite length current carrying wire is kept
parallel to another infinite length current carrying wire, it can suspend freely in air as shown below
Movable l Y Fixed i1
X i2 h Movable X
h
Fixed i1 i2
Y
l
In both the situations for equilibrium of XY it's downward weight = upward magnetic force i.e. mg = μ0 . 2i1i 2 .l
4π h
Note : ≅ In the first case if wire XY is slightly displaced from its equilibrium position, it executes SHM and it’s time
period is given by T = 2π h .
g
≅ If direction of current in movable wire is reversed then it’s instantaneous acceleration produced is 2g ↓.
Case 3 : Current carrying wire and circular loop : If a current carrying straight wire is placed in the
magnetic field of current carrying circular loop.
i1 i1
i2 i2
l
d
Wire is placed in the perpendicular magnetic field wire is placed along the axis of coil so magnetic field
due to coil at it's centre, so it will experience a produced by the coil is parallel to the wire. Hence it
will not experience any force.
µ 0 i1
maximum force F = Bil = 2r × i2l
Case 4 : Current carrying spring : If current is passed through a spring, then it will contract because
current will flow through all the turns in the same direction.
Spring Spring +
m –
K
Hg
If current makes to flow through spring, If switch is closed then current start flowing,
then spring will contract and weight lift up spring will execute oscillation in vertical plane
Case 5 : Tension less strings : In the following figure the value and direction of current through the
conductor XY so that strings becomes tensionless?
Strings becomes tensionless if weight of conductor XY balanced by magnetic force (Fm ) .
String
×××× ×× ×→ Fm
B Y
×××× ×× i
× X mg
T ×X × l× × Y×
m T
× ××
×
××××
×
×× ××
Hence direction of current is from X → Y and in balanced condition Fm = mg ⇒ Bil = mg ⇒ i= mg
Bl
Case 6 : A current carrying conductor floating in air such that it is making an angle θ with the direction of
magnetic field, while magnetic field and conductor both lies in a horizontal plane.
Fm In equilibrium mg = B i l sinθ ⇒ i= mg
θ B l sinθ
mg
Case 7 : Sliding of conducting rod on inclined rails : When a conducting rod slides on conducting rails.
X B F cosθ R
Fθ
Insulated i i
stand v θ mg sinθ
θ mg θ
Y
i
+
–
In the following situation conducting rod (X, Y) slides at constant velocity if
F cosθ = mg sinθ ⇒ B i l cosθ = mg sinθ ⇒ B = mg tanθ
il
Concepts
Electric force is an absolute concept while magnetic force is a relative concept for an observer.
The nature of force between two parallel charge beams decided by electric force, as it is dominator. The nature of force between
two parallel current carrying wires decided by magnetic force.
i1 i2 ++ –+
++ –+
++ –+
++ –+
Fnet = Fm only Fe → repulsion Fe → attraction
Fm → attraction Fm → repulsion
Fnet → repulsion (Due to this Fnet → attraction (Due to this
force these beams diverges) force these beams converges)
Example
Example: 46 A vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed
Solution : (d) towards north. The wire will experience a force directed towards
(a) North (b) South (c) East (d) West
By applying Flemings left hand rule, direction of force is found towards west.
Fi BH
W N
S E
Example: 47 3 A of current is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a magnetic
Solution : (c)
Example: 48 field of strength 500 gauss and makes an angle of 30o with the direction of the field. It experiences a force of
Solution : (c) magnitude [MP PET 1993]
Example: 49
(a) 3 × 104 N (b) 3 × 102 N (c) 3 × 10– 2 N (d) 3 × 10–4 N
Solution : (c)
By using F = Bil sinθ ⇒ F = (500 × 10–4) × 0.4 × sin 30o ⇒ 3 × 10–2 N.
Example: 50
Wires 1 and 2 carrying currents t1 and t2 respectively are inclined at an angle θ to each other. What is the
force on a small element dl of wire 2 at a distance of r from 1 (as shown in figure) due to the magnetic field of
wire 1
[AIEEE 2002]
(a) µ0 i1 , i2dl tanθ
2πr
(b) µ0 i1 , i2 dl sinθ i1 i2
2πr r dl
(c) µ0 i1 , i2 dl cosθ θ
2πr
(d) µ0 i1, i2 dl sinθ
4πr
Length of the component dl which is parallel to wire (1) is dl cos θ, so force on it
F = µ0 ⋅ 2i1i2 (dl cosθ ) = µ0i1i2dl cosθ .
4π r 2πr
A conductor PQRSTU, each side of length L, bent as shown in the figure, carries a current i and is placed in a
uniform magnetic induction B directed parallel to the positive Y-axis. The force experience by the wire and its
direction are Z R →
(a) 2iBL directed along the negative Z-axis i Q B
(b) 5iBL directed along the positive Z-axis S P
(c) iBL direction along the positive Z-axis T Y
(d) 2iBL directed along the positive Z-axis X U
As PQ and UT are parallel to Q, therefore FPQ = FUT = 0
The current in TS and RQ are in mutually opposite direction. Hence, FTS − FRQ = 0
Therefore the force will act only on the segment SR whose value is Bil and it’s direction is +z.
Alternate method :
The given shape of the wire can be replaced by a Z R → → →
i B F B
straight wire of length l between P and U as shown S P
Q P⇒ i
below
Y
Hence force on replaced wire PU will be F = Bil
and according to FLHR it is directed towards +z-axis T U U
X
A conductor in the form of a right angle ABC with AB = 3cm and BC = 4 cm carries a current of 10 A. There
is a uniform magnetic field of 5T perpendicular to the plane of the conductor. The force on the conductor will
be [MP PMT 1997]
(a) 1.5 N (b) 2.0 N (c) 2.5 N (d) 3.5 N
Solution : (c) According to the question figure can be drawn as shown below. A ⊗→B A 10 F
Example: 51 Force on the conductor ABC = Force on the conductor AC 10
Solution : (b)
Example: 52 = 5 × 10 × (5 × 10–2) 3⇒
Solution : (b) = 2.5 N
B 4 CB C
A wire of length l carries a current i along the X-axis. A magnetic field exists which is given as B = B0
( ˆi + ˆj + kˆ) T. Find the magnitude of the magnetic force acting on the wire
(a) B0il (b) B0i l × 2 (c) 2B0i l (d) 1 × B0i l
2
By using F = i( l × B) ⇒ F = i[l ˆi × B0 (ˆi + ˆj + kˆ)] = B0il[ˆi × (ˆi + ˆj + kˆ)]
⇒ F = B0il[ˆi × ˆi + ˆi × ˆj + ˆi × kˆ] = B0il[kˆ − ˆj] {ˆi × ˆi = 0,ˆi × ˆj = kˆ,ˆi × kˆ = −ˆj}
It's magnitude F = 2B0il
A conducting loop carrying a current i is placed in a uniform magnetic field pointing into the plane of the
paper as shown. The loop will have a tendency to
[IIT-JEE (Screening) 2003]]
(a) Contract (b) Expand
(c) Move towards + ve x-axis (d) Move towards – ve x-axis
Net force on a current carrying loop in uniform magnetic field is zero. Hence the loop can't translate. So,
options (c) and (d) are wrong. From Flemings left hand rule we can see that if iY → ⊗
magnetic field is perpendicular to paper inwards and current in the loop is Fm
clockwise (as shown) the magnetic force Fm on each element of the loop is X
radially outwards, or the loops will have a tendency to expand.
Example: 53 A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre of the
Solution : (d) loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is B.
Example: 54
Find the magnetic force on the wire iY
(a) π i a B
(b) 4π i a B aB
(c) Zero
(d) 2π i a B
The direction of the magnetic force will be vertically downwards at each element of the wire.
Thus F = Bil = Bi (2πa) = 2πiaB.
A wire abc is carrying current i. It is bent as shown in fig and is placed in a uniform magnetic field of magnetic
induction B. Length ab = l and ∠ abc = 45o. The ratio of force on ab and on bc is
(a) 1 b
2
→
(b) 2
B
(c) 1 l 45o i
ic
a
(d) 2
3
Solution : (c) Force on portion ab of wire F1 = Bil sin 90o = Bil
Example: 55
Force on portion bc of wire F2 = Bi l sin 45 o = Bil . So F1 = 1.
2 F2
Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a
point P on the right bisector of the angle XOY at a distance r from O is
(a) µ0i Y
πr
P
(b) 2µ 0 i ir
πr
(c) µ0i ( 2 + 1) 45o X
4πr O
(d) µ0 . 2i ( 2 + 1)
4π r
Solution : (d) By using B = µ0 . i (sin φ1 + sin φ 2 ) , from figure d = r sin 45o = r Y
Example: 56 4π r 2
Solution : (a)
Example: 57 Magnetic field due to each wire at P B= µ0 . i (sin 45o sin 90o) d P
4π (r / 2) 45o
Solution : (a) + i
r 45o d
= µ0 . i ( 2 + 1) 45o X
4π r
O
i i
Hence net magnetic field at P Bnet = 2× µ0 . r ( 2 + 1) = µ0 . r ( 2 + 1)
4π 2π
A long wire A carries a current of 10 amp. Another long wire B, which is parallel to A and separated by 0.1 m
from A, carries a current of 5 amp. in the opposite direction to that in A. What is the magnitude and nature of
the force experienced per unit length of B [ µ0 = 4π × 10−7 weber/amp – m] [MP PET 2000]
(a) Repulsive force of 10 −4 N / m (b) Attractive force of 10 −4 N / m
(c) Repulsive force of 2π × 10 −5 N / m (d) Attractive force of 2π × 10 −5 N / m
By using F = µ0 . 2i1i2
l 4π a
10 A 5A
⇒ F = 10−7 × 2 × 10 × 5 = 10−4 N
l 0.1
Wires are carrying current in opposite direction so the force will be repulsive. 0.1 m
Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force
experienced by 10 cm length of wire Q is [MP PET 1997]
RQ P
2cm
(a) 1.4×10–4 N towards the right 10cm
(b) 1.4×10–4 N towards the left
(c) 2.6 × 10–4 N to the right 20 A 10 A 30 A
(d) 2.6×10–4 N to the left
Force on wire Q due to R ; FR = 10 −7 × 2 × 20 × 10 × (10 × 10 −2 ) = 2 × 10–4 m (Repulsive)
(2 × 10 −2 )
Force on wire Q due to P ; FP = 10 −7 × 2 × 10 × 30 × (10 × 10 −2 ) = 0.6 × 10–4 N (Repulsive)
(10 × 10 −2
)
Hence net force Fnet = FR – FP = 2 × 10–4 – 0.6 × 10–4 = 1.4 × 10–4 N (towards right i.e. in the direction of FR .
Example: 58 What is the net force on the coil [DCE 2000]
Solution : (a)
Example: 59 (a) 25 × 10 −7 N moving towards wire 10 cm
(b) 25 × 10 −7 N moving away from wire
Solution : (a) (c) 35 × 10−7 N moving towards wire 2A 15 cm
Example: 60 (d) 35 × 10−7 N moving away from wire 1A
Force on sides BC and CD cancel each other.
Solution : (c) 2 cm
Force on side AB FAB = 10 −7 × 2× 2×1 ×15 ×10−2 = 3 × 10−6 N 2A 1A B 10 cm C
2 × 10−2
15 cm
Force on side CD FAB = 10−7 2× 2×1 ×15 ×10−2 0.5 ×10−6 N FAB FCD
12 ×10−2
× = 2 cm
A
D
Hence net force on loop = FAB – FCD = 25 × 10–7 N (towards the wire).
A long wire AB is placed on a table. Another wire PQ of mass 1.0 g and length 50 cm is set to slide on two
rails PS and QR. A current of 50A is passed through the wires. At what distance above AB, will the wire PQ be
in equilibrium SR
(a) 25 mm PQ
(b) 50 mm
(c) 75 mm BA
(d) 100 mm 50 A
Suppose in equilibrium wire PQ lies at a distance r above the wire AB
Hence in equilibrium mg = Bil ⇒ mg = µ0 2i × il ⇒ 10 −3 ×10 = 10−7 × 2 × (50)2 = 0.5 ⇒ r = 25 mm
4π r r
An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable straight
wire CD of finite length and carrying current is held perpendicular to it and released. Neglect weight of the
wire A i1
(a) The rod CD will move upwards parallel to itself
(b) The rod CD will move downward parallel to itself C D i2
(c) The rod CD will move upward and turn clockwise at the same time B
(d) The rod CD will move upward and turn anti –clockwise at the same time
Since the force on the rod CD is non-uniform it will experience force and torque. From the left hand side it can
be seen that the force will be upward and torque is clockwise.
A D i2
i1
C
B
Tricky example: 5
A current carrying wire LN is bent in the from shown below. If wire carries a current of 10 A and it is placed
in a magnetic field a 5T which acts perpendicular to the paper outwards then it will experience a force
(a) Zero N
(b) 5 N 4 cm
10A
(c) 30 N 4 cm 6 cm
(d) 20 N L
Solution : (b) The given wire can be replaced by a straight wire as shown below
N 10A N N
10A 6 cm 4 cm 10 cm
L
L 4 cm 6 cm
L
Hence force experienced by the wire F = Bil = 5 × 10 × 0.1 = 5 N
Tricky example: 6
A wire, carrying a current i, is kept in X − Y plane along the curve y = A sin 2π x . A magnetic field
λ
B exists in the Z-direction find the magnitude of the magnetic force on the portion of the wire between
x = 0 and x = λ
(a) iλB (b) Zero (c) iλB (d) 3 / 2iλB
2
Solution : (a) The given curve is a sine curve as shown below.
The given portion of the curved wire may be treated as a straight wire AB of length λ which experiences
a magnetic force Fm = Biλ Y
i
A B X
x=0 x=λ
Zλ
Current Loop As a Magnetic Dipole.
A current carrying circular coil behaves as a bar magnet whose magnetic moment is M = NiA; Where N = Number
of turns in the coil, i = Current through the coil and A = Area of the coil
Magnetic moment of a current carrying coil is a vector and it's direction is given by right hand thumb rule
Current →
SN Magnetic M
moment
Specific examples
A given length constant current carrying straight wire moulded into different shaped loops. as shown
Linear Square Equilateral Circle
i i
li
a i
l = 4a
l = 3a l = 2πr
A = πr2
A = a2 A = 3 a2
4
M = ia 2 = il 2 M = i 3 a2 = 3 il 2 M = i (πr 2 ) = il 2 ← max.
16 4 36 4π
Note : ≅ For a given perimeter circular shape have maximum area. Hence maximum magnetic
moment.
≅ For a any loop or coil B and M are always parallel.
B,M B,M
Behaviour of Current loop In a Magnetic Field.
(1) Torque
Consider a rectangular current carrying coil PQRS having N turns and area A, placed in a uniform field B, in
snionrθm.aVl e(ncˆt)otroialtlhyeτco=ilMm×akBes
such a way that the an angle θ with the direction of B. the coil experiences a torque
given by τ = NBiA
P B→
(i) τ is zero when θ = 0, i.e., when the plane of the coil is perpendicular to the field. θ
S ^n PQ = RS
(ii) τ is maximum when θ = 90o , i.e., the plane of the coil is parallel to the field. =i
⇒ τ max = NBiA Q
The above expression is valid for coils of all shapes. R
(2) Workdone
If coil is rotated through an angle θ from it's equilibrium position then required work. W = MB(1 − cosθ ). It is
maximum when θ = 180o ⇒ Wmax = 2 MB
(3) Potential energy
Is given by U = – MB cosθ ⇒ U = M.B
Note : ≅Direction of M is found by using Right hand thumb rule according to which curl the fingers of right
hand in the direction of circulation of conventional current, then the thumb gives the direction of M .
≅Instruments such as electric motor, moving coil galvanometer and tangent galvanometers etc. are
based on the fact that a current-carrying coil in a uniform magnetic field experiences a torque (or couple).
Moving coil galvanometer.
In a moving coil galvanometer the coil is suspended between the pole pieces of a strong horse-shoe magnet.
The pole pieces are made cylinderical and a soft iron cylinderical core is placed within the coil without touching it.
This makes the field radial. In such a field the plane of the coil always remains parallel to the field. Therefore
θ = 90o and the deflecting torque always has the maximum value.
4 501 2
3
2 3 4 F
1 5
Core
Spring S Moving S
N coil N
Magnet F
(fixed) F=nBil
Moving coil Galvanometer
τ def = NBiA ......(i)
coil deflects, a restoring torque is set up in the suspension fibre. If α is the angle of twist, the restoring torque is
τ rest = Cα .....(ii) where C is the torsional constant of the fibre.
When the coil is in equilibrium.
NBiA = Cα ⇒ i= C α ⇒ i= Kα ,
NBA
Where K= C is the galvanometer constant. This linear relationship between i and α makes the moving
NBA
coil galvanometer useful for current measurement and detection.
Current sensitivity : The current sensitivity of a galvanometer is defined as the deflection produced in the
galvanometer per unit current flowing through it.
Si = α = NBA
i C
Thus in order to increase the sensitivity of a moving coil galvanometer, N, B and A should be increased and C
should be decreased.
Quartz fibres can also be used for suspension of the coil because they have large tensile strength and very low
value of k.
Voltage sensitivity (SV) : Voltage sensitivity of a galvanometer is defined as the deflection produced in the
galvanometer per unit applied to it.
SV = α = α = Si = NBA
V iR R RC
Concepts
The field in a moving coil galvanometer radial in nature in order to have a linear relation between the current and the deflection.
A rectangular current loop is in an arbitrary orientation in an external magnetic field. No work required to rotate the loop about an axis
perpendicular to it's plane.
Moving coil galvanometer can be made ballistic by using a non-conducting frame (made of ivory or bamboo) instead of a metallic frame.
Example
Example: 61 A circular coil of radius 4 cm and 20 turns carries a current of 3 ampere. It is placed in a magnetic field of 0.5
Solution : (c) T. The magnetic dipole moment of the coil is
Example: 62 [MP PMT 2001]
Solution : (c) (a) 0.60 A-m2 (b) 0.45 A-m2 (c) 0.3 A-m2 (d) 0.15 A-m2
Example: 63 M = niA ⇒ M = 20 × 3 × π ( 4 × 10–2)2 = 0.3 A-m2.
Solution : (b) A steady current i flows in a small square loop of wire of side L in a horizontal plane. The loop is now folded
Example: 64
Solution : (a) about its middle such that half of it lies in a vertical plane. Let µ1 and µ 2 respectively denote the magnetic
Example: 65
moments due to the current loop before and after folding. Then [IIT-JEE 1993]
(a) µ 2 = 0 (b) µ1 and µ 2 are in the same direction
(c) | µ1 | = 2 (d) | µ1 | = 1
|µ2 | |µ2 | 2
Initially Finally
i L
M
M2
L/2 L M
L L/2 L
µ1 = iL2
M = magnetic moment due to each part = i L × L = iL2 = µ1
2 2 2
∴ µ2 = M 2 = µ1 × 2 = µ1
2 2
A coil of 50 turns is situated in a magnetic field b = 0.25weber/m2 as shown in figure. A current of 2A is
flowing in the coil. Torque acting on the coil will be A →B B
(a) 0.15 N
NS
(b) 0.3 N 12
(c) 0.45 N D 10 C
(d) 0.6 N
Since plane of the coil is parallel to magnetic field. So θ = 90o
Hence τ = NBiA sin 90o = NBiA = 50 × 0.25 × 2 × (12 × 10–2 × 10 × 10–2) = 0.3 N.
A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to
the plane of the loop. The torque on the loop due to the magnetic field is
(a) Zero (b) 10–4 N-m (c) 10–2 N-m (d) 1 N-m
τ = NBiA sinθ ; given θ = 0 so τ = 0.
A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field
of 0.1 weber/m2. The amount of work done in rotating it through 180o from its equilibrium position will be
[CPMT 1977]
(a) 0.1 J (b) 0.2 J (c) 0.4 (d) 0.8 J
Solution : (a) Work done in rotating a coil through an angle θ from it's equilibrium position is W = MB(1 – cosθ) where θ = 180o
Example: 66 and M = 50 × 2 × π (4 × 10–2) = 50.24 × 10–2 A-m2. Hence W = 0.1 J
Solution : (d)
Example: 67 A wire of length L is bent in the form of a circular coil and current i is passed through it. If this coil is placed in
a magnetic field then the torque acting on the coil will be maximum when the number of turns is
Solution : (b)
(a) As large as possible (b) Any number (c) 2 (d) 1
Example: 68
Solution : (d) τ max = MB or τ max = niπa2B . Let number of turns in length l is n so l = n(2πa) or a = l
Example: 69 2πn
Solution : (a)
⇒ τ max = niπBl 2 l 2iB ⇒ τ max ∝ 1 ⇒ nmin = 1
4π 2n2 = 4πnmin nmin
A square coil of N turns (with length of each side equal L) carrying current i is placed in a uniform magnetic
field B = B0ˆj as shown in figure. What is the torque acting on the coil i
(a) + B0 NiL2kˆ B = B0 ˆj
(b) − B0 NiL2kˆ kˆ
(c) + B0 NiL2ˆj L ˆi ˆj
(d) − B0 NiL2ˆj
The magnetic field is B = B0ˆj and the magnetic moment m = i A = −i(NL2ˆi ) i
The torque is given by τ = m × B
= −iNL2ˆi × B0ˆj = −iNB0 L2ˆi × ˆj
= −iNB0 L2kˆ X
The coil of a galvanometer consists of 100 turns and effective area of 1 square cm. The restoring couple is 10– 8 N-m rad.
The magnetic field between the pole pieces is 5 T. The current sensitivity of this galvanometer will be
[MP PMT 1997]
(a) 5 × 104 rad/µ amp (b) 5 × 10– 6 per amp (c) 2 × 10– 7 per amp (d) 5 rad./µ amp
Current sensitivity (Si) = θ = NBA ⇒ θ = 100 × 5 × 10 −4 = 5 rad /µ amp .
i C i 10 −8
The sensitivity of a moving coil galvanometer can be increased by [SCRA 2000]]
(a) Increasing the number of turns in the coil (b) Decreasing the area of the coil
(c) Increasing the current in the coil (d) Introducing a soft iron core inside the coil
Sensitivity (Si) = NBA ⇒ Si ∝ N.
C
Tricky example: 7
The square loop ABCD, carrying a current i, is placed in uniform magnetic field B, as shown. The loop
can rotate about the axis XX'. The plane of the loop makes and angle θ (θ < 90°) with the direction of
B. Through what angle will the loop rotate by itself before the torque on it becomes zero
X CB
θ
(a) θ B Z
(b) 90°– θ i
(c) 90° + θ D
(d) 180°– θ A X′
Solution : (c) In the position shown, AB is outside and CD is inside the plane of the paper. The Ampere force on AB
acts into the paper. The torque on the loop will be clockwise, as seen from above. The loop must rotate
through an angle (90o + θ) before the plane of the loop becomes normal to the direction of B and the
torque becomes zero.
Magnetism
The molecular theory of magnetism was given by Weber and modified later by Ewing. According to this theory.
Every molecule of a substance is a complete magnet in itself. However, in an magnetised substance the molecular
magnets are randomly oriented to give zero net magnetic moment. On magnetising, the molecular magnets are
realigned in a specific direction leading to a net magnetic moment.
(Unmagnetised) (Magnetised)
Note : ≅ On heating/hammering the magnetism of magnetic substance reduces.
Bar Magnet.
A bar magnet consist of two equal and opposite magnetic pole separated by a small distance. Poles are not
exactly at the ends. The shortest distance between two poles is called effective length (Le) and is less then its
geometric length (Lg). 2R
SN SN
Le = 2l
Lg
Le = 5 × Lg Le = 2R and Lg = πR
6
(1) Directive properties : When a magnet suspended freely it stays in the earth’s N-S direction (in magnetic
meridian).
Magnetic axis
N
S Magnetic meridian
(2) Monopole concept : If a magnet is Broken into number of pieces, each piece becomes a magnet. This in
turn implies that monopoles do not exist. (i.e., ultimate individual unit of magnetism in any magnet is called dipole).
SN
SN SN SN
(3) For two rods as shown, if both the rods attract in case (i) and doesn’t attract in case (ii) then, B is a
magnetic and A is simple iron rod. Repulsion is sure test of magnetism.
A B
B A
(i) (ii)
(4) Pole strength (m) : The strength of a magnetic pole to attract magnetic materials towards itself is known
as pole strength.
(i) It is a scalar quantity.
(ii) Pole strength of N and S pole of a magnet is conventionally represented by +m and –m respectively.
(iii) It's SI unit is amp × m or N/Tesla and dimensions are [LA].
(iv) Pole strength of the magnet depends on the nature of material of magnet and area of cross section. It
doesn't depends upon length.
SN SSSSSSS A – less N
SS NN m – less NN NN
SS A – more N N
S m – more N N
SN
(5) Magnetic moment or magnetic dipole moment (M ) : It represents the strength of magnet.
Mathematically it is defined as the product of the strength of either pole and effective length. i.e. M = m(2l)
(i) It is a vector quantity directed from south to north. –m S N +m
(ii) It's S.I. unit amp×m2 or N-m/Tesla and dimensions [AL2] L = 2l
(iii) Magnetic moment in various other situations. M
Current carrying coil Magnetic moment M = NiA
M
N = number of turns,
i iM i= current through the coil, A= Area of the coil
Combination of bar magnet
M2 M net = M 2 + M 2 + 2M1M 2 cos θ
S 1 2
Nθ M net tan α = M 2 sinθ
S M1 + M 2 cosθ
α
M1
N
2M M N 2M N
S 90o M S 60o SN M Mnet = 2M
M NS 3M N S SN M
M
Revolving charge
(a) Orbital electron : In an atom electrons revolve around the nucleus in circular orbit and it is equivalent to the flow
of current in the orbit. Thus the orbit of electrons is considered as tiny current loop with magnetic moment.
M = eνA = eωr 2 = 1 evr = e L = eh ; where, ω = angular speed, ν = frequency, v = linear speed and
2 2 2m 4πm
L= Angular moment Iω.
(b) For geometrical symmetrical charged rotating bodies : The magnetic moment given by M = QL = QIω ; where
2m 2m
m = mass of rotating body, Q = charge on body, I = moment of inertia of rotating body about axis of rotation.
ω ω ω
+ +++ + ++R+++ + Q + ++ + ++ +
+ + ++ + R +Q +
++ + +
+ + +
+ ++ Q+ L
Disc Ring Rod
I = mR 2 , M = 1 QωR 2 I = MR 2 , M = 1 QωR 2 I = mL2 , M = 1 Qω L2
2 4 2 12 24
Note : ≅Bohr magneton µ B = eh = 9.27×10–24 A/m2 . It serves as natural unit of magnetic moment. Bohr
4πm
magneton can be defined as the orbital magnetic moment of an electron circulating in inner most orbit.
≅ Magnetic moment of straight current carrying wire is zero.
≅ Magnetic moment of toroid is zero.
≅ If a magnetic wire of magnetic moment (M) is bent into any shape then it's M decreases as
it's length (L) always decreases and pole strength remains constant.
L/3
L L/2 L/3 L/3 R
– m +m L = πR
L/2
M = mL M' = M / 2 M' = M / 3 M' = 2M / π M′ = 0
(6) Cutting of a bar magnet : Suppose we have a rectangular bar magnet having length, breadth and mass
are L, b and w respectively if it is cut in n equal parts along the length as well as perpendicular to the length
simultaneously as shown in the figure then
b b′
L′
L
Length of each part L' = L , breadth of each part b' = b , Mass of each part w' = w , pole strength of each
n n n
part m' = m , Magnetic moment of each part M' = m' L' = m × L = M
n n n n
If initially moment of inertia of bar magnet about the axes passing from centre and perpendicular to it’s length
is I w L2 +b 2 then moment of inertia of each part I' = I
12 n2
=
Note : ≅For short bar magnet b = 0 so L' = L ,w' = w , m' = m, M' = M and I' = I
n n n n3
≅ Commonly asked question
SN S N A/2 SN ⇒S N
L
⇒ S N A/2 L L/2 L/2
(m, M) m′ = m, M′ = M/2
L
(m, M) m′ = m/2, M′ = M/2
Various Terms Related to Magnetism.
(1) Magnetic field and magnetic lines of force : Space around a magnetic pole or magnet or current
carrying wire within which it's effect can be experienced is defined magnetic field. Magnetic field can be represented
with the help of a set of lines or curves called magnetic lines of force.
NS SN ×××× ••••
Magnetic dipole ×××× ••••
Isolated north pole Isolated south pole ×××× ••••
×××× ••••
×××× ••••
Inward magnetic field Outward magnetic field
(2) Magnetic flux (φ) and flux density (B)
(i) The number of magnetic lines of force passing normally through a surface is defined as magnetic flux (φ). It's
S.I. unit is weber (wb) and CGS unit is Maxwell.
Remeber 1 wb = 108 maxwell.
(ii) When a piece of a magnetic substance is placed in an external magnetic field the substance becomes
magnetised. The number of magnetic lines of induction inside a magnetised substance crossing unit area normal to
their direction is called magnetic induction or magnetic flux density (B). It is a vector quantity.
It's SI unit is Tesla which is equal to
wb = N = J = volt × sec
m2 amp × m amp × m2 m2
SN
and CGS unit is Gauss. Remember 1 Tesla = 104 Gauss.
Note : ≅ Magnetic flux density can also be defined in terms of force
experienced by a unit north pole placed in that field i.e. B= F .
m0
(3) Magnetic permeability : It is the degree or extent to which magnetic lines of force can enter a substance
and is denoted by µ. or
Characteristic of a medium which allows magnetic flux to pass through it is called it's permeability. e.g.
permeability of soft iron is 1000 times greater than that of air.
In air In soft iron
Also µ = µ0 µr ; where µ0 = absolute permeability of air or free space = 4π × 10−7 tesla × m / amp.
and µr = Relative permeability of the medium = B = flux density in material .
B0 flux density in vacuum
(4) Intensity of magnetising field (H) (magnetising field) : It is the degree or extent to which a magnetic
field can magnetise a substance. Also H = B .
µ
It's SI unit is A / m. = m2 N N = m3 J = m J It's CGS unit is Oersted. Also 1oersted = 80 A/m
× Tesla = wb × Tesla × wb
(5) Intensity of magnetisation (I) : It is the degree to which a substance is magnetised when placed in a
magnetic field.
It can also be defined as the pole strength per unit cross sectional area of the substance or the induced dipole
moment per unit volume.
Hence I= m = M . It is a vector quantity, it's S.I. unit is Amp/m.
A V
(6) Magnetic susceptibility (χm) : It is the property of the substance which shows how easily a substance
can be magnetised. It can also be defined as the ratio of intensity of magnetisation (I) in a substance to the magnetic
intensity (H) applied to the substance, i.e. χm = I . It is a scalar quantity with no units and dimensions.
H
(7) Relation between permeability and susceptibility : Total magnetic flux density B in a material is the
sum of magnetic flux density in vacuum B0 produced by magnetising force and magnetic flux density due to
magnetisation of material Bm . i.e. B = B0 + Bm
⇒ B = µ 0 H + µ 0 I = µ 0 (H + I) = µ 0 H(1 + χ m ) . Also µr = (1 + χ m )
Note : ≅ In CGS B = H + 4πI and µ = 1 + 4πχm.
Force and Field.
(1) Coulombs law in magnetism : The force between two magnetic poles of strength m1 and m2 lying at a
m1m2
distance r is given by F = k. r2 . In S.I. units k = µ0 = 10−7 wb / Amp × m , In CGS units k =1
4π
(2) Magnetic field
(i) Magnetic field due to an imaginary magnetic pole (Pole strength m) : Is given by B= F also B = µ0 . m
m0 4π d2
(ii) Magnetic field due to a bar magnet : At a distance r from the ceBntre of magnet
(a) On axial position
Be e g
Ba µ0 2Mr ; Equatorial line
4π (r 2 − l 2)2
= θ θ+α
If l<<r then Ba = µ0 2M SN a Ba
4π r3 2l
M Axial line
(b) On equatorial position : Be = µ0 (r 2 M ; If l <<r ; then Be = µ0 M
4π + l 2)3 / 2 4π r3
(c) General position : In general position for a short bar magnet Bg = µ0 M (3 cos 2 θ + 1)
4π r3
(3) Bar magnet in magnetic field : When a bar magnet is left free in an uniform magnetic field, if align it
self in the directional field.
(i) Torque : τ = MB sinθ ⇒ τ = M × B
(ii) Work : W = MB(1 − cosθ )
(iii) Potential energy : U = MB cosθ = −M . B ; (θ = Angle made by the dipole with the field)
Note : ≅ For more details see comparative study of electric and magnetic dipole in electrostatics.
∫(4) Gauss's law in magnetism : Net magnetic flux through any surface is always zero i.e. B.ds = 0
Concepts
The property of magnetism is materials is on account of magnetic moment in the material.
Atoms which have paired electron have the magnetic moment zero.
Magnetostriction : The length of an iron bar changes when it is magnetised, when an iron bar magnetised it's length increases
due to alignment of spins parallel to the field. This increase is in the direction of magnetisation. This effect is known as
magnetostriction.
A current carrying solenoid can be treated as the arrangement of small magnetic dipoles placed in line with each other as shown.
The number of such small magnetic dipoles is equal to the number of such small magnetic dipoles is equal to the number of
turns in the in the solvent
≡ ≡S N
When a magnetic dipole of moment M moves from unstable equilibrium to stable equilibrium position in a magnetic field B, the
kinetic energy by it will be 2 MB.
Intensity of magnetisation (I) is produced in materials due to spin motion of electrons.
For protecting a sensitive equipment from the external magnetic field it should be placed inside an iron cane. (magnetic shielding)
B=0
Example
Example: 1 The work done in turning a magnet of magnetic moment M by an angle of 90° form the meridian, is n times
Solution : (a) the corresponding work done to turn it through an angle of 60°. The value of n is given by
[MP PET 2003]
(a) 2 (b) 1 (c) 1/2 (d) 1/4
( )W = MB(1 − cosθ ) ⇒ W0o →90o = nx W0o →60o ⇒ MB(1 − cos 90o ) = n × MB(1 − cos 60o ) ⇒ n = 2
Example: 2 The magnetic susceptibility of a material of a rod is 499, permeability of vacuum is 4π × 10−7 H / m .
Solution : (b)
Example: 3 Permeability of the material of the rod in henry/metre is [EAMCET 2003]
Solution : (a) (a) π × 10 −4 (b) 2π × 10 −4 (c) 3π × 10 −4 (d) 4π × 10 −4
Example: 4
µr = (1 + γ m)⇒ µr = (1 + 499) = 500 Also µ = µ0µr = 4π ×10−7 × 500 = 2π ×10−4
Solution : (b)
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque
required to maintain the needle in this position will be [MNR 1991; KCET 1994; MP PET 1996; AIEEE 2003]
(a) 3 W (b) – W (c) 3 W (d) 2W
2
τ= MB sinθ and W = MB(1 − cos θ )⇒ W = MB(1 − cos 60o) = MB . Hence τ = MB sin 60o = 3MB = 3W
2 2
An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment
will be
[CPMT 1984; MP Board 1986; NCERT 1975; MP PET/PMT 1988; EAMCET (Med.) 1995;
Manipal MEE 1995; RPMT 1996; BHU 1995; MP PMT 2002]
(a) M (b) 2M (c) M (d) Mπ
π π
On bending a rod it's pole strength remains unchanged where as it's magnetic moment changes
New magnetic moment M' = m(2R) = m 2L = 2M S L N L' = 2R N
π π
⇒S
Example: 5 A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. It the
Solution : (d) magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is : (Horizontal component of
Example: 6 earth's magnetic field = BH) [EAMCET (Engg.) 2000]
Solution : (d)
Example: 7 (a) 0 (b) 2 BH (c) 5 BH (d) 5 BH
Initially 2
N BH
E N
W P
S SB
Neutral point is obtained on equatorial line and at neutral point | BH |=| Be |
Where BH = Horizontal component of earth's magnetic field and Be = Magnetic field due to bar magnet on
it's equatorial line N BH
Finally E
W SN P Ba
S
Point P comes on axial line of the magnet and at P, net magnetic field
B = BH2 + Ba2 = (2Be)2 + (BH )2 = (2BH )2 + BH2 = 5BH
A bar magnet of magnetic moment 3.0 Amp × m is placed in a uniform magnetic induction field of 2 × 10−5 T.
If each pole of the magnet experiences a force of 6 ×10−4 N the length of the magnet is [EAMCET (Med.) 2000]
(a) 0. 5 m (b) 0. 3 m (c) 0.2 m (d) 0.1 m
M = mL and F = mB ,⇒ F = M × B⇒ 6 × 10−4 = 3 × 2 × 10−5 ⇒ L = 0.1m
L L
Force between two identical bar magnets whose centres are r metre apart is 4.8 N when their axes are in the
same line. If the separation is increases to 2r metre, the force between them is reduced to
[AIIMS 1995; Pb. CET 1997]
(a) 2.4 N (b) 1.2 N (c) 0.6 N (d) 0.3 N
Solution : (d) Force between two bar magnet F ∝ 1 ⇒ F1 = d2 4 ⇒ 4.8 = 2r 2 ⇒ F2 = 0.3 N Where d= separation
Example: 8 d4 F2 d1 F2 r
Solution : (b) between magnets.
Example: 9
Solution : (c) Two identical magnetic dipoles of magnetic moments 1.0 A-m2 each, placed at a separation of 2m with their
Example: 10 axis perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is
Solution : (d)
Example: 11 [Roorkee 1995]
Solution : (b)
Example: 12 (a) 5 × 10−7 T (b) 5 × 10−7 T (c) T (d) None of these
2
Solution : (c)
B1 = µ0 . 2M = 10 −7 × 2×1 ; B2 = µ0 M = 10 −7 × 1 ; Bnet = (2 × 10 −7 )2 + (10 −7 )2 = 5 × 10 −7 T
Example: 13 4π 4π
r 3 2 3 r 3 2 3
Solution : (b) 2 2 2 2
A magnet of magnetic moment 20 C.G.S. units is freely suspended in a uniform magnetic field of intensity 0.3
C.G.S. units. The amount of work done in deflecting it by an angle of 30° in C.G.S. units is
[MP PET 1991]
(a) 6 (b) 3 3 (c) 3(2 − 3) (d) 3
W = MB(1 − cosθ ) ⇒ W = 20 × 0.3(1 − cos 30o) = 3(2 − 3)
The magnetic field at a point X on the axis of a small bar magnet is equal to the field at a point Y on the
equator of the same magnet. The ratio of the distance of X and Y from the centre of the magnet is
[MP PMT 1990]
(a) 2−3 (b) 2−1 / 3 (c) 23 (d) 21 / 3
Suppose distances of points X and Y from magnet are x and y respectively then According to question
Baxial = Bequatorial ⇒ µ0 . 2M = µ0 . M ⇒ x = 21 / 3
4π x3 4π y3 y 1
A magnetising field of 2000 A/m produces a flux 6.28 ×10−4 weber in a rod. If the area of cross-section is
2×10−5m2. Then the relative permeability of the substance is
(a) 0.75 × 10−2 (b) 1.25 × 10 4 (c) 0.25 (d) 1.01
By using B = µ0µr H and B= φ , ⇒ µr = φ = 6.28 × 10−4 = 1.25 × 104
A Aµ0 H 2 × 10−5 × 4π × 10−7 × 2000
Due to a small magnet intensity at a distance x in the end on position is 9 Gauss. What will be the intensity at
a distance x on broad side on position
2
(a) 9 Gauss (b) 4 Gauss (c) 36 Gauss (d) 4.5 Gauss
In C.G.S. Baxial =9= 2M .....(i) Bequaterial = M = 8M .....(ii)
x3 x3
x 3
2
From equation (i) and (ii) Bequaterial = 36 Gauss.
The magnetic moment produced in a substance of 1gm is 6 ×10−7ampere − metre2. If its density is 5 gm / cm3,
then the intensity of magnetisation in A/m will be
(a) 8.3 × 106 (b) 3.0 (c) 1.2 × 10−7 (d) 3 × 10−6
I = M = M , given mass = 1gm = 10−3kg , and density = 5gm / cm3 = 5 × 10−3kg = 5 × 103kg / m3
V mass/density (10−2)3 m3
Hence I = 6 × 10−7 × 5 × 103 = 3
10 − 3
Example: 14 The distance between the poles of a horse shoe magnet is 0.1 m and its pole strength is 0.01 amp-m. The
Solution : (c) induction of magnetic field at a point midway between the poles will be
Example: 15 (a) 2 × 10−5 T (b) 4 × 10−6 T (c) 8 × 10−7 T (d) Zero
Solution : (b)
Net magnetic field at mid point P, B = BN + BS 0.1 m
Example: 16
Solution : (b) where BN = magnetic field due to N- pole BS = magnetic field due to S- pole S P N
Example: 17 BN Bs
Solution : (d)
Example: 18 BN = BS = µ0 m = 10−7 × 0.01 = 4 × 10−7 T ∴ Bnet = 8 × 10−7 T.
4π r2
0.1 2
2
A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of
5.30 × 103Amp/m3. What its magnetic dipole moment
(a) 1 × 10−2 J / T (b) 2.08 × 10−2 J / T (c) 3.08 × 10−2 J / T (d) 1.52 × 10−2 J / T
Relation for dipole moment is, M = I × V , Volume of the cylinder V = πr 2l, Where r is the radius and l is the
length of the cylinder, then dipole moment,
M = I × πr 2l = (5.30 × 103) × 22 × (0.5 × 10−2)2 (5 × 10−2) = 2.08 × 10−2 J / T
7
A bar magnet has a magnetic moment of 2.5JT −1 and is placed in a magnetic field of 0.2T. Work done in
turning the magnet from parallel to anti-parallel position relative to field direction is
(a) 0.5J (b) 1J (c) 2J (d) 0J
Work done, W = −MB(cosθ2 − cosθ1) = −MB(cos180o − cos 0o ) = −MB(−1− 1) = 2MB = 2 × 2.5 × 0.2 = 1J
A bar magnet with it's poles 25 cm apart and of pole strength 24 amp×m rests with it's centre on a frictionless
pivot. A force F is applied on the magnet at a distance of 12 cm from the pivot so that it is held in equilibrium
at an angle of 30° with respect to a magnetic field of induction 0.25 T. The value of force F is
(a) 5.62 N (b) 2.56 N (c) 6.52 N (d) 6.25 N
In equilibrium
Magnetic torque = Deflecting torque ⇒ MB sin θ = F.d or F = mlB sin θ 24 × 0.25 × 0.25 sin 30o = 6.25N
d 0.12
Two identical bar magnets with a length 10 cm and weight 50 gm – weight are arranged freely with their like
poles facing in a arranged vertical glass tube. The upper magnet hangs in the air above the lower one so that
the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be
(a) 6.64 amp × m S
(b) 2 amp × m N
(c) 10.25 amp × m
N
S
(d) None of these
Solution : (a) The weight of upper magnet should be balanced by the repulsion between the two magnet
∴ µ . m2 = 50gm − wt ⇒ 10−7 × (9 m2 ) = 50 × 10−3 × 9.8 ⇒ m = 6.64amp × m
4π r2 × 10−6
Tricky Example: 1
A bar magnet of magnetic moment 2.0 A-m2 is free to rotate about a vertical axis passing through its
centre. The magnet is released form rest from east–west position. Then the kinetic energy of the
magnet as it takes north-south position is (Horizontal component of earth's field is 25µ T)
(a) 25µJ (b) 50µJ (c) 100µJ [EAMCET (Engg.) 1996]
(d) 12.5µJ
Solution : (b) When a bar magnet suspended freely in earth's magnetic field, it always align it self in the direction of
field (i.e. along N – S direction)
BH
So by using U = −MBH cos θ ; where θ = angle between M and BH
U = −M × B cos 0 = −2 × 25 = −50µJ SN M
Earth’s magnetic Field (Terrestrial Magnetism).
As per the most established theory it is due to the rotation of the earth where by the various charged ions
present in the molten state in the core of the earth rotate and constitute a current.
Magnetic Geographic
axis axis
Sm Ng Equator
θ
Sg Nm
(1) The magnetic field of earth is similar to one which would be obtained if a huge magnet is assumed to be
buried deep inside the earth at it's centre.
(2) The axis of rotation of earth is called geographic axis and the points where it cuts the surface of earth are called
geographical poles (Ng, Sg). The circle on the earth's surface perpendicular to the geographical axis is called equator.
(3) A vertical plane passing through the geographical axis is called geographical meridian.
(4) The axis of the huge magnet assumed to be lying inside the earth is called magnetic axis of the earth. The
points where the magnetic axis cuts the surface of earth are called magnetic poles. The circle on the earth's surface
perpendicular to the magnetic axis is called magnetic equator.
(5) Magnetic axis and Geographical axis don't coincide but they makes an angle of 17.5° with each other.
(6) Magnetic equator divides the earth into two hemispheres. The hemisphere containing south polarity of
earth's magnetism is called northern hemisphere while the other, the southern hemisphere.
(7) The magnetic field of earth is not constant and changes irregularly from place to place on the surface of the
earth and even at a given place in varies with time too.
(8) Direction of earth's magnetic field is from S (geographical south) to N (Geographical north).
Elements of Earth's Magnetic Field.
The magnitude and direction of the magnetic field of the earth at a place are completely given by certain.
quantities known as magnetic elements.
(1) Magnetic Declination (θ) : It is the angle between geographic and the magnetic meridian planes.
BH θ θ oW N θ oE
φ θθ E
BV
Geographical B Magnetic W
meridian meridian
S
Declination at a place is expressed at θ o E or θ oW depending upon whether the north pole of the compass
needle lies to the east or to the west of the geographical axis.
(2) Angle of inclination or Dip (φ) : It is the angle between the direction of intensity of total magnetic field
of earth and a horizontal line in the magnetic meridian.
(3) Horizontal component of earth's magnetic field (BH) : Earth's magnetic field is horizontal only at the
magnetic equator. At any other place, the total intensity can be resolved into horizontal component (BH) and vertical
component (BV).
Also BH= B cos φ ...... (i) and BV = B sin φ ....... (ii)
By squaring and adding equation (i) and (ii) B = BH2 + BV 2
Dividing equation (ii) by equation (i) tan φ = BV
BH
Note : ≅ At equator θ = 0 ⇒ BH = B, BV = 0 while at poles φ = 90o ⇒ BH = 0, BV = B.
Magnetic Maps and Neutral Points.
(1) Magnetic maps (i.e. Declination, dip and horizontal component) over the earth vary in magnitude from place
to place. It is found that many places have the same value of magnetic elements. The lines are drawn joining all place
on the earth having same value of a magnetic elements. These lines forms magnetic map.
(i) Isogonic lines: These are the lines on the magnetic map joining the places of equal declination.
(ii) Agonic line: The line which passes through places having zero declination is called agonic line.
(iii) Isoclinic lines : These are the lines joining the points of equal dip or inclination.
(iv) Aclinic line : The line joining places of zero dip is called aclinic line (or magnetic equator)
(v) Isodynamic lines : The lines joining the points or places having the same value of horizontal component of
earth's magnetic field are called isodynamic lines.
(2) Neutral points : At the neutral point, magnetic field due to the bar magnet is just equal and opposite to
the horizontal component of earth's magnetic field.
(i) Magnet is placed horizontally in a horizontal plane.
N- pole of magnet is facing N- pole of earth N - pole of magnet is facing N- pole of earth
BH BH
N1
B
N BH N BH N BH Sd BH
E N1 N2 W B
W S B B
S B BH d
E Nd
d S
BH
N2
B
Two neutral points N1 and N2 are obtained on equatorial Two neutral points N1 and N2 are obtained on axial line
line of bar magnet as shown and at Neutral points of B or magnet and at neutral points B = BH i.e.
B = BH ⇒ µ0 M = BH µ0 . 2M = BH
4π d3 4π d3
(ii) Magnet is placed vertically in a horizontal plane
N- pole of magnet is the horizontal plane S- pole of magnet is the horizontal plane
S BH N BH
Pθ
BH BN N BH BS BN W N
BH E BS S E
BS N BN W S BH
θ S BS
BN BH
P BN = Magnetic field due to N-pole BH
BN BS = Magnetic field due to S-pole BS
M = Pole strength of each pole of the magnet
At neutral point P : BN – BS cosθ = BH (BS < BN) At neutral point P : BS – BN cosθ = BH (BS < BN)
If suppose effect of S-pole is neglected : As seen from top If suppose effect of N-pole is neglected : As seen from top
only one neutral point is obtained as shown and at
only one neutral point is obtained as shown and at
neutral point BN = BH ⇒ µ0 m = BH neutral point BS = BH ⇒ µ0 m = BH
4π r2 4π r2
BN BH BH
BH BH N P N
N BN WE WE
BH BS BH
BN r BH S S
P BS S BS
BN r BH
BS
Concepts
Apparent dip : In a vertical plane inclined at an angle β to the magnetic meridian, vertical component of earth's magnetic field
remains uncharged while in the new inclined plane horizontal component B' H = BH cos β
φ' = apparent angle of dip
BV BV BH Magnetic meridian
BH' BH cos β
and tan φ' = = β BH cosβ
φ′
⇒ tanφ' = tan φ BV B′
cos β
Inclined plane
If at any place the angle of dip is θ and magnetic latitude is λ then tan θ = 2tanλ
At the poles and equator of earth the values of total intensity are 0.66 and 0.33 Oersted respectively.
Example
Example: 19 If the angles of dip at two places are 30° and 45° respectively, Then the ratio of horizontal components of
earth's magnetic field at the two places will be [MP PET 1989]
(a) 3 : 2 (b) 1: 2 (c) 1: 3 (d) 1 : 2
Solution : (a) By using BH = B cos φ ⇒ (BH )1 = (cos φ)1 = cos 30 = 3
Example: 20 (BH )2 (cos φ)2 cos 45 2
Solution : (d) At a place the earth's horizontal component of magnetic field is 0.38 × 10−4 weber / m2. If the angle of dip at
Example: 21 that place is 60°, then the vertical component of earth's field at that place in weber/m2 will be approximately
Solution : (c)
[MP PMT 1985]
Example: 22
(a) 0.12 × 10 −4 (b) 0.24 × 10 −4 (c) 0.40 × 10 −4 (d) 0.62 × 10 −4
By using tan φ = BV ⇒ tan 60o = BV ⇒ BV = 0.38 × 10 −4 × 3 = 0.62 × 10 −4.
BH 0.38 × 10 −4
A dip circle is so set that it moves freely in the magnetic meridian. In this position, the angle of dip is 40°. Now
the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic
meridian. In this position, the needle will dip by the angle [Roorkee 1983]
(a) 40° (b) 30° (c) More than 40° (d) Less than 40°
By using tanφ ′ = tan φ ; where φ = 40o , β = 30 o
cos β
As cos 30o < 1 1 >1
⇒ cos 30o
Hence tanφ ′ > 1⇒ tan φ ′ > tan φ = φ′ > φ or φ ′ > 40o .
tan φ
Earth's magnetic field may be supposed to be due to a small bar magnet located at the centre of the earth. If
the magnetic field at a point on the magnetic equator is 0.3×10–4 T. Magnet moment of bar magnet is
(a) 7.8 × 108 amp × m2 N Magnetic
(b) 7.8 ×1022amp × m2 R equator
(c) 6.4 × 1022amp × m2 S
(d) None of these
Solution : (b) When a magnet is freely suspended in earth’s magnetic field, it's north pole points north, so the magnetic field
Example: 23 of the earth may be suppose to be due to a magnetic dipole with it's south pole towards north and as
equatorial point is on the broad side on position of the dipole.
Solution : (c)
Example: 24 Be = µ0 . M ⇒ 0.3 × 10 −4 = 10 −7 × (6.4 M ⇒ M = 7.8 × 10 22 A-m2.
4π r3 × ×10 6 )3
Solution : (b)
A short bar magnet is placed with its south pole towards geographical north. The neutral points are situated at
a distance of 20 cm from the centre of the magnet. If BH = 0.3 × 10−4 wb / m2 then the magnetic moment of
the magnet is
(a) 9000 ab- amp × cm2 N E N1
(b) 900 ab − amp × cm2 W S S r = 20 cm
(c) 1200 ab − amp × cm2
N r = 20 cm
(d) 225ab − amp × cm2 N2
At neutral point magnetic field due to magnet = Horizontal component of earth's magnetic field
⇒ µ0 . 2M = BH ⇒ 10−7 × 2× M × 1 = 0.3 × 154 ⇒ M = 1.2amp × m2 = 1200ab − amp × cm2.
4π r3 (0.2)3
Two magnets of equal mass are joined at right angles to each other as shown the magnet 1 has a magnetic
moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal
plane. In equilibrium what angle will the magnet 1 subtend with the magnetic meridian
(a) tan−1 1
2
m1 θ N m2
N (2)
1 (1)
(b) tan −1 3
S 90° S
(c) tan−1(1)
(d) 0°
For equilibrium of the system torques on M1 and M 2 due to BH must counter balance each other i.e.
M 1 × B H = M 2 × B H . If θ is the angle between M1 and BH then the angle between M2 and BH will be
(90 – θ) ; so M1BH sinθ = M2BH sin(90 − θ )
⇒ tanθ = M2 = M = 1 ⇒θ = tan−1 1
M1 3M 3 3
Tricky Example: 2
A compass needle whose magnetic moment is 60 amp × m2 pointing geographical north at a certain
place, where the horizontal component of earth's magnetic field is 40 µω b/m2, experiences a torque
1.2 ×10−3 N × m. What is the declination at this place [EAMCET (Engg.) 1996]
(a) 30° (b) 45° (c) 60° (d) 25°
Solution : (a) As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian,
so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal
component of earth's magnetic field i.e. τ = MBH sin θ Magnetic mBH
Where θ = angle between geographical meridian
θ
and magnetic meridians called angle of declination M
So, sinθ = 1.2 × 10−3 = 1 θ = 30o mBH
60 × 40 × 10−6 2⇒
Tangent Law and it's Application.
When a small magnet is suspended in two uniform magnetic fields B and BH which are at right angles to each
other, the magnet comes to rest at an angle θ with respect to BH such that B = BH tanθ . This is called tangent law.
BH mBH mB Circular coil
N Compass box
θ
N S
mB S B Terminals
mBH Leveling screws
Tangent galvanometer : It is an instrument which can detect/measure very small electric currents. It is also
called as moving magnet galvanometer. It consists of three circular coils of insulated copper wire wound on a
vertical circular frame made of nonmagnetic material as ebonite or wood. A small magnetic compass needle is
pivoted at the centre of the vertical circular frame. This needle rotates freely in a horizontal plane inside a box made
of nonmagnetic material. When the coil of the tangent galvanometer is kept in magnetic meridian and current
passes through any of the coil then the needle at the centre gets deflected and comes to an equilibrium position
under the action of two perpendicular field : one due to horizontal component of earth and the other due to field set
up by the coil due to current (B).
In equilibrium B = BH tanθ where B= µ 0 ni ; n = number of turns, r = radius of coil, i = the current to be
2r
measured, θ = angle made by needle from the direction of BH in equilibrium.
Hence µ0 Ni = BH tanθ ⇒ i = k tanθ where k = 2rBH is called reduction factor.
2r µ0 N
Note : ≅ Principle of moving coil galvanometer is i ∝ tanθ . Since i ∝ tanθ so it’s scale is not
uniform.
≅When θ = 45o , reduction factor equals to current flows through coil.
≅ Sensitivity of this galvanometer is maximum at θ = 45o.
≅ This instrument is also called moving magnet type galvanometer.
Magnetic Instruments.
Magnetic instruments are used to find out the magnetic moment of a bar magnet, find out the horizontal
component of earth's magnetic field, compare the magnetic moments of two bar magnets.
(1) Deflection magnetometer
It's working is based on the principle of tangent law. It consist of a small compass needle, pivoted at the centre
of a circular box. The box is kept in a wooden frame having two meter scale fitted on it's two arms. Reading of a
scale at any point directly gives the distance of that point from the centre of compass needle.
0o
90o 90o
0o
Different position of deflection magnetometer : Deflection magnetometer can be used according to two
following positions.
Tan A position Tan B position
Arms of magnetometer are placed along E-W direction Arms of magnetometer are placed along N-S direction
such that magnetic needle is acted upon by only such that magnetic needle align itself in the direction of
horizontal component of earth's magnetic field (BH ) as earth's magnetic field (i.e. BH ) as shown.
shown
N
BH BH N W E
90θ W E
2l S
SN 00
90 B S BH
r 90θ B
00
90
r
It a bar magnet is placed on one arm with it's length NS
parallel to arm, so magnetic needle comes under the
influence of two mutual perpendicular magnetic field (i) If a bar magnet is placed on one arm with it's length
BH and (ii) Axial magnetic field of experimental bar perpendicular to arm, so magnetic needle comes under
magnet. the influence of two mutual perpendicular magnetic
fields (i) BH and (ii) equatorial magnetic field of
In equilibrium B = BH tanθ ⇒ µ0 2M = BH tanθ
4π r3 experimental bar magnet.
(M= Magnetic moment of experimental bar magnet) In equilibrium B = BH and ⇒ µ0 . M = BH tanθ
4π r3
Note : ≅Deflection magnetometer also used to compare the magnetic moments either by deflection method or by null
3
deflection method. Deflection method : M1 = tanθ1 , Null deflection method : M1 = d1
M2 tanθ 2 M2 d2
where d1 and d2 are the position of two bar magnet placed simultaneously on each arm.
(2) Vibration magnetometer
Vibration magnetometer is used for comparison of magnetic moments and magnetic fields. This device works
on the principle, that whenever a freely suspended magnet in a uniform magnetic field, is disturbed from it's
equilibrium position, it starts vibrating about the mean position.
Time period of oscillation of experimental bar magnet (magnetic moment M) in Torsion head
earth's magnetic field (BH ) is given by the formula. T = 2π I
MBH
Where, I = moment of inertia of short bar magnet = wL2 (w = mass of bar NS
12
magnet)
(3) Use of vibration magnetometer
(i) Determination of magnetic moment of a magnet :
The experimental (given) magnet is put into vibration magnetometer and it's time period T is determined. Now
T = 2π I ⇒M = 4π 2 I
MBH BH .T 2
(ii) Comparison of horizontal components of earth's magnetic field at two places.
T = 2π I ; since I and M the magnet are constant, so T2 ∝ 1 ⇒ (BH )1 = T22
MBH BH (BH )2 T12
(iii) Comparison of magnetic moment of two magnets of same size and mass.
T = 2π I ; Here I and BH are constants. So M ∝ 1 ⇒ M1 = T22
M.BH T2 M2 T12
(iv) Comparison of magnetic moments of two magnets of unequal sizes and masses (by sum and difference
method) :
In this method both the magnets vibrate simultaneously in two following position.
Sum position : Two magnets are placed such that their magnetic moments are additive
Net magnetic moment Ms = M1 + M2 M1
Net moment of inertia Is = I1 + I2
Time period of oscillation of this pair in earth's magnetic field (BH) M2
Ts = 2π Is = 2π I1 + I2 .....(i)
Ms BH (M1 + M 2 )BH
Frequency νs = 1 M s (BH ) SN
2π Is SN
Difference position : Magnetic moments are subtractive
Net magnetic moment Md = M1 + M2 M2 N S M1
Net moment of inertia Id = I1 + I2 SN
and Td = 2π Id = 2π I1 + I2 .......(ii)
Md BH (M1 − M 2 )BH
and νd = 1 (M1 + M 2 ) BH
2π (I1 + I 2 )
From equation (i) and (ii) we get Ts = M1 − M2 ⇒ M1 = Td2 + Ts2 = ν 2 + ν 2
Td M1 + M2 M2 Td2 − Ts2 s d
ν 2 − ν 2
s d
(v) To find the ratio of magnetic field : Suppose it is required to find the ratio B where B is the field created
BH
by magnet and BH is the horizontal component of earth's magnetic field.
To determine B a primary (main) magnet is made to first oscillate in earth's magnetic field (BH) alone and
BH
it's time period of oscillation (T) is noted.
T = 2π I BH
M BH
and frequency ν = 1 M BH M NS
2π I
Now a secondary magnet placed near the primary magnet so primary magnet oscillate in a new field with is
the resultant of B and BH and now time period, is noted again.
There are two important possibilities for placing secondary magnet
Possibility 1
New field increases so time period of oscillation of primary magnet decreases
BH
Sec. Pri. or or Pri
or NS
NS NS NS NS
Sec d
dd SN
Now time period T' = 2π I or new frequency ν ' = 1 M(M + BH ) SN
M(B + BH ) 2π I d
Also ν ' 2 = B + BH ⇒ ν ' 2 = B +1 ⇒ B = ν ' 2 − 1 SN
ν BH ν B1 BH ν
Possibility 2
Net field decreases so time period of oscillation of primary magnet increases
BH Pri. Sec. Sec. Pri.
S N
NS N or SN S
d
d
T' = 2π I (BH > B) and ν ' = 1 M(BH − B)
M(BH − B) 2π I
Also ν ' 2 = BH − B ⇒ ν ' 2 = 1 − B ⇒ B = 1 − ν ' 2
ν BH ν B1 BH ν
Concepts
Remember time period of oscillation in difference position is greater than that in sum position Td > Ts .
If a rectangular bar magnet is cut in n equal parts then time period of each part will be 1 times that of complete magnet (i.e.
n
T' = T ) while for short magnet T' = T . If nothing is said then bar magnet is treated as short magnet.
n n
Suppose a magnetic needle is vibrating in earth’s magnetic field. With temperature rise M decreases hence time period (T)
increases but at 770oC (Curie temperature) it stops vibrating.
Example
Example: 25 Two magnets are held together in a vibration magnetometer and are allowed to oscillate in the earth's
magnetic field. With like poles together 12 oscillations per minute are made but for unlike poles together only
4 oscillations per minute are executed. The ratio of their magnetic moments is [MP PMT 1996]
(a) 3 : 1 (b) 1 : 3 (c) 3 : 5 (d) 5 : 4
Solution : (d) By using M1 = Td2 + Ts2 ; where Ts = 60 = 5sec and Td = 60 = 15sec ∴ M1 = (15)2 + (5)2 = 5
Example: 26 M2 Td2 − Ts2 12 4 M2 (15)2 − (5)2 4
Solution : (b) The magnetic needle of a tangent galvanometer is deflected at an angle 30° due to a magnet. The horizontal
Example: 27 component of earth's magnetic field 0.34×10–4 T is along the plane of the coil. The magnetic intensity is
Solution : (c) [KCET 1999; AFMC 1999, 2000; BHU 2000; AIIMS 2000, 02]
Example: 28
Solution : (b) (a) 1.96 × 10−4 T (b) 1.96 × 10−5 T (c) 1.96 × 104 T (d) 1.96 × 105 T
Example: 29 B = BH tanθ ⇒ B = 0.34 × 10−4 tan 30o = 1.96 × 10−5 T
Solution : (b) A magnet freely suspended in a vibration magnetometer makes 10 oscillations per minute at a place A and 40
Example: 30
Solution : (d) oscillations per minute at a place B. If the horizontal component of earth's magnetic field at A is 36 ×10−6T,
Example: 31
Solution : (c) then its value at B is [EAMCET 2001]
(a) 36 × 10−6 T (b) 72 × 10−6 T (c) 144 × 10−6 T (d) 288 × 10−6 T
By using T = 2π I ⇒T∝ 1 ⇒ TA = (BH )B ⇒ 60 / 10 = (BH )B ⇒ (BH )B = 144 × 10 −6 T.
MBH BH TB (BH )A 60 / 20 36 × 10 −6
The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. By doing
this the periodic time of the magnetometer will [MP PMT 2000]
(a) Increase by 19% (b) Increase by 11% (c) Decrease by 19% (d) Decrease by 21%
T = 2π I ⇒T∝ 1 ⇒ T1 = M2
MBH M T2 M1
If M1 = 100 then M2 = (100 − 19) = 81 . So, T1 = 81 = 9 ⇒ T2 = 10 T1 = 11%T1
T2 100 10 9
A magnet makes 40 oscillations per minute at a place having magnetic field intensity BH = 0.1 × 10 −5 . At
another place, it takes 2.5 sec to complete one-vibration. The value of earth's horizontal field at that place
[CPMT 1999; AIIMS 2000]
(a) 0.25 × 10−6 T (b) 0.36 × 10−6 T (c) 0.66 × 10−8 T (d) 1.2 × 10−6 T
By using T = 2π I ⇒ T1 = (BH )2 ⇒ 60 / 40 = (BH )2 ⇒ (BH )2 = 0.36 × 10 −6 T .
MBH T2 (BH )1 2,5 0.1 × 10 −5
When 2 amp. current is passed through a tangent galvanometer, it gives a deflection of 30°. For 60° deflection,
The current must be
[MP PET 2000]
(a) 1 amp. (b) 2 3 amp. (c) 4 amp. (d) 6 amp.
By using i ∝ tanθ ⇒ i1 = tanθ1 ⇒ 2 = tan 30o = 1 ⇒ i2 =6 amp.
i2 tanθ 2 i2 tan 60o 3
In vibration magnetometer the time period of suspended bar magnet can be reduced by [CBSE PMT 1999]
(a) Moving it towards south pole (b) Moving it towards north pole
(c) Moving it toward equator (d) Any one them
As we move towards equator BH increases and it becomes maximum at equator. Hence T = 2π I , we
MBH
can say that according to the relation T decreases as BH ↑ increases (i.e. as we move towards equator).
Example: 32 The time period of a freely suspended magnet is 2 sec. If it is broken in length into two equal parts and
Solution : (d) one part is suspended in the same way, then its time period will be
Example: 33 [MP PMT 1999]
Solution : (b)
Example: 34 (a) 4 sec (b) 2 sec (c) 2 sec (d) 1 sec
Solution : (b) T = 2π I ; When a bar magnet is broken in n equal parts so magnetic moment of each part become 1 times
Example: 35 MBH n
Solution : (b)
and moment of inertia becomes of each part becomes 1 times. Hence time period becomes 1 times i..e. T'= T
n3 n 4
In this question n = 2 so, T' = T = 2 = 1sec
2 2
A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations
per minute at a place where dip angle is 30o and 15 oscillations per minute at a place where dip angle
is 60o. The ratio of total earth's magnetic field at the two places is [MP PMT 1991; BHU 1997]
(a) 3 3 : 8 (b) 16 : 9 3 (c) 4 : 9 (d) 2 3 : 9
By using T = 2π I = 2π I
MBH MB cos φ
⇒ T∝ 1 ⇒ T1 = B2 × cos φ 2 ⇒ 60 / 20 = B2 × cos 60 ⇒ B1 = 16 .
B cosφ T2 B1 cos φ1 60 / 15 B1 cos 30 B2 93
If θ1 and θ2 are the deflections obtained by placing small magnet on the arm of a deflection
magnetometer at the same distance from the compass box in tan A and tan B positions of the
magnetometer respectively then the value of tanθ1 will be approximately [MP PMT 1992]
tanθ 2
(a) 1 (b) 2 (c) 1 (d) 2
2
In tan A position µ0 . 2M = BH tanθ1 ......(i)
4π d3
In tan B position µ0 . M = BH tanθ2 .....(ii)
4π d3
Dividing equation (i) by equation (ii) tanθ1 = 2 .
tanθ 2 1
In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of
earth's magnetic field is 2 sec. When a magnet is brought near and parallel to it, the time period
reduces to 1 sec. The ratio H/F of the horizontal component H and the field F due to magnet will be
[MP PMT 1990]
(a) 3 (b) 1 (c) 3 (d) 1
3 3
Time period decreases i.e. field due to magnet (F) assist the horizontal component of earth's magnetic field
(see theory)
Hence by using B = T − 1 ⇒ F = 2 2 − 1 = 3 ⇒ H = 1 .
BH T' H 1 F 3
Example: 36 A certain amount of current when flowing in a properly set tangent galvanometer, produces a
Solution : (b)
Example: 37 deflection of 45o. If the current be reduced by a factor of 3, the deflection would
Solution : (d)
(a) Decrease by 30o (b) Decreases by 15o (c) Increase by 15o (d) Increase by 30o
Example: 38
Solution : (c) By using i ∝ tanθ ⇒ i1 = tanθ1 ⇒ i1 3 = tan 45o ⇒ 3 tanθ2 = 1 ⇒ tanθ2 = 1 ⇒ θ2 = 30o
i2 tanθ 2 i1 / tanθ 2 3
So deflection will decrease by 45o − 30o = 15o .
The angle of dip at a place is 60o. A magnetic needle oscillates in a horizontal plane at this place with
period T. The same needle will oscillate in a vertical plane coinciding with the magnetic meridian with
a period
(a) T (b) 2T (c) T (d) T
2 2
When needle oscillates in horizontal plane
Then it's time period is T = 2π I ......(i)
MBH
When needle oscillates in vertical plane i.e. It oscillates in total earth's total magnetic field (B)
Hence T' = 2π I ......(ii)
M
Dividing equation (ii) by (i) T' = BH = B cosφ = cos 60 = 1 ⇒ T'= T
T B B 2 2
A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of
vibration is found to be 2 seconds. The same needle is then allowed to vibrate in the horizontal plane
and the time period is again found to be 2 seconds. Then the angle of dip is
(a) 0o (b) 30o (c) 45o (d) 90o
In vertical plane perpendicular to magnetic meridian.
T = 2π I ......(i)
MBV
In horizontal plane T = 2π I .....(ii)
MBH
Equation (i) and (ii) gives BV = BH
Hence by using tan φ = BV ⇒ tanφ =1⇒φ = 45o
BH
Tricky Example: 3
A magnet is suspended horizontally in the earth's magnetic field. When it is displaced and then
released it oscillates in a horizontal plane with a period T. If a place of wood of the same
moment of inertia (about the axis of rotation) as the magnet is attached to the magnet what
would the new period of oscillation of the system become
(a) T (b) T (c) T (d) T 2
3 2 2
Solution : (d) Due to wood moment of inertia of the system becomes twice but there is no change magnetic moment
of the system.
Hence by using T = 2π I ⇒T∝ I ⇒ T' = 2T
MBH
Magnetic Materials.
(1) Types of magnetic material : On the basis of mutual interactions or behaviour of various materials in
an external magnetic field, the materials are divided in three main categories.
(i) Diamagnetic materials : Diamagnetism is the intrinsic property of every material and it is generated due to
mutual interaction between the applied magnetic field and orbital motion of electrons.
(ii) Paramagnetic materials : In these substances the inner orbits of atoms are incomplete. The electron spins
are uncoupled, consequently on applying a magnetic field the magnetic moment generated due to spin motion align
in the direction of magnetic field and induces magnetic moment in its direction due to which the material gets feebly
magnetised. In these materials the electron number is odd.
(a) (b)
When no field is applied On application of field (B)
(iii) Ferromagnetic materials : In some materials, the permanent atomic magnetic moments have strong
tendency to align themselves even without any external field.
These materials are called ferromagnetic materials.
In every unmagnetised ferromagnetic material, the atoms form domains inside the material. The atoms in any
domain have magnetic moments in the same direction giving a net large magnetic moment to the domain. Different
domains, however, have different directions of magnetic moment and hence the materials remain unmagnetised.
On applying an external magnetic field, these domains rotate and align in the direction of magnetic field.
Unmagnetised Magnetised
(2) Curie Law : The magnetic susceptibility of paramagnetic substances in inversely to its absolute
temperature i.e. χ ∝ 1 ⇒ χ ∝ C
T T
where C = Curie constant, T = absolute temperature
On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa.
The magnetic susceptibility of ferromagnetic substances does not change according to Curie law.
(i) Curie temperature (Tc) : The temperature above which a ferromagnetic material behaves like a
paramagnetic material is defined as Curie temperature (Tc).
or
The minimum temperature at which a ferromagnetic substance is converted into paramagnetic substance is
defined as Curie temperature.
For various ferromagnetic materials its values are different, e.g. for Ni, TCNi = 358o C
for Fe, TCFe = 770o C
for CO, TCCO = 1120o C
At this temperature the ferromagnetism of the substances suddenly vanishes.
(ii) Curie-weiss law : At temperatures above Curie temperature the magnetic susceptibility of ferromagnetic
materials is inversely proportional to (T – Tc) i.e. χ ∝ 1 χ
T − Tc
⇒ χ = (T C Here Tc = Curie temperature
− Tc )
χ-T curve is shown (for Curie-Weiss Law) TC T
(3) Comparative study of magnetic materials
Property Diamagnetic Paramagnetic Ferromagnetic
Cause of magnetism substances substances substances
Orbital motion of electrons
Spin motion of electrons Formation of domains
Explanation of magnetism On the basis of orbital On the basis of spin and On the basis of domains
motion of electrons orbital motion of electrons formed
Behaviour In a non- These are repelled in an These are feebly attracted These are strongly attracted
uniform magnetic field external magnetic field i.e. in an external magnetic in an external magnetic
have a tendency to move field i.e., have a tendency field i.e. they easily move
from high to low field to move from low to high from low to high field
region. field region region
State of magnetisation These are weekly These get weekly These get strongly
magnetised in a direction magnetised in the direction magnetised in the direction
opposite to that of applied of applied magnetic field of applied magnetic field
magnetic field
When the material in the Liquid level in that limb Liquid level in that limb Liquid level in that limb
form of liquid is filled in the gets depressed rises up S rises up very much
U-tube and placed between
pole pieces. NS N NS
Liquid Liquid Liquid
On placing the gaseous The gas expands at right The gas expands in the The gas rapidly expands in
materials between pole angles to the magnetic direction of magnetic field. the direction of magnetic
pieces field. field
The value of magnetic B < B0 B > B0
induction B B >> B0
Low and negative |χ| ≈ 1
Magnetic susceptibility χ where B0 is the magnetic induction in vacuum
Low but positive χ ≈ 1 Positive and high χ ≈ 102
Dependence of χ on Does not depend on Inversely proportional to χ ∝ 1 or
temperature temperature (except Bi at − Tc
low temperature) temperature χ ∝ 1 or T
T
χ = C This is called
C T − Tc
χ = T . This is called
Curie law, where C = Curie Weiss law.
Curie constant Tc = Curie temperature
Dependence of χ on H Does not depend Does not depend Does not depend
independent independent independent
Relative µr < 1 µr > 1 µr >> 1
permeability (µr) I is in a direction opposite I is in the direction of H but µr = 102
Intensity of magnetisation to that of H and its value is value is low
(I) very low I is in the direction of H
and value is very high.
I-H curves + Is
H
– I H Hs H
Magnetic moment (M) The value of M is very low The value of M is very low The value of M is very high
(≈ 0 and is in a direction and is in the direction of H and is in the direction of H
Transition of materials (at opposite to H.)
Curie temperature) These do not change. On cooling, these get These get converted into
converted to ferromagnetic paramagnetic materials
χ materials at Curie temperature above Curie temperature
χ χ
T T TC T
The property of magnetism Diamagnetism is found in Paramagnetism is found in Ferro-magnetism is found
those materials the atoms those materials the atoms in those materials which
Examples of which have even of which have majority of when placed in an external
Nature of effect number electrons electron spins in the same magnetic field are strongly
direction magnetised
Cu, Ag, Au, Zn, Bi, Sb, Al, Mn, Pt, Na, CuCl2, O2 Fe, Co, Ni, Cd, Fe3O4 etc.
and crown glass
NaCl, H2O air and
diamond etc.
Distortion effect Orientation effect Hysteresis effect
(4) Hysteresis : For ferromagnetic materials, by removing external magnetic field i.e. H = 0. The magnetic
moment of some domains remain aligned in the applied direction of previous magnetising field which results into a
residual magnetism.
The lack of retracibility as shown in figure is called hysteresis and the curve is known as hysteresis loop.
I or (B) B
C
A
D OG H
EF
(i) When magnetising field (H) is increased from O, the intensity of magnetisation I increases and becomes
maximum. This maximum value is called the saturation value.
(ii) When H is reduced, I reduces but is not zero when H = 0. The remainder value OC of magnetisation when
H = 0 is called the residual magnetism or retentivity.
The property by virtue of which the magnetism (I) remains in a material even on the removal of magnetising
field is called Retentivity or Residual magnetism.
(iii) When magnetic field H is reversed, the magnetisation decreases and for a particular value of H, denoted
by Hc, it becomes zero i.e., Hc = OD when I = 0. This value of H is called the corecivity.
(iv) So, the process of demagnetising a material completely by applying magnetising field in a negative
direction is defined Corecivity. Corecivity assesses the softness or hardness of a magnetic material. Corecivity
signifies magnetic hardness or softness of substance :
Magnetic hard substance (steel) → High corecvity
Magnetic soft substance (soft iron) → Low corecivity
(v) When field H is further increased in reverse direction, the intensity of magnetisation attains saturation value
in reverse direction (i.e. point E)
(vi) When H is decreased to zero and changed direction in steps, we get the part EFGB.
Thus complete cycle of magnetisation and demagnetisation is represented by BCDEFGB.
Note : ≅ The energy loss (or hysteresis energy loss) in magnetising and demagnetising a specimen is
proportional to the area of hysteresis loop.
(vii) Comparison between soft iron and steel :
Soft iron Steel
II
HH
The area of hysteresis loop is less (low energy loss) The area of hysteresis loop is large (high energy loss)
Less relativity and corecive force More retentivity and corecive force
Magnetic permeability is high Magnetic permeability is less
Magnetic susceptibility (χ) is high χ is low
Intensity of magnetisation (I) is high I is low
It magnetised and demagnetised easily Magnetisation and demagnetisation is complicated
Used in dynamo, transformer, electromagnet tape Used for making permanent magnet.
recorder and tapes etc.
Concepts
An iron cored coil and a bulb are connected in series with an ac generator. If an iron rod is introduced inside a coil, then the intensity of
bulb will decrease, because some energy lest in magnetising the rod.
Hysteresis energy loss = Area bound by the hysteresis loop = VAnt Joule
Where , V = Volume of ferromagnetic sample, A = Area of B – H loop P, n = Frequency of alternating magnetic field and t = Time.
Example
Example: 39 A ferromagnetic substance of volume 10–3 m3 is placed in an alternating field of 50 Hz. Area of hysteresis
curve obtained is 0.1 M.K.S. unit. The heat produced due to energy loss per second in the substance will be
Solution : (c)
Example: 40 (a) 5 J (b) 5 × 10–2 cal (c) 1.19 × 10–3 cal (d) No loss of energy
By using heat loss = VAnt ; whre V = volume = 10–3 m3; A = Area = 0.1m2, n = frequency = 50 Hz and t = time = 1sec
Heat loss = 10–3 × 0.1 × 50 × 1 = 5 × 10–3 J = 1.19 × 10–3 cal
A magnetising field of 1600 A-m–1 produces a magnetic flux of 2.4 × 10–5 Wb in an iron bar of cross-sectional
area 0.2 cm2. The susceptibility of an iron bar is [BHU 2002]
(a) 298 (b) 596 (c) 1192 (d) 1788
Solution : (b) By using B= µH = µ0µr H and µr = (1 + χ m ) ⇒ µr = B = φ
µ0H µ0 HA
Example: 41
µr = (4π × 10 −7 2.4 × 10 −5 × 10 −4 ) = 596.8. Hence χm = 595.8 ≈ 596
Solution : (c) ) × 1600 × (0.2
Example: 42
For iron it's density is 7500 kg/m3 and mass 0.075 kg. If it's magnetic moment is 8 × 10–7 Amp × m2, it's
Solution : (a) intensity of magnetisation is
Example: 43
(a) 8 Amp/m (b) 0.8 Amp/m (c) 0.08 Amp/m (d) 0.008 Amp/m
Solution : (c)
Example: 44 I = M = Md = 8 × 10 −7 × 7500 = 0.08 Amp / m
Solution : (a) V m 0.075
Example: 45
Solution : (c) The dipole moment of each molecule of a paramagnetic gas is 1.5 × 10–23 Amp × m2. The temperature
of gas is 27oC and the number of molecules per unit volume in it is 2 × 1026 m–3. The maximum
possible intensity of magnetisation in the gas will be
(a) 3 × 103 Amp/m (b) 4 × 10–3 Amp/m (c) 5 × 105 Amp/m (d) 6 × 10–4 Amp/m
I = M = µN = 1.5 × 10 −23 × 2 × 10 26 = 3 × 10 3 Amp / m
V V 1
The coereivity of a small bar magnet is 4 × 103 Amp/m. It is inserted inside a solenoid of 500 turns and
length 1 m to demagnetise it. The amount of current to be passed through the solenoid will be
(a) 2.5 A (b) 5 A (c) 8 A (d) 10 A
H = ni ⇒ i= H = 4 × 10 3 = 8A
n 500
The units for molar susceptibility
(a) m3 (b) kg-m–3 (c) kg–1 m3 (d) No units
Molar susceptibility = Volume susceptibility × molecular weight = I/H ×M = I/H ×M
Density of material ρ M/V
So it's unit is m3.
The ratio of the area of B-H curve and I-H curve of a substance in M.K.S. system is
(a) µ02 (b) 1 (c) µ0 (d) 1
µ02 µ0
Area of B-H loop = µ0 (Area of I-H)
Heating & chemical effects of current
Joules Heating.
When some potential difference V is applied across a resistance R then the work done by the electric field on
charge q to flow through the circuit in time t will be W = qV = Vit = i2Rt = V 2t Joule . i R
R B
A
This work appears as thermal energy in the resistor.
Heat produced by the resistance R is H = W = Vit = i 2 Rt = V 2t Cal. This relation is called joules heating.
J 4⋅2 4⋅2 4⋅ 2R
Some important relations for solving objective questions are as follow :
Condition Graph
If R and t are constant
H
H ∝ i 2 and H ∝ V 2 i (or V)
H
If i and t are constant (series grouping)
H∝R
If V and t are constant (Parallel R
grouping) H
H ∝ 1 R
R H
If V, i and R constant
H∝t
t
Electric Power.
The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.
P = W = Vi = i2R = V2
t R
(1) Units : It’s S.I. unit is Joule/sec or Watt 40W
220V
Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt
(2) Rated values : On electrical appliances (Bulbs, Heater ……. etc.)
Wattage, voltage, ……. etc. are printed called rated values e.g. If suppose we have a bulb of 40 W, 220 V then
rated power (PR) = 40 W while rated voltage (VR) = 220 V. It means that on operating the bulb at 220 volt, the power
dissipated will be 40 W or in other words 40 J of electrical energy will be converted into heat and light per second.
(3) Resistance of electrical appliance : If variation of resistance with temperature is neglected then
resistance of any electrical appliance can be calculated by rated power and rated voltage i.e. by using R = VR2 e.g.
PR
Resistance of 100 W, 220 volt bulb is R = 220 × 220 = 484 Ω
100
(4) Power consumed (illumination) : An electrical appliance (Bulb, heater, …. etc.) consume rated power
(PR) only if applied voltage (VA) is equal to rated voltage (VR) i.e. If VA = VR so Pconsumed = PR. If VA < VR then
Pconsumed = V A2 also we have R = VR2 so Pconsumed (Brightness) = V 2 . PR
R PR A
VR2
110 2
220
e.g. If 100 W, 220 V bulb operates on 110 volt supply then Pconsumed = × 100 = 25 W
Note : ≅If VA < VR then % drop in output power = (PR − Pconsumed ) × 100
PR
≅ For the series combination of bulbs, current through them will be same so they will consume
power in the ratio of resistance i.e., P ∝ R {By P = i2R) while if they are connected in parallel i.e. V
is constant so power consumed by them is in the reverse ratio of their resistance i.e. P ∝ 1 .
R
(5) Thickness of filament of bulb : We know that resistance of filament of bulb is given by R= VR2 , also
PR
R=ρ l , hence we can say that A ∝ PR ∝ 1 i.e. If rated power of a bulb is more, thickness of it’s filament
A R
(Thickness )
is also more and it’s resistance will be less.
If applied voltage is constant then P(consumed) ∝ 1 (By P = V A2 ). Hence if different bulbs (electrical appliance)
R R
1
operated at same voltage supply then Pconsumed ∝ PR ∝ thickness ∝ R
Note : ≅Different bulbs 25W 100W 1000W
220V 220V 220V
⇒ Resistance R25 > R100 > R1000
⇒ Thickness of filament t1000 > t100 > t40
⇒ Brightness B1000 > B100 > B25
(6) Long distance power transmission : When power is transmitted through a power line of resistance R,
power-loss will be i2 R
Now if the power P is transmitted at voltage V P = Vi i.e. i = (P / V) So, Power loss = P2 × R
V2
Now as for a given power and line, P and R are constant so Power loss ∝ (1 / V 2 )
So if power is transmitted at high voltage, power loss will be small and vice-versa. e.g., power loss at 22 kV
is 10–4 times than at 220 V. This is why long distance power transmission is carried out at high voltage.
(7) Time taken by heater to boil the water : We know that heat required to raise the temperature ∆θ of
any substance of mass m and specific heat S is H = m.S.∆θ
Here heat produced by the heater = Heat required to raise the temp. ∆θ of water.
i.e. p × t = J × m.S.∆θ ⇒ t = J(m.S.∆θ ) {J = 4.18 or 4.2 J/cal)
p
for m kg water t = 4180 ( or 4200)m ∆θ {S = 1000 cal/kgoC)
p
Note : ≅ If quantity of water is given n litre then t = 4180(4200)n ∆θ
p
Electricity Consumption.
(1) The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of
electrical power.
(2) The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour
(KWH) or board of trade unit (B.T.U.) or simple unit.
(3) 1 KWH or 1 unit is the quantity of electrical energy which dissipates in one hour in an electrical circuit
when the electrical power in the circuit is 1 KW thus 1 KW = 1000 W × 3600 sec = 3.6 × 106 J.
(4) Important formulae to calculate the no. of consumed units is n = Total watt × Total hours
1000
Concepts
When some potential difference applied across the conductor then collision of free electrons with ions of the lattice result’s in
conversion of electrical energy into heat energy
If a heating coil of resistance R, (length l) consumed power P, when voltage V is applied to it then by keeping V is constant if it is
cut in n equal parts then resistance of each part will be R and from Pconsumed ∝ 1, power consumed by each part P' = nP .
n R
Joule’s heating effect of current is common to both ac and dc.
Example
Example: 1 The approximate value of heat produced in 5 min. by a bulb of 210 watt is (J = 4.2 joule/calorie)
Solution : (a)
(a) 15,000 (b) 1,050 (c) 63,000 [MP PET 2000; MNR 1985]
(d) 80,000
By using H = P×t = 210 × 5 × 60 = 15000 Cal
4.2 4.2
Example: 2 A heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the heat
Solution : (a)
produced by this half coil to that by the original coil is [NCERT 1972]
Example: 3
Solution : (b) (a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
Example: 4
If suppose resistance of the coil is R so resistance of it’s half will be R . Hence by using H = V 2t ⇒H∝ 1
Solution : (b) 2 R R
Example: 5
⇒ H Half = R Full R2
Solution : (b) H Full R Half = R/2 = 1
Example: 6
Solution : (a) Note : ≅ In general if coil is divided in n equal parts then heat produced by each part will be n times of the
Example: 7
Solution : (d) heat produced by coil it self i.e. H′ = nH
If current in an electric bulb changes by 1%, then the power will change by [AFMC 1996]
(a) 1% (b) 2% (c) 4% (d) 1 %
2
By using P = i2R ⇒ P ∝ i2 ⇒ ∆P = 2 ∆i ⇒ change in power = 2%
P i
A constant voltage is applied on a uniform wire, then the heat is produced. The heat so produced will be
doubled, if [IIT-JEE 1980; MP PET 1987; SCRA 1994; MP PMT 1999]
(a) The length and the radius of wire are halved (b) Both length and radius are doubled
(c) Only the length is doubled (d) Only the radius is doubled
By using H = V 2t and R = ρ l = ρl ⇒ H = V 2tπ r 2 ⇒H ∝ r2 ; on doubling both r and l heat will be
R A π r2 ρl l
doubled.
An electric heater of resistance 6 ohm is run for 10 minutes on a 120 volt line. The energy liberated in this
period of time is [MP PMT 1996]
(a) 7.2 × 10 3 J (b) 14.4 × 105 J (c) 43.2 × 10 4 J (d) 28.8 × 10 4 J
By using H = V 2t ⇒ H = (120)2 × 10 × 60 = 14.4 × 105 J
R 6
An electric bulb of 100 W is designed to operate on 220 V. Resistance of the filament is
(a) 484 Ω (b) 100 Ω (c) 22000 Ω [EAMCET 1981, 82; MP PMT 1993, 97]
(d) 242 Ω
By using P = V2 ⇒R= V2 = (220)2 = 484 Ω
R P 100
An electric bulb is rated 220 V and 100 W. Power consumed by it when operated on 110 volt is
[AFMC 2000; MP PMT 1986, 94; CPMT 1986]
(a) 50 W (b) 75 W (c) 90 W (d) 25 W
By using Pconsumed = VA 2 × PR ⇒ PConsumed = 110 2 × 100 = 25 W
VR 220
Example: 8 A 500 watt heating unit is designed to operate from a 115 Volt line. If the line voltage drops to 110 volt, the
Solution : (c) percentage drop in heat output will be [ISM Dhanbad 1994]
Example: 9 (a) 10.20% (b) 8.1% (c) 8.6% (d) 7.6%
Solution : (d)
Example: 10 VA 2 110 2
Solution : (d) VR 115
Example: 11 By using Pconsumed = × PR ⇒ PConsumed = × 500 = 456.6 Watt
Solution : (b)
Example: 12 So % drop in heat output = PActual − PConsumed × 100 = (500 − 456.6) × 100 = 8.6%
Solution : (d) PActual 500
Example: 13 An electric lamp is marked 60 W, 230 V. The cost of 1 kilowatt hour of power is Rs. 1.25. The cost of using
Solution : (d)
this lamp for 8 hours is [KCET 1994]
Example: 14
(a) Rs. 1.20 (b) Rs. 4.00 (c) Rs. 0.25 (d) Rs. 0.60
By using consumed unit (n) or KWH = Total Watt × Total time ⇒ n= 60 × 8 = 12
1000 1000 25
So cost = 12 × 1.25 = 0.60 Rs
25
How much energy in Kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a
month (30 days) [Pb PMT 2000; CPMT 1991; NCERT 1978]
(a) 1500 (b) 15.000 (c) 15 (d) 150
By using n= Total Watt × Total time ⇒ n = (50 × 10) × (10 × 30) = 150
1000 1000
An immersion heater is rated 836 watt. It should heat 1 litre of water from 20o C to 40o C in about
(a) 200 sec (b) 100 sec (c) 836 sec (d) 418 sec
By using t = 4180 × n × ∆θ ⇒ t = 4180 × 1 × (40 − 20) = 100 sec
P 836
The power of a heater is 500 watt at 800o C. What will be its power at 200o C if α = 4 × 10 −4 per °C
(a) 484 W (b) 672 W (c) 526 W (d) 611 W
By using P = i2R = V2 ⇒ P ∝ 1 ⇒ P1 = R2 = (1 + α t2 ) ⇒ 500 = (1 + 4 × 10 −4 × 200)
R R P2 R1 (1 + α t1 ) P2 (1 + 4 × 10 −4 × 800)
⇒ 500 = 1.08 ⇒ 611W
P2 1.32
A heater of 220 V heats a volume of water in 5 minute time. A heater of 110 V heats the same volume of
water in [AFMC 1993]
(a) 5 minutes (b) 8 minutes (c) 10 minutes (d) 20 minutes
By using H = V2 t . Here volume of water is same. So same heat is required in both cases. Resistance is also
R
constant so V2t = constant ⇒ t ∝ 1 ⇒ t1 = V2 2 ⇒ 5 = 110 2 = 1 ⇒ t 2 = 20 min
V2 t2 V1 t2 220 4
Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to
2/3 of its initial value, then the same amount of water will boil with the same supply voltage in [MP PMT 1994]
(a) 15 minutes (b) 12 minutes (c) 10 minutes (d) 8 minutes
Solution : (c) By using H = V 2t where R=ρ l ⇒H = V 2t A . Since volume is constant so H is also constant so t ∝ l
R A ρl
which gives t2 l2 t2 2 l1 t2 = 10 min
t1 l1 15 3
= ⇒ = ⇒
l1
Tricky Example: 1
If resistance of the filament increases with temperature, what will be power dissipated in a 220 V – 100 W
lamp when connected to 110 V power supply
(a) 25 W (b) < 25 W (c) > 25 W (d) None of these
VA 2 . PR 110 2
VR 220
Solution : (b) If resistance do not varies with temperature PConsumed = = × 100 = 25 W. But
actually resistance is increasing with temperature so consumed power will be lesser then 25W.
Combination of Bulbs (or Electrical Appliances).
Bulbs (Heater etc.) are in series Bulbs (Heater etc.) are in parallel
(1) Total power consumed P1 P2 (1) Total power consumed P1
Supply Ptotal = P1 + P2 + P3 ...... + Pn
1 = 1 1 + ...... P2
Ptotal P1 + P2 Supply
(2) If ‘n’ bulbs are identical, Ptotal = P (2) If ‘n’ identical bulbs are in parallel. Ptotal = nP
N
(3) Pconsumed (Brightness) ∝ V ∝ R ∝ 1 i.e. in series (3) Pconsumed (Brightness) ∝ PR ∝ i ∝ 1 i.e. in parallel
Prated R
combination bulb of lesser wattage will give more bright combination bulb of greater wattage will give more bright
light and p.d. appeared across it will be more. light and more current will pass through it.
Some Standard Cases for Series and Parallel Combination.
(1) If n identical bulbs first connected in series so PS = P and then connected in parallel. So PP = nP hence PP = n2 .
n PS
(2) To operate a bulb on voltage which is more then it’s rated voltage, a proper resistance is connected in series
with it. e.g. to glow a bulb of 30 W, 6 V with full intensity on 126 volt required series resistance calculated as follows
Bulb will glow with it’s full intensity if applied voltage on it is 6 V i.e. 120 V appears across the series resistance
R current flows through bulb = current flows through resistance R
120 V
i = 30 = 5 amp i
6 6V
Hence for resistance V = iR i.e. 120 = 5 × R ⇒ 5 × R ⇒ R = 24 Ω 126 V
Note : ≅ If you want to learn Short Trick then remember Series resistance = Voperating − VR × VR
PR
(3) An electric kettle has two coils when one coil is switched on it takes time t1 to boil water and when the
second coil is switched on it takes time t2 to boil the same water.
If they are connected in series If they are connected in parallel
1 11 PP = P1 + P2
PS = P1 + P2
⇒ HP = H1 + H2
1 11 tp t1 t2
⇒ HS / tS = H1 / t1 + H2 / t2
H S = H1 = H 2 so t s = t1 + t 2 Hp = H1 = H2 so 1 = 1 1
tp t1 + t2
i.e. time taken by combination to boil the same i.e. time taken by parallel combination to boil the same
quantity of water tS = t1 + t2
quantity of water tp = t1t2
t1 + t2
(4) If three identical bulbs are connected in series as shown in figure then on closing the switch S. Bulb C short
circuited and hence illumination of bulbs A and B increases
AB C
Reason : Voltage on A and B increased.
+ S
V–
(5) If three bulbs A, B and C are connected in mixed combination as shown, then illumination of bulb A
decreases if either B or C gets fused
Reason : Voltage on A decreases.
A
V C
B
(6) If two identical bulb A and B are connected in parallel with ammeter A and key K as shown in figure.
It should be remembered that on pressing key reading of ammeter becomes twice.
AK Reason : Total resistance becomes half.
V
AB
Concepts Heater
When a heavy current appliance such us motor, heater or geyser is switched on, it will draw a heavy current from the source so
that terminal voltage of source decreases. Hence power consumed by the bulb decreases, so the light of bulb becomes less.
r
~
K
If the source is ideal i.e. r = 0, there will be no change in the brightness of the bulb.
Example
Example: 15 An electric kettle has two heating coils. When one of the coils is connected to ac source the water in the kettle
Solution : (d) boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected
Example: 16 in parallel, the time taken by the same quantity of water to boil will be [CBSE PMT 2003]
Solution : (c)
Example: 17 (a) 4 min (b) 25 min (c) 15 min (d) 8 min
By using the formula tp = t1t 2 (as we discussed in theory) ⇒ tp = 10 × 40 = 8 min
t1 + t2 (10 + 40)
Note : ≅ In this question if coils are connected in series then the time taken by the same quantity of water to
boil will be ts = t1 + t2 = 10 + 40 = 50 min
If a 30 V, 90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be connected in
series with the bulb [KCET 2003; MP PMT 2002]
(a) 10 ohm (b) 20 ohm (c) 30 ohm (d) 40 ohm
By using Series resistance R = Voperating − VR × VR (As we discussed in theory) ⇒ R = (120 − 30) × 30 = 30 Ω
PR 90
In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat
produced in 4 ohm resistance is [BHU 2003; RPMT 1994; IIT 1981]
(a) 1 cal/sec 4Ω 6Ω
5Ω
(b) 2 cal/sec
(c) 3 cal/sec
(d) 4 cal/sec
Solution : (b) Ratio of currents i1 10 = 2 by using H = i 2 Rt i2 4Ω 6Ω Line (2)
i2 =5 1 i Line (1)
⇒ H1 = i1 2 × R1 ⇒ 10 = 2 2 × 5 ⇒ H2 = 2cal / sec 5Ω
H2 i2 R2 H2 1 4 i1
Example: 18 Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced
Solution : (d)
Example: 19 in the two cases is [MP PET 2002, 1999; MP PMT 2001, 2000, 1996; AIIMS 2000; MNR 1987; DCE 1997, 94]
Solution : (b)
Example: 20 (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4
Solution : (d) Both the wires are of equal length so they will have same resistance and by using H = V 2t ⇒ H ∝ 1
Example: 21 R R
Solution : (b) ⇒ Hs = RP ; ⇒ Hs = R/2 = 1
Example: 22 HP Rs HP 2R 4
Solution : (b)
Example: 23 If two bulbs of wattage 25 and 100 respectively each rated at 220 volt are connected in series with the supply
of 440 volt, then which bulb will fuse [MP PET 2000; MNR 1988]
(a) 100 watt bulb (b) 25 watt bulb (c) None of them (d) Both of them
In series VA ∝ 1 i.e. voltage appear on 25W bulb will be more then the voltage appears on 100 W bulb. So
PR
bulb of 25 W will gets fused.
Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt. If the same
resistors are connected in parallel across the same e.m.f., then the power dissipated will be
[KCET 1999; DCE 1998; CBSE 1998; MP PAT 1996]
(a) 10 W (b) 30 W (c) 10/3 W (d) 90 W
In series consumed power Ps P while in parallel consumed power Pp = nP ⇒ PP = n2. Ps
=n
⇒ PP = (3)2 × 10 = 90W
Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused, the remaining 39 are
connected again in series across the same supply. The illumination will be [Haryana CEE 1996; NCERT 1972]
(a) More with 40 bulbs than with 39 (b) More with 39 bulbs than with 40
(c) Equal in both the cases (d) In the ratio of 402 : 392
Illumination = PConsumed = V2 . Initially there were 40 bulbs in series so equivalent resistance was 40 R, finally
R
39 bulbs are in series so equivalent resistance becomes 39 R. Since resistance decreases so illumination
increases with 39 bulbs.
Two bulbs of 100 watt and 200 watt, rated at 220 volts are connected in series. On supplying 220 volts, the
consumption of power will be
(a) 33 watt (b) 66 watt (c) 100 watt (d) 300 watt
In series PConsumed = P1P2 ⇒ PConsumed = 100 × 200 = 66W
P1 + P2 300
Two wires ‘A’ and ‘B’ of the same material have their lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. The
two wires are connected in parallel across a battery. The ratio of the heat produced in ‘A’ to the heat
produced in ‘B’ for the same time is [MNR 1998]
(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1
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