Chapter 13 Moving Charge and Magnetism ( Magnetic effects of Current Part 3) free study material by TEACHING CARE online tuition and coaching classes
Chapter 13 Moving Charge and Magnetism ( Magnetic effects of Current Part 3) free study material by TEACHING CARE online tuition and coaching classes
Motion of Charged Particle in a Magnetic Field.
If a particle carrying a positive charge q and moving with velocity v enters a magnetic field B then it experiences a force F which is given by the expression
´ ´ ´ ´ ´ ® ´

F = r ´ B) Þ
F = qvB sinq
B
´ ´ ´ ´ ´
q, m
´ ´ ´ ´ ´
Here v =
velocity of the particle, B =
magnetic field
´ ´ ´ ´ ´ ´
 Zero force
Force on charged particle will be zero (i.e. F = 0) if
 No field e. B = 0 Þ F = 0
 Neutral particle e. q = 0 Þ F = 0
 Rest charge e. v = 0 Þ F = 0
 Moving charge e. q = 0^{o} or q = 180^{o} Þ F = 0
(2) Direction of force
q = 0^{o}
q = 180^{o}
The force F is always perpendicular to both the velocity v and the field B in accordance with Right Hand Screw Rule, through v and B themselves may or may not be perpendicular to each other.
Direction of force on charged particle in magnetic field can also be find by Flemings Left Hand Rule (FLHR). Here, First finger (indicates) ® Direction of magnetic field
Middle finger ® Direction of motion of positive charge or direction,
opposite to the motion of negative charge. ^{B}
Thumb ® Direction of force
(3) Circular motion of charge in magnetic field ^{v}
Consider a charged particle of charge q and mass m enters in a uniform magnetic field B with an initial velocity v perpendicular to the field.
q = 90^{o}, hence from F = qvB sinq particle will experience a maximum magnetic force F_{max} = qvB which act’s in a direction perpendicular to the motion of charged particle. (By Flemings left hand rule).
 Radius of the path : In this case path of charged particle is circular and magnetic force provides the
necessary centripetal force i.e. qvB = mv2
r
Þ radius of path r = mv
qB
If p = momentum of charged particle and K = kinetic energy of charged particle (gained by charged particle
after accelerating through potential difference V) then p = mv = =
So r = mv = p = 2mK = 1
qB qB qB B
r µ v µ p µ i.e. with increase in speed or kinetic energy, the radius of the orbit increases.
Note : @ Less radius (r) means more curvature (c) i.e. c µ 1
r
Less : r
More : c
More : r
Less : c
r = ¥
c = 0
(ii) Direction of path : If a charge particle enters perpendicularly in a magnetic field, then direction of path described by it will be
Type of charge Direction of magnetic field Direction of it’s circular motion  
Negative  Outwards ¤ 
– q
– q
+ q
+ q  r ¤ B
r Ä B
r Ä B
r ¤ B  
Anticlockwise  
Negative  Inward Ä  
Clockwise  
Positive  Inward Ä  
Anticlockwise  
Positive  Outward ¤  
Clockwise 
 Time period : As in uniform circular motion v = rw, so the angular frequency of circular motion, called
cyclotron or gyrofrequency, will be given by w = v = qB and hence the time period, T = 2p = 2p m
r m w qB
i.e., time period (or frequency) is independent of speed of particle and radius of the orbit and depends only on

the field B and the nature, i.e., specific charge æ q ö , of the particle.
m
è ø
(4) Motion of charge on helical path
When the charged particle is moving at an angle to the field (other than 0^{o}, 90^{o}, or 180^{o}).
In this situation resolving the velocity of the particle along and perpendicular to the field, we find that the particle moves with constant velocity v cosq along the field (as no force acts on a charged particle when it moves parallel to the field) and at the same time it is also moving with velocity v sinq perpendicular to the field due to
which it will describe a circle (in a plane perpendicular to the field) of radius. r = m(vsinθ)
qB
®
B
®v
q, m q
Time period and frequency do not depend on velocity and so they are given by T = 2p m
qB
and n =
qB
2p m
So the resultant path will be a helix with its axis parallel to the field B as shown in figure in this situation.
The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by
p = T(v cosq ) = 2p m (v cosq )
qB
Note : @ 1 rotation º 2p º T and 1 pitch º 1 T
@ Number of pitches º Number of rotations º Number of repetition = Number of helical turns
@ If pitch value is p, then number of pitches obtained in length l given as
Number of pitches = l
p
Some standard results
and time reqd. t =
l
v cosq
Ratio of radii of path described by proton and aparticle in a magnetic field (particle enters perpendicular to the field)
Particle motion between two parallel plates (v ^ B)
(i)

To strike the opposite plate it is essential that d < r .
 Does not strike the opposite plate d >
(iii) To touch the opposite plate d = r .
 To just not strike the opposite plate d ³ r .
 To just strike the opposite plate d £ r .
(5) Lorentz force
When the moving charged particle is subjected simultaneously to both electric field E and magnetic field B ,

r
the moving charged particle will experience electric force Fe = qE and magnetic force Fm = ´ B); so the net
force on it will be F = q[E + (v ´ B)] . Which is the famous ‘Lorentzforce equation’.
Depending on the directions of v, E and B following situations are possible
(i) When
r
v, E
and B all the three are collinear : In this situation as the particle is moving parallel or
antiparallel to the field, the magnetic force on it will be zero and only electric force will act and so
r

a = m
= qE
m
The particle will pass through the field following a straight line path (parallel field) with change in its speed. So in this situation speed, velocity, momentum kinetic energy all will change without change in direction of motion as
r
shown
q
(ii) When E is parallel to B and both these fields are perpendicular to v then : Fe is
perpendicular to Fm
and they cannot cancel each other. The path of charged particle is curved in both these fields.
r r
E B
v
q
 v , E
and B are mutually perpendicular : In this situation if E and B are such that
y r
F = Fe
 Fm = 0
i.e.,
r
a = (F / m) = 0
E
F_{e}
+ q v + q x
as shown in figure, the particle will pass through the field with same velocity.
r F
And in this situation, as F = F i.e., qE = qvB v = E / B ^{B} m
e m z
This principle is used in ‘velocityselector’ to get a charged beam having a specific velocity.
Note : @ From the above discussion, conclusion is as follows
@ If E = 0, B = 0, so F = 0.
@ If E = 0, B ¹ 0, so F may be zero (if q = 0^{o} or 180^{o} ).
@ If E ¹ 0, B ¹ 0, so F = 0 (if  Fe = Fm  and their directions are opposite)
@ If E ¹ 0, B = 0, so F ¹ 0 (because v ¹ constant ).
Cyclotron.
Cyclotron is a device used to accelerated positively charged particles (like, aparticles, deutrons etc.) to acquire enough energy to carry out nuclear disintegration etc. t is based on the fact
that the electric field accelerates a charged particle and the magnetic field keeps it revolving in circular orbits of constant frequency. Thus a small potential difference would impart if enormously large velocities if the particle is made to traverse the potential difference a number of times.
It consists of two hollow Dshaped metallic chambers D_{1} and D_{2}
called dees. The two dees are placed horizontally with a small gap separating them. The dees are connected to the source of high frequency electric field. The dees are enclosed in a metal box containing a gas at a
low pressure of the order of 10^{–3} mm mercury. The whole apparatus is placed between the two poles of a strong electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.
Note : @ The positive ions are produced in the gap between the two dees by the ionisation of the gas.
To produce proton, hydrogen gas is used; while for producing alphaparticles, helium gas is used.
 Cyclotron frequency : Time taken by ion to describe q semicircular path is given by t = pr = p m
If T = time period of oscillating electric field then T = 2t = 2p m
qB
v
the cyclotron frequency n = 1 =
T
qB Bq 2pm
æ q 2 B2 ö 2
 Maximum energy of position : Maximum energy gained by the charged particle
where r_{0} = maximum radius of the circular path followed by the positive ion.
Note : @Cyclotron frequency is also known as magnetic resonance frequency.
Emax = ç
è
÷ r
2m ø
@ Cyclotron can not accelerate electrons because they have very small mass.
Hall effect : The Phenomenon of producing a transverse emf in a current carrying conductor on applying a magnetic field perpendicular to the direction of the current is called Hall effect.
Hall effect helps us to know the nature and number of charge carriers in a conductor.
Example: 28 Electrons move at right angles to a magnetic field of
 ´ 10 ^{–}^{2} Tesla with a speed of
6 ´ 10^{27}
m / s.
If the
specific charge of the electron is 1.7 ´ 10^{11} Coul/kg. The radius of the circular path will be [BHU 2003]
(a) 2.9 cm (b) 3.9 cm (c) 2.35 cm (d) 3 cm
mv v
6 ´ 10^{27} _{2}
Solution : (c)
r = qB
Þ (q / m). B = 17 ´ 10^{11} ´ 1.5 ´ 10 ^{2}
= 2.35 ´ 10
m = 2.35 cm.
Example: 29 An electron (mass
= 9 ´ 10 ^{31} kg.
charge
= 1.6 ´ 10 ^{19} coul. ) whose kinetic energy is
7.2 ´ 10 ^{18}
joule is
moving in a circular orbit in a magnetic field of 9 ´ 10^{5} weber / m^{2}. The radius of the orbit is [MP PMT 2002]
(a) 1.25 cm (b) 2.5 cm (c) 12.5 cm (d) 25.0 cm
Solution : (d)
r = 2mK =
qB
2 ´ q ´ 10 ^{31} ´ 7.2 ´ 10 ^{8}
1.6 ´ 10 ^{19} ´ q ´ 10 ^{5}
= 0.25 cm = 25 cm .
Example: 30 An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic energy. Which of the following is true [MP PET 1999]
 Trajectory of electron is less curved (b) Trajectory of proton is less curved
 Both trajectories are equally curved (d) Both move on straight line path
Solution : (b) By using r =
2mk qB
; For both particles q ® same, B ® same, k ® same
Hence r µ Þ
re =
rp
Q mp
 me
so rp
 re
Since radius of the path of proton is more, hence it’s trajectory is less curved.
Example: 31 A proton and an a – particles enters in a uniform magnetic field with same velocity, then ratio of the radii of path describe by them
(a)
1 : 1
(b)
1 : 2
(c)
2 : 1
 None of these
Solution : (b) By using r = mv ; v ® same, B ® same Þ r µ m Þ
rp = mp ´ qa
= mp ´ 2qp = 1
qB 2
ra ma qp
4mp qp 2
Example: 32 A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What
should be the energy of a–particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius [BHU 1997]
 1 MeV (b) 4 MeV (c) 2 MeV (d) 5 MeV
Solution : (a) By using r =
2mK qB
; r ® same, B ® same Þ
K µ q 2
m
K æ q ö 2 m æ 2q ö 2 m
Hence
a = ç a ÷ ´ p = ç
p ÷ ´
^{p} 1 Þ
Ka = Kp = 1meV.
Kp ç qp ÷ ma
ç qp ÷
4mp
è ø è ø
Example: 33 A proton and an a – particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25m sec to make 5 revolutions, then the periodic time for the a – particle would be [MP PET 1993]
(a)
50m sec
25m sec
10m sec
5m sec
Solution : (c) Time period of proton Tp = 25 = 5m sec
5
By using T = 2p m Þ
qB
Ta = ma Tp mp
 qp
qa
= 4mp ´
mp
qp Þ
2qp
Ta = 2Tp
= 10m sec .
Example: 34 A particle with 10^{–11} coulomb of charge and 10^{–7} kg mass is moving with a velocity of 10^{8} m/s along the yaxis.
A uniform static magnetic field B = 0.5 Tesla is acting along the xdirection. The force on the particle is
[MP PMT 1997]
(a) 5 ´ 10^{–11} N along ˆi (b) 5 ´ 10^{3} N along kˆ (c) 5 ´ 10^{–11} N along – ˆj (d) 5 ´ 10^{–4} N along – kˆ
Solution : (d) By using
F = q(v ´ B); where v
= 10ˆj
and B = 0.5ˆi
Þ F = 10 ^{–}^{11}(10^{8} ˆj ´ 0.5ˆi ) = 5 ´ 10 ^{–}^{4} (ˆj ´ ˆi ) = 5 ´ 10 ^{–}^{4} (kˆ) i.e., 5 ´ 10 ^{–}^{4} N along – kˆ.
Example: 35 An electron is moving along positive xaxis. To get it moving on an anticlockwise circular path in xy plane, a
magnetic filed is applied  [MP PMT 1999]  
Solution : (a)  (a) Along positive yaxis (c) Along negative yaxis The given situation can be drawn as follows  (b) Along positive zaxis (d) Along negative zaxis 
According to figure, for deflecting electron in xy plane, force must be acting an it towards yaxis. Hence according to Flemings left hand rule, magnetic field directed along positive y – axis.
y xy plane
e–
e–
x
z
Example: 36 A particle of charge – 16 ´ 10^{–}^{18}
coulomb moving with velocity 10 m/s along the xaxis enters a region where
a magnetic field of induction B is along the yaxis, and an electric field of magnitude 10^{4} V/m is along the negative zaxis. If the charged particle continuous moving along the xaxis, the magnitude of B is [AIEEE 2003]
(a)
10^{–}^{3} Wb / m^{2}
(b)
10^{3} Wb / m^{2}
(c)
10^{5} Wb / m^{2}
(d)
10^{16} Wb / m^{2}
Solution : (b) Particles is moving undeflected in the presence of both electric field as well as magnetic field so it’s speed
v = E
B
Þ B = E
v
= 10^{4}
10
= 10^{3} Wb / m^{2}.
Example: 37 A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is
[IITJEE (Screening) 2002]
(a) qbB/m (b) q(b – a)B/m (c) qaB/m (d) q(b+a)B/2m
Solution : (b) As shown in the following figure, the z – axis points out of the paper and the magnetic fields is directed into
the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the

magnetic field. It can just enter the region x > b for r ³ (b – q) Y
Now r = mv ³ (b – a) qb
Þ v ³
q(b – a)B m
Þ vmin =
q(b – a)B . O X
m
Example: 38 At a certain place magnetic field vertically downwards. An electron approaches horizontally towards you and enters in this magnetic fields. It’s trajectory, when seen from above will be a circle which is
(a) Vertical clockwise (b) Vertical anticlockwise
(c) Horizontal clockwise (d) Horizontal anticlockwise
Solution : (c) By using Flemings left hand rule.
Example: 39 When a charged particle circulates in a normal magnetic field, then the area of it’s circulation is proportional to
(a) It’s kinetic energy (b) It’s momentum
(c) It’s charge (d) Magnetic fields intensity
Solution : (a)
r = 2mK
qB
and
A = Aq ^{2} Þ
A = p (2mK)
q 2b 2
Þ A µ K.
Example: 40 An electron moves straight inside a charged parallel plate capacitor at uniform charge density s . The space
between the plates is filled with constant magnetic field of induction electron in the capacitor is
 B. Time of straight line motion of the
(a)
(b)
(c)
(d)
es
e _{0}lB
e _{0}lB
s
es
e _{0} B
e _{0} B es
Solution : (b) The net force acting on the electron is zero because it moves with constant velocity, due to it’s motion on straight line.
Þ F = F e + F m = 0 Þ  F e = F m  Þ e E = evB Þ
ve = E = s
éE = s ù
net
\ The time of motion inside the capacitor t = l
v


= e _{0}lB .
s
B e _{0} B
ê e o ú
Example: 41 A proton of mass 1.67 ´ 10 ^{–}^{27} kg and charge 1.6 ´ 10^{–}^{19}
C is projected with a speed of
2 ´ 10^{6} m / s
at an
angle of 60^{0} to the Xaxis. If a uniform magnetic field of 0.104 Tesla is applied along Yaxis, the path of proton is [IITJEE 1995]
 A circle of radius = 2 m and time period p ´ 10^{–}^{7} s
 A circle of radius = 1 m and time period 2p ´ 10^{–}^{7} s
 A helix of radius = 1 m and time period 2p ´ 10^{–}^{7} s
 A helix of radius = 2 m and time period 4p ´ 10^{–}^{7} s
Solution : (b) By using r = mv sinq Þ
qB
r = 1.67 ´ 15^{27} ´ 2 ´ 10^{6} ´ sin 30°
1.6 ´ 10^{19} ´ 0.104
Y
®
= 0.1m B
and it’s time period T = 2pm =
qB
2 ´ p ´ 9.1´ 10^{31}
1.6 ´ 10^{19} ´ 0.104
X
= 2p ´ 10^{7} sec .
Example: 42 A charge particle, having charge q accelerated through a potential difference V enter a perpendicular magnetic field in which it experiences a force F. If V is increased to 5V, the particle will experience a force
(a) F (b) 5F (c)
F (d) 5 F
5
Solution : (d)
1 mv^{2} = qV Þ v =
2
. Also F = qvB
Þ F = qB
hence F µ which gives F‘ =
5F.
Example: 43 The magnetic field is downward perpendicular to the plane of the paper and a few charged particles are projected in it. Which of the following is true [CPMT 1997]
 A represents proton and B and electron
 Both A and B represent protons but velocity of A is more than that of B
 Both A and B represents protons but velocity of B is more than that of A
 Both A and B represent electrons, but velocity of B is more than that of A
Solution : (c) Both particles are deflecting in same direction so they must be of same sign.(i.e., both A and B represents protons)
By using r = mv
qB
Þ r µ v
From given figure radius of the path described by particle B is more than that of A. Hence vB > v A .
Example: 44 Two very long straight, particle wires carry steady currents i and – i respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the
plane of the wires. It’s instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is [IITJEE 1998]
(a)
m _{0} iqv
2pd
(b)
m _{0} iqv
pd
(c)
2m _{0}iqv
pd
 Zero
Solution : (d) According to gives information following figure can be drawn, which shows that direction of magnetic field is along the direction of motion of charge so net on it is zero.
Example: 45 A metallic block carrying current i is subjected to a uniform magnetic induction B as shown in the figure. The moving charges experience a force F given by ……. which results in the lowering of the potential of the face
……. Assume the speed of the carriers to be v [IITJEE 1996]
 eVBkˆ, ABCD
 eVBkˆ, ABCD
 – eVBkˆ, ABCD
 – eVBkˆ, EFGH
Solution : (c) As the block is of metal, the charge carriers are electrons; so for current along positive xaxis, the electrons are
moving along negative xaxis, i.e. v
= – vi y
and as the magnetic field is along the yaxis, i.e.
B = Bˆj x
z
so F = q(v ´ B) for this case yield F = (e)[vˆi ´ Bˆj]
i.e., F = evBkˆ [As ˆi ´ ˆj = kˆ]
As force on electrons is towards the face ABCD, the electrons will accumulate on it an hence it will acquire lower potential.
Force on a Current Carrying Conductor in Magnetic Field.
In case of current carrying conductor in a magnetic field force experienced by its small length element is
dF = idl ´ B ; idl = current element dF = l(dl ´ B)
Total magnetic force
F = ò dF = ò i(dl ´ B)
If magnetic field is uniform i.e., B = constant
r éò rù r r¢ r
F = iêë dl úû ´ B = i(L ´ B)
r r
ò dl = L = vector sum of all the length elements from initial to final point. Which is in accordance with the
law of vector addition is equal to length vector L¢ joining initial to final point.
 Direction of force : The direction of force is always perpendicular to the plane containing
idl
and B
r
and is same as that of crossproduct of two vectors (A ´ B) with
r
A = i dl .
The direction of force when current element determined by applying either of the following rules
i dl
and B are perpendicular to each other can also be
 Force on a straight wire : If a current carrying straight conductor (length l) is placed in an uniform magnetic field (B) such that it makes an angle q with the direction of field then force experienced by it is F = Bil sinq
If q = 0^{o} , F = 0
If q = 90^{o} ,
Fmax
= Bil
 Force on a curved wire
The force acting on a curved wire joining points a and b as shown in the figure is the same as that on a straight
wire joining these points. It is given by the expression F = i L ´ B
Specific Example
The force experienced by a semicircular wire of radius R when it is carrying a current i and is placed in a uniform magnetic field of induction B as shown.

r
L = 2Ri
and
r

B = Bi
r ¢ = 2Rˆi and
r

B = Bj
r ¢ = 2Rˆi and
r

B = B(k)

So by using

F = i(L‘ ´ B) force on ˆ ˆ
\ r ˆ
F = i´ 2BR(i ´ j)
F = i´ 2BR(+ j)

the wire r
r
F = 2BiR (along Yaxis)
r = ˆ
ˆ Þ F = 0
F = 2BiR k i.e. F = 2BiR
F i(2R)(B)(i ´ i )
(perpendicular to paper outward)
Force Between Two Parallel Current Carrying Conductors.
When two long straight conductors carrying currents i_{1}
and
i_{2} placed parallel to each other at a distance ‘a’
from each other. A mutual force act between them when is given as
F = F = F = m 0 ×
1 2 4p
2i1i2 ´ l a
i_{1} i_{2}
where l is the length of that portion of the conductor on which force is to be calculated. a
Hence force per unit length
F = m _{0} × 2i_{1}i_{2} N or F = 2i_{1}i_{2} dyne
l 4p a m l a cm
Direction of force : If conductors carries current in same direction, then force between them will be attractive. If conductor carries current in opposite direction, then force between them will be repulsive.
 2×
i_{1} i_{2}
 ×


 F F ×
 ×
 ×
 ×
1 × ×
i_{1}
× ×

× ×
1 F_{2}
× ×
× ×
× ×
Note : @ If a = 1m and in free space
F = 2 ´ 10 ^{7} N / m l
then
i1 = i2
= 1Amp
in each identical wire. By this
concept S.I. unit of Ampere is defined. This is known as Ampere’s law.
Force Between Two Moving Charges.
If two charges q_{1} and q_{2} are moving with velocities v_{1} and v_{2} respectively and at any instant the distance between them is r, then
Stationary charges
Moving charges
Magnetic force between them is F
= m0 . q1q2 v1v2
….. (i)
^{m} 4p r ^{2}
and Electric force between them is F
= 1 . q_{1}q_{2}
….. (ii)

^{e} 4pe r ^{2}
F 1
F æ v ö ^{2}
From equation (i) and (ii)
^{ } ^{m} = m_{0}e _{0}v ^{2} but m_{0}e _{0} = ; where c is the velocity light in vacuum. So ^{m} = ç ÷
Fe c ^{2} Fe è c ø
If v << c then F_{m} << F_{e}
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.
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
dq Þ dq
2T sin 2 = dF = Bidl 2T sin 2
= BiRdq
If dq is small so, sin dq
2
» dq
2
Þ 2T. dq
2
= BiRdq BiL
T = BiR , if
2pR = L
so T = 2p
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
In both the situations for equilibrium of XY it’s downward weight = upward magnetic force i.e. mg =
μ0 . 2i1i2 .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 = 2p .
@ 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.
Wire is placed in the perpendicular magnetic field due to coil at it’s centre, so it will experience a

maximum force F = Bil = m 0 i1 ´ i l
wire is placed along the axis of coil so magnetic field produced by the coil is parallel to the wire. Hence it will not experience any force.
2r
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.
If current makes to flow through spring, then spring will contract and weight lift up
If switch is closed then current start flowing, 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 ) .

Hence direction of current is from X ® Y and in balanced condition
Fm = mg
Þ Bil = mg
Þ i = mg
Bl
Case 6 : A current carrying conductor floating in air such that it is making an angle q with the direction of magnetic field, while magnetic field and conductor both lies in a horizontal plane.
In equilibrium mg = Bil sinq
Þ i = mg
Bl sinq
Case 7 : Sliding of conducting rod on inclined rails : When a conducting rod slides on conducting rails.
X F cosq
B
In the following situation conducting rod (X, Y) slides at constant velocity if
F cosq = mg sinq
Þ Bil cosq = mg sinq
Þ B = mg tanq
i l
Example: 46 A vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed towards north. The wire will experience a force directed towards
 North (b) South (c) East (d) West
Solution : (d) By applying Flemings left hand rule, direction of force is found towards west.
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 field of strength 500 gauss and makes an angle of 30^{o} with the direction of the field. It experiences a force of magnitude [MP PET 1993]
(a) 3 ´ 10^{4} N (b) 3 ´ 10^{2} N (c) 3 ´ 10^{–} ^{2} N (d) 3 ´ 10^{–4} N
Solution : (c) By using
F = Bil sinq
Þ F = (500 ´ 10^{–4}) ´ 0.4 ´ sin 30^{o} Þ 3 ´ 10^{–2} N.
Example: 48 Wires 1 and 2 carrying currents t_{1}
and t2
respectively are inclined at an angle q 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)
m0 i1,i2dl tanq 2 r

m0 i ,i

2pr ^{2}

m
dl sinq
0 i1,i2 dl cosq 2 r

m0 i1,i2 dl sinq
4 r
Solution : (c) Length of the component dl which is parallel to wire (1) is dl cos q, so force on it
F = m _{0} × 2i1i2 (dl cosq ) = m _{0}i1i2 dl cosq .
4p r 2pr
Example: 49 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 Yaxis. The force experience by the wire and its direction are
 2iBL directed along the negative Zaxis
 5iBL directed along the positive Zaxis
 iBL direction along the positive Zaxis
 2iBL directed along the positive Zaxis
Solution : (c) 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 ®

straight wire of length l between P and U as shown i P B
below Þ
Y
Hence force on replaced wire PU will be F = Bil T U U
X
and according to FLHR it is directed towards +zaxis
Example: 50 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
Force on the conductor ABC = Force on the conductor AC 10
3
= 5 ´ 10 ´ (5 ´ 10^{–2})
A
10 F
Þ
= 2.5 N
B 4 C B C
Example: 51 A wire of length l carries a current i along the Xaxis. A magnetic field exists which is given as
( ˆi + ˆj + kˆ) T. Find the magnitude of the magnetic force acting on the wire
B = B0
(a)
B0il
(b)
B0i l ´
(c)
2B0i l
(d)
1 ´ B0i l
2
Solution : (b) By using
F = i( l ´ B) Þ F = i [l ˆi ´ B0 (ˆi + ˆj + kˆ)] = B0 il[ˆi ´ (ˆi + ˆj + kˆ)]
Þ F = B0 il[ˆi ´ ˆi + ˆi ´ ˆj + ˆi ´ kˆ] = B0 il[kˆ – ˆj]
{ˆi ´ ˆi = 0,ˆi ´ ˆj = kˆ,ˆi ´ kˆ = ˆj}
It’s magnitude F =
2B0 il
Example: 52 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 [IITJEE (Screening) 2003]]
 Contract (b) Expand
(c) Move towards + ve xaxis (d) Move towards – ve xaxis
Solution : (b) 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 magnetic field is perpendicular to paper inwards and current in the loop is
Y ®

i F_{m}
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 twodimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire
 p i a B
 4p i a B
 Zero
 2p i a B
Solution : (d) The direction of the magnetic force will be vertically downwards at each element of the wire.
Thus F = Bil = Bi (2pa) = 2piaB.
Example: 54 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 Ð abc = 45^{o}. The ratio of force on ab and on bc is
(a)
(b)
 1
 23
Solution : (c) Force on portion ab of wire F_{1} = Bil sin 90^{o} = Bil
æ l ö _{o} F_{1}
Force on portion bc of wire F_{2} = Bi ç
è
÷ sin 45
2 ø
= Bil . So
= 1 .

2
Example: 55 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)
(b)
(c)
(d)
m _{0}i
pr
2m _{0}i
pr
m _{0}i ( 4pr

m 2i
. (
+ 1)
+ 1)
4p
Solution : (d) By using
r
B = m _{0} i
f + sin f
) , from figure d = r sin 45 ^{o} = r
4p . r (sin 1 2 Y
m i
Magnetic field due to each wire at P
B = ^{0} .
4p
(sin 45^{o} + sin 90^{o}) i

m i
. (
+ 1)
4p r O X
Hence net magnetic field at P
B = 2 ´ m 0 i
+ = m _{0} i
+ 1)
net
4p . r (
1) 2p . r (
Example: 56 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 [ m_{0} = 4p ´ 10 ^{7} weber/amp – m] [MP PET 2000]
(a) Repulsive force of 10 ^{4} N / m
(c) Repulsive force of 2p ´ 10 ^{5} N / m
(b) Attractive force of 10 ^{4} N / m
(d) Attractive force of 2p ´ 10 ^{5} N / m
Solution : (a) By using
F = m _{0} . 2i_{1}i_{2}
l 4p a
10 A 5 A
Þ 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.
 m
Example: 57 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]
 4×10^{–4} N towards the right
 4×10^{–4} N towards the left
 6 × 10^{–4} N to the right
 6×10^{–4} N to the left
Solution : (a) 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 F_{net} = F_{R} – F_{P} = 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]
(a)
(b)
(c)
(d)
25 ´ 10 ^{7} N
25 ´ 10 ^{7} N
35 ´ 10^{7} N
35 ´ 10^{7} N
moving towards wire moving away from wire moving towards wire moving away from wire
Solution : (a) Force on sides BC and CD cancel each other.
Force on side AB
FAB = 10^{7} ´ 2 ´ 2 ´ 1 ´ 15 ´ 10^{2} = 3 ´ 10^{6} N
2 ´ 10^{2}
2A 1A
B 10 cm
C
15 cm
Force on side CD
FAB = 10^{–}^{7}
´ 2 ´ 2 ´ 1 ´ 15 ´ 10^{2} 12 ´ 10^{2}
= 0.5 ´ 10^{6} N
FAB
2 cm
A
FCD
D
Hence net force on loop = F_{AB} – F_{CD} = 25 ´ 10^{–7} N (towards the wire).
Example: 59 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
 25 mm
 50 mm
 75 mm
 100 mm
Solution : (a) Suppose in equilibrium wire PQ lies at a distance r above the wire AB

Hence in equilibrium mg = Bil Þ mg = m0 æ 2i ö ´ il Þ
4p r
10^{3} ´ 10 = 10^{7} ´ 2 ´(50)2 = 0.5 Þ r = 25 mm
r
è ø
Example: 60 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
 The rod CD will move upwards parallel to itself
 The rod CD will move downward parallel to itself
 The rod CD will move upward and turn clockwise at the same time
 The rod CD will move upward and turn anti –clockwise at the same time
Solution : (c) Since the force on the rod CD is nonuniform 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
i_{2}
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
 Zero
 5 N
 30 N
 20 N
4 cm
L
10A
6 cm
N
4 cm
Solution : (b) The given wire can be replaced by a straight wire as shown below
N
10A
L
4 cm
L
6 cm
6 cm
N

4 cm
N
10 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 sinæ 2p x ö. A magnetic field

ç ÷
è ø
B exists in the Zdirection find the magnitude of the magnetic force on the portion of the wire between
x = 0 and x = l
(a)
ilB
(b) Zero (c)
ilB
2
(d)
3 / 2ilB
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 l which experiences
a magnetic force Fm = Bil
Y
i
A
x = 0
Z l
B X
x = l
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
Magnetic ^{®}
moment M
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.
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
such a way that the normal (nˆ) to the coil makes an angle q with the direction of B. the coil experiences a torque
given by t = NBiA sinq . Vectorially tr = M ´ B
(i) t is zero when q = 0, i.e., when the plane of the coil is perpendicular to the field.
 t is maximum when q = 90^{o} , e., the plane of the coil is parallel to the field.
Þ t max
= NBiA
The above expression is valid for coils of all shapes.
(2) Workdone
If coil is rotated through an angle q from it’s equilibrium position then required work. W = MB(1 – cosq ). It is maximum when q = 180^{o} Þ W_{max} = 2 MB
(3) Potential energy
Is given by U = – MB cosq Þ 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 currentcarrying 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 horseshoe 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
q = 90^{o} and the deflecting torque always has the maximum value.
t def
= NBiA
……(i)
coil deflects, a restoring torque is set up in the suspension fibre. If a is the angle of twist, the restoring torque is
t rest
= Ca
…..(ii) where C is the torsional constant of the fibre.
When the coil is in equilibrium.
NBiA = Ca Þ i = C a
NBA
Þ i = Ka ,
Where
K = C NBA
is the galvanometer constant. This linear relationship between i and a makes the moving
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.
S = a = NBA
^{i} 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 (S_{V}) : Voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit applied to it.

a
SV V
= a = Si iR R
= NBA
RC
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
 The magnetic dipole moment of the coil is [MP PMT 2001]
(a) 0.60 Am^{2} (b) 0.45 Am^{2} (c) 0.3 Am^{2} (d) 0.15 Am^{2}
Solution : (c) M = niA Þ M = 20 ´ 3 ´ p ( 4 ´ 10^{–2})^{2} = 0.3 Am^{2}.
Example: 62 A steady current i flows in a small square loop of wire of side L in a horizontal plane. The loop is now folded
about its middle such that half of it lies in a vertical plane. Let m_{1}
and m _{2}
respectively denote the magnetic
moments due to the current loop before and after folding. Then [IITJEE 1993]
(a)
m _{2} = 0
(b)
m_{1} and m _{2} are in the same direction
 m_{1} 
 m_{1} æ 1 ö
(c)
=
 m _{2} 
(d)
= ç ÷
 m _{2}  è 2 ø
Solution : (c) Initially Finally
i
M
L
m_{1} = iL^{2}
L/2
M
L/2 L
M = magnetic moment due to each part = i æ L ö ´ L = iL2 = m1
ç 2 ÷ 2 2
\ m _{2} = M
= m1 ´
2
è ø
= m_{1}
Example: 63 A coil of 50 turns is situated in a magnetic field b = 0.25weber/m^{2} as shown in figure. A current of 2A is flowing in the coil. Torque acting on the coil will be
(a) 0.15 N
(b) 0.3 N
(c) 0.45 N
(d) 0.6 N
Solution : (b) Since plane of the coil is parallel to magnetic field. So q = 90^{o}
Hence t = NBiA sin 90^{o} = NBiA = 50 ´ 0.25 ´ 2 ´ (12 ´ 10^{–2} ´ 10 ´ 10^{–2}) = 0.3 N.
Example: 64 A circular loop of area 1 cm^{2}, 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} Nm (c) 10^{–2} N–m (d) 1 N–m
Solution : (a) t = NBiA sinq ; given q = 0 so t = 0.
Example: 65 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/m^{2}. The amount of work done in rotating it through 180^{o} from its equilibrium position will be
(a) 0.1 J (b) 0.2 J (c) 0.4 (d) 0.8 J
[CPMT 1977]
Solution : (a) Work done in rotating a coil through an angle q from it’s equilibrium position is W = MB(1 – cosq) where q = 180^{o} and M = 50 ´ 2 ´ p (4 ´ 10^{–2}) = 50.24 ´ 10^{–2} A–m^{2}. Hence W = 0.1 J
Example: 66 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
(a) As large as possible (b) Any number (c) 2 (d) 1
Solution : (d)
t _{max} = MB
or t _{max} = nipa^{2}B . Let number of turns in length l is n so
l = n(2pa) or
a = l
2pn
Þ t = nipBl ^{2} max 4p 2n2
= l ^{2}iB
4pnmin
Þ t max
µ 1
nmin
Þ nmin = 1
Example: 67 A square coil of N turns (with length of each side equal L) carrying current i is placed in a uniform magnetic
field B = B0 ˆj
as shown in figure. What is the torque acting on the coil
 + B0 NiL^{2}kˆ
 – B0 NiL^{2}kˆ
 B0
NiL^{2}ˆj
 – B0 NiL^{2}ˆj
Solution : (b) The magnetic field is B = B0 ˆj and the magnetic moment m = i A = –i(NL^{2}ˆi )
The torque is given by t = m ´ B
= –iNL^{2}ˆi ´ B0 ˆj = –iNB0 L^{2}ˆi ´ ˆj
= –iNB0 L^{2}kˆ
Example: 68 The coil of a galvanometer consists of 100 turns and effective area of 1 square cm. The restoring couple is 10^{–} ^{8} Nm rad.
The magnetic field between the pole pieces is 5 T. The current sensitivity of this galvanometer will be
[MP PMT 1997]
(a) 5 ´ 10^{4} rad/m amp (b) 5 ´ 10^{–} ^{6} per amp (c) 2 ´ 10^{–} ^{7} per amp (d) 5 rad./m amp
q
Solution : (d) Current sensitivity (S_{i}) = i =
NBA Þ q
C i
= 100 ´ 5 ´ 10 ^{4}
10 ^{8}
= 5 rad /m amp .
Example: 69 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
Solution : (a) Sensitivity (S ) = NBA Þ S
i C i
µ N .