Chapter
17
Moving Charges and Magnetism
Solutions
SECTION - A
Objective Type Questions (One option is correct)
[Magnetic Force]
1.
A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when
subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle
completes one full circle is
(1) BQv 2R
(2)
⎛ Mv 2 ⎞
⎜
⎟ 2R
⎝ R ⎠
(3)
Zero
(4)
BQ 2R
Sol. Answer (3)
Work done by magnetic field on a moving charge particle is zero.
2.
A charged particle, having charge q1 accelerated through a potential difference V enters a perpendicular
magnetic field in which it experiences a force F. If V is increased to 5 V, the particle will experience a force
(1) F
(2)
5F
(3)
F
5
(4)
5F
Sol. Answer (4)
KE 
1
mv 2  qV
2
 v 
2q
m
V
Fv
 F V
Thus
F

F
5v
v
 F   5F
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2
Moving Charges and Magnetism
Solutions of Assignment
[Field Due to Current Element]
3.
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance
of 4 cm from the centre is 54 T. What will be its value at the centre of the loop?
(1) 125 T
(2)
150 T
(3)
250 T
(4)
75 T
Sol. Answer (3)
 0 iR 2
2(R 2  x 2 )3 2
B
 0 iR 2
 (25  10 4 )3 2  54
2
B0 
 0 iR 2 (25  10 4 )3 2

 54
2(R )3
(3  10 2 )3
 B0 
4.
125
 54 = 250 T
27
A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the
axis of the coil located at a distance r from the centre of the coil, such that r >> R, varies as
(1) 1/r
(2)
1/r 3/2
(3)
1/r 2
(4)
1/r 3
Sol. Answer (4)
A circular current carrying loop behaves as a dipole.
5.
A current flows in a conductor from east to west. The direction of the magnetic field at a point above the
conductor is
(1) Towards west
(2)
Towards east
(3)
Towards south
(4)
Towards north
Sol. Answer (4)
N
W
y
x
E
S
Current is along (–x) axis then at any point on the positive z-axis B is towards north.
6.
The earth’s magnetic field at a given point is 0.5 × 10–5 Wb m–2. This field is to be cancelled by magnetic
induction at the centre of a circular conducting loop of radius 5.0 cm. The current required to be flown in the
loop is nearly
(1) 0.2 A
(2)
0.4 A
(3)
4A
(4)
40 A
Sol. Answer (2)
B
0i
2R
5
 0.5  10 
 i
4  10 7  i
2  5  10 –2
5
 0.4 A
4
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
7.
Moving Charges and Magnetism
3
A long straight wire carrying current of 30 A is placed in an external uniform magnetic field of induction
4 × 10–4 T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant
magnetic induction (in tesla) at a point 2.0 cm away from the wire is (0 = 4 ×10–7 H/m)
(1) 10–4
(2)
3 × 10–4
5 × 10–4
(3)
(4)
6 × 10–4
Sol. Answer (3)
i = 30 A
B = 4 × 10–4 T
Magnetic field due to the wire =
=
0i
2x
4  10 –7  30
2    2  10 2
= 30 × 10–5 = 3 × 10–4

Resultant magnetic field = 42  10 –8  32  10 8

12
= 5 × 10–4
8.
Two long straight wires are set parallel to each other. Each carries a current in the same direction and the
separation between them is 2r. The intensity of the magnetic field midway between them is
(1) 0i/r
(2)
40i/r
(3)
Zero
(4)
0i/4r
Sol. Answer (3)
Direction of magnetic field due to both the wires are opposite to each other, and in magnitude both are equal
thus cancel each other.
i
9.
r
i
r
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and
the number of turns per cm is halved, the new value of the magnetic field is
(1) B
(2)
2B
(3)
4B
(4)
B/2
Sol. Answer (1)
B = 0ni
 Bn
 B i
⎛ n⎞
2i
B  ⎜⎝ 2 ⎟⎠

B
ni
 B = B
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
4
Moving Charges and Magnetism
Solutions of Assignment
10. A long solenoid has 200 turns per cm and carries a current of 2.5 ampere. The magnetic field at its centre
is [0 = 4 ×10–7 weber/m2]
(1) 3.14 × 10–2 Wb/m2
(2)
6.28 × 10–2 Wb/m2
(3) 9.42 × 10–2 Wb/m2
(4)
12.56 × 10–2 Wb/m2
Sol. Answer (2)
B = 0 ni
4 × 10–7 × 200 × 100 × 2.5
6.28 × 10–2 Wb/m2
11. If a long hollow copper pipe carries a current, the produced magnetic field will be
(1) Inside the pipe only
(2)
Outside the pipe only
(3) Both inside and outside the pipe
(4)
Neither inside nor outside the pipe
Sol. Answer (2)
Magnetic field inside a hollow current carrying point is zero.
12. A long solenoid has 800 turns per metre length of solenoid. A current of 1.6 A flows through it. The magnetic
induction at the end of the solenoid on its axis is
(1) 16 × 10–4 tesla
(2)
8 ×10–4 tesla
(3) 32 × 10–4 tesla
(4)
4 × 10–4 tesla
Sol. Answer (2)
B
 0 ni 4  10 7  800  1.6

 8  10 4 tesla.
2
2
1
A is flowing through a toroid. It has 1000 number of turns per meter then value of magnetic
4
field (in Wb/m2) along its axis is
13. A current of
(1) 10–2
(2)
10–3
(3)
10–4
(4)
10–7
Sol. Answer (3)
i
1
A
4
B = 0 ni
–7
= 4 × 10 × 1000 ×
1
4
= 10–4
14. Two long conductors, separated by a distance d carry current i1 and i2 in the same direction. They exert a
force F on each other. Now the current in one of them is increased to two times and its direction is reversed.
The distance is also increased to 3 d. The new value of the force between them is
2F
(1) 
3
(2)
F
3
(3)
2F
(4)
F

3
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
5
Sol. Answer (1)
F
 0 i1i 2L
2d
F  ⎛ i  d ⎞ ⎛ 2⎞
⎜
⎟ ⎜ ⎟
F ⎝ i d  ⎠ ⎝ 3⎠
2 F 
F
3
15. A and B are two conductors carrying a current i in the same direction. x and y are two electron beams moving
in the same direction. There will be
A
B
x
y
(1) Repulsion between A and B, attraction between x and y
(2) Attraction between A and B, repulsion between x and y
(3) Repulsion between A and B and also x and y
(4) Attraction between A and B and also x and y
Sol. Answer (2)
Parallel currents attract each other and parallel electron beams dominantly repel each other.
16. A, B and C are parallel conductors of equal length carrying currents i, i and 2i respectively. Distance between
A and B is x. Distance between B and C is also x. F1 is the force exerted by B on A. F2 is the force exerted
by C on A. Choose the correct answer
(1) F 1  2F 2
(2)
A
B
C
i
i
2i
F 2  2F 1
x
x
(3)
F1  F 2
(4)
F 1  F 2
Sol. Answer (4)
A and B will attract whereas A and C will repel.
F
 0 i1i 2
2d

F1 
0i 2
2 x
 0 2i 2
 i2
 0
2(2 x ) 2x
Thus F 1  – F 2
F2 
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
6
Moving Charges and Magnetism
Solutions of Assignment
17. The force between two parallel conductors, each of length 50 m and distance 20 cm apart, is 10–2 N. If the
current in one conductor is double that in another one, then their values will respectively be
(1) 10 A and 20 A
(2)
50 A and 100 A
(3)
5 A and 10 A
(4)
25 A and 50 A
Sol. Answer (1)
F
 0 i1i 2L
2d
–2
 10 
4  10 –7  i  2i
 50
2  0.2
 i = 10 A
18. A conductor PQ, carrying a current i is placed perpendicular to a long conductor xy carrying a current i. The
direction of force on PQ will be
Y
P
i
Q
i
L
X
(1) Towards right
(2)
Towards left
(3)
Upwards
(4)
Downwards
Sol. Answer (4)
Magnetic field at PQ is into the plane and magnetic force is downwards.
19. A circular wire ABC and a straight conductor ADC are carrying current i and are kept in the magnetic field
B then considering points A and C
B
C
i
D
A

(1) Force on ABC is more than that on ADC
(2)
Force on ABC is less than that on ADC
(3) Force on ABC is equal to that on ADC
(4)
Data insufficient
Sol. Answer (3)
Effective length of both ABC and ADC is same thus force on them due to magnetic field is same.
20. There are two concentric circular loops. One loop of radius r is made up of only one turn. The other loop has
a radius 2r and has two turns of wire. They are arranged such that they have a common centre and their planes
are perpendicular to each other. When the same current i is passed through both the coils, the magnetic field
at the common centre is
(1)
0i
2r
(2)
0i
r
(3)
0i
2
2r
(4)
Zero
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
7
Sol. Answer (3)
When current i flow in a ring of radius r then at the centre of the ring magnetic field is given by B 
0 i
2r
0 i n
.
2r
If the ring has n–turns then B =
So for the ring having radius r, B1 =
0 i
2r
…(1)
For the ring having radius 2r and 2 turns
 0 2i
0 i
B2 = 2  2r  =
2r
…(2)
Since B1 and B2 are  to each other
 Net field at the common centre will be
B=
0 i ⎞
⎛
B12  B22 = ⎜⎝ 2 2r ⎟⎠
21. A long current carrying wire is bent as shown. The magnetic field induction at O is
90°
i
L
(1) Zero
(2)
2 2 0 i
L
O
L
(3)
0i
L
(4)
2 0i
L
Sol. Answer (3)
–

4
L
2

4
0
i
Magnetic field induction at O is
B=
0 i
⎛ L ⎞
4 ⎜
⎝ 2 ⎟⎠
0 i

⎤
⎡
⎢cos 4  cos 4 ⎥ 
⎣
⎦ 4 ⎛ L ⎞
⎜⎝
⎟
2⎠
=

⎤
⎡
⎢cos 4  cos 4 ⎥
⎣
⎦
 i
0 i 2
2 2 = 0
L
4 L
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
8
Moving Charges and Magnetism
Solutions of Assignment
22. A short current carrying conductor is placed perpendicular to the axis of a current carrying ring as shown. The
direction of force acting of the conductor is
y-axis
R
I
x
x-axis
I
z-axis
(1) Along positive x-axis
(2)
Along negative y-axis
(3) Along positive z-axis
(4)
Along negative z-axis
Sol. Answer (4)
The direction of magnetic field is along the x-axis i.e., along the axis of the ring.
Here the current in the short conductor is along the (+ve) y-axis.
 F = i l B
 
= i  l jˆ  B iˆ  = i l B  ˆj  iˆ 
= i l B  – kˆ  along (–ve) z-axis
23. Figure shows an arrangement in which long parallel wires carrying equal currents into or out of the page of
paper at the corners of a square. The magnetic field at the centre of square is
I
I
r
r
r
O
r
I
I
(1)
 0I
2r
2
(2)
2 0I
2
2r
(3)
4 0I
2r
(4)
Zero
Sol. Answer (4)
At the centre ‘O’. The magnetic field due to M and R has equal magnitude and opposite direction. Also due
to S and N has equal magnitude and opposite direction as shown in figure.
I
M
I
N
BN
O
BR
I S
BM
BS
R
I
 Net field at the centre ‘O’ is zero.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
9
24. The magnetic field at the centre of dotted circle in the arrangement shown is
I
R
R
I
(1)
 0I
 I
 0
4R 2R
(2)
O
I
R
I
I
 0I
4R
(3)
 0I
2R
(4)
Zero
Sol. Answer (3)
i
A
C
O
M
T
D
P
N
S
The magnetic field at the centre ‘O’ due to circular section has equal magnitude and opposite direction.
⎛ 0 i ⎞
Magnetic field due to MA is B1 = ⎜⎝ 4 R ⎟⎠
⎛ 0 i ⎞
Magnetic field due to CN is B2 = ⎜⎝ 4 R ⎟⎠
O
O
Magnetic field due to TD = 0 and also due to PS = 0
 Net field at O is B = B1 + B2
0 i
2R
25. What is the magnetic field | B | at the point P due to a current carrying wire of length AB (having a current of 2 A)
=
shown in figure?
B
2A
A
(1) 2 μT
(2)
25 μT
5m
P
4m
(3)
2.5 nT
(4)
30 nT
Sol. Answer (4)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
10
Moving Charges and Magnetism
Solutions of Assignment
26. Which of the following properties of magnetic line of force differs them from electrostatic lines of force?
(1) These are closed curves
(2) These can emit or terminate at any angle to the surface
(3) Total flux linked with a close surface is always zero
(4) All of these
Sol. Answer (4)
27. What is 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 in figure?
B
I
(1) Zero
(2)
I(2R)B
(3)
I(R)B
(4)
B(I)πR
Sol. Answer (1)
28. Keeping current per unit length of arc constant, the variation of magnetic field at the centre of an arc (B) with the
angle subtended by the arc at the centre () can be best represented by
B
B
(1)
B
B
(2)
(3)

(4)



Sol. Answer (3)
29. The magnetic field at point P in the hollow cylindrical wire carrying a current i shown in the figure at a distance
r from axis (a < r < b) is given by (Taking current density to be uniform across its cross-section)
i
a
b
r
(1)
 0 i (r 2  a 2 )
2
2
2 (b  a )r
(2)
0i r
2
2
2 ( b  a )
(3)
P
 0 i ab
2 ( b 2  a 2 ) r
(4)
Zero
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
11
Sol. Answer (1)
Let us consider a cylindrical shell of radius x and thickness dx. Current in this cylindrical shell is
i  2 xdx 

di =
 b2 – a2

 Total current inside the circle of radius r is
I1 =
b
2i
2
– a2
r
∫
a
x dx =
b
i
2
– a2

(r2
–
a2)

 ⎫⎪⎬
 ⎪⎭
⎧ i r 2 – a2
⎪
= ⎨ 2
2
⎪⎩ b – a

 Magnetic field at P is
0 i (r 2 – a2 )
0I1
B=
=
2 r
2r (b 2 – a 2 )
[Motion of Charged Particle]
30. An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent
motion will be
(1) Straight line along the x-direction
(2)
A circle in the xz-plane
(3) A circle in the yz-plane
(4)
A circle in the xy-plane
Sol. Answer (2)
Magnetic force F  q(v  B ). F is perpendicular to the plane of v and B. Then the force is along z-axis and
the particle will move in a circle in (x-z) plane.
31. A charged particle is at rest in the region where magnetic field and electric field are parallel. The particle will
move in a
(1) Straight line
(2)
Circle
(3)
Ellipse
(4)
None of these
Sol. Answer (1)
Straight line : As electric field and magnetic field are parallel and particle is at rest.
32. Which particle will have minimum frequency of revolution when projected with the same velocity perpendicular
to a magnetic field?
(1) Li+
(2)
Electron
(3)
Proton
(4)
He+
Sol. Answer (1)
f 
qB
2m
f 
1
m
fmin 
1
mmax
Thus Li+ will have minimum frequency of revolution.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
12
Moving Charges and Magnetism
Solutions of Assignment
33. He+ and O 2+ ions (mass of He+ = 4 amu and that of O2+ = 16 amu) they pass a region of constant
perpendicular magnetic field. If kinetic energy of all the ions is same then
(1) He+ ions will be deflected more than those of O2+
(2) He+ ions will be deflected less than those of O2+
(3) All the ions will be deflected equally
(4) No ions will be deflected
Sol. Answer (3)
r 
p

qB
r
m
q
rHe

rO2
=
2m KE
qB
mHe
mO2
qO 2
qHe
rHe
4 2
1

=
16 1 rO2
Thus both will be deflected equally.
34. A beam of protons is moving horizontally towards you. As it approaches, it passes through a magnetic field
directed downward. The beam deflects
(1) To your left side
(2)
To your right side
(3)
Does not deflect
(4)
Nothing can be said
Sol. Answer (2)
F  q(v  B )
Let the proton be moving towards z-axis and the magnetic field is along  jˆ.
y
Thus F  qvB(k   j )
B
 F  qvBi
x
35. There is a magnetic field acting in a plane perpendicular to this sheet of paper, downward into the paper as
shown in the figure. Particles in vacuum move in the plane of the paper from left to right. The path indicated
by the arrow could be travelled by
(1) Proton
(2)
Neutron
(3)
Electron
(4)
None of these
Sol. Answer (3)
The path shown can be travelled by a negative charge particle hence that is electron.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
(O2–)
13
(Li+)
36. Doubly ionized oxygen atoms
and singly-ionized lithium atoms
are travelling with the same speed,
perpendicular to a uniform magnetic field. The relative atomic masses of oxygen and lithium are 16 and 7
respectively. The ratio
(1) 16 : 7
radius of O2– orbit
is
radius of Li+ orbit
(2)
8:7
(3)
7:8
(4)
7 : 16
Sol. Answer (2)
rO2
rLi
⎛ mO
 ⎜⎜ 2
⎝ mLi
⎞ ⎛ qLi
⎟⎟ ⎜⎜
⎠ ⎝ qO2
⎞
16
1
8
⎟ = ⎛⎜ ⎞⎟ 
=
⎟
⎝
⎠
7
2
7
⎠
37. A long straight wire is kept along x-axis. It carries a current i in the positive x-direction. A proton and an electron
are placed at (0, a, 0) and (0, –a, 0) respectively. The proton is imparted an initial velocity v along +z-axis and
the electron is imparted an initial velocity v along + x-axis. The magnetic forces experienced by the two particles
at the instant are
(1)
 0 iev ˆ  0 iev ˆ
i,
i
2a
2a
(2)
0,
 0 iev ˆ
j
2a
0,
(3)
 0 iev ˆ
j
2a
(4)
0, 0
Sol. Answer (2)
Force acting on a charge particle is given by F  q v  B

⎛ i
 Force on proton is FP  e v kˆ  ⎜ 0
⎝ 2 a
 

Y
P
⎞
kˆ ⎟
⎠
(0, a, 0)
i
= 0 [∵ k̂ × k̂ = 0]
 i
Force on electron, Fe  – e v iˆ  0 – kˆ
2 a
 
=
e (0, –a, 0)
–  0 i ev ˆ
j
2 a
38. A charged particle of mass m is moving with a speed u in a circle of radius r. If the magnetic field induction
at the centre is B, the charge on the particle is
(1)
4r 2B
 0u
(2)
2r 2B
 0u
(3)
mu
Br
(4)
2mu
Br
Sol. Answer (1)
⎛ 2r ⎞
The time period of the charge particle is T = ⎜⎝
⎟
u ⎠
Current corresponding to the charge particle is i =
Magnetic field at the centre is B =
qu
q
=
2r
T
 0 qu
4 r 2B
0 i
=
2  q =
0 u
2r
4 r
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
14
Moving Charges and Magnetism
Solutions of Assignment
39. In a region of space, both electric and magnetic field are present simultaneously in opposite directions. A
positively charged particle is projected with certain speed an angle  (< 90°) with magnetic field. It will move
in a
(1) Helical path of uniform pitch
(2) Helical path of increasing pitch
(3) Helical path of decreasing pitch
(4) Helical path, whose pitch first decreases and then increases
Sol. Answer (4)
B
The component of the velocity is perpendicular to the magnetic field and opposite
to the electric field.
v cos 
v

v sin 
 The component of the velocity opposite to the electric field starts decreasing,
attains zero and again starts increasing in the direction of electric field
E
 Pitch of the path first decreases and then increases.
40. An electron is fired parallel to uniform electric and uniform magnetic field acting simultaneously and in the same
direction. The electron
(1) Moves along a circular path
(2)
Moves along a parabolic path
(3) Loses K.E.
(4)
Gains K.E.
Sol. Answer (3)
[Magnetic Moment]
41. Due to the flow of current in a circular loop of radius R, the magnetic induction produced at the centre of
the loop is B. The magnetic moment of the loop is (0 = permeability constant)
BR 3
20
(1)
(2)
2BR 3
0
(3)
BR 2
20
(4)
2BR 2
0
Sol. Answer (2)
B
0i
2R
i
2BR
0
M = iR2 =
2B R 3
0
42. A straight wire carrying current i is turned into a circular loop. If the magnitude of magnetic moment associated
with it in M.K.S. unit is M, the length of wire will be
(1)
4
M
(2)
4M
i
(3)
4i
M
(4)
M
4i
Sol. Answer (2)
iR2 = M
R
M
i
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
15
2R = L
R
L
2
L

2
M
i
L2
M

i
4M 
i
43. A rectangular coil 20 cm × 20 cm has 100 turns and carries a current of 1 A. It is placed in a uniform magnetic
field B = 0.5 T with the direction of magnetic field parallel to the plane of the coil. The magnitude of the torque
required to hold this coil in this position is
(1) Zero
(2)
200 Nm
(3)
2 Nm
(4)
10 Nm
Sol. Answer (3)
 = N/AB sin
= 100 × 1 × 400 × 10–4 × 0.5 × sin90° = 2 Nm
44. A length l of a wire is bent to form a circular coil of some turns. A current i is then established in the coil
and it is placed in a uniform magnetic field B. The maximum torque that acts on the coil is
(1) iBl2
(2)
4iBl2
(3)
iBl 2
4
(4)
Zero
Sol. Answer (3)
n(2r) = l
  in
max 
=
l2
B
4 2n 2
il 2B
4
(for n = 1)
il 2B
4
45. A ring carrying current I lies in x-z plane as shown. A short magnetic dipole with dipole moment directed along
x-axis is fixed at origin. If the ring is free to move anywhere, then it will
y
M
z
x
I
(1) Rotate clockwise about z-axis
(2)
Rotate anticlockwise about z-axis
(3) Move along z-axis
(4)
Move along –y-axis
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
16
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (1)
The magnetic field at the centre of the ring is along the (+ve) y-axis and magnetic dipole moment of the short
magnet is along the x-axis.
 Torque acting on the short magnet is
 =
 m iˆ  B jˆ 
= m B Kˆ
As the short magnet is fixed, it will not rotate but the ring itself starts rotating about z-axis.
46. If plane of coil and uniform magnetic field B(4T) is same then torque on the current carrying coil is
I
R
I
(1) I(πR2)4
(2)
30°
I ( R 2 ).4
2
(3)
I R 2 .4
8
(4)
Zero
Sol. Answer (1)
47. Calculate the torque acting upon the following structure carrying current I due to the magnetic field B. Both the
magnetic field and the structure are in the plane of paper as shown
B
2R
30°
R
(1) Zero
(2)
5 2
R IB
4
(3)
5 2
R IB
2
(4)
5R2IB
Sol. Answer (3)
48. Two galvanometers A and B require 3 mA and 5 mA respectively to produce the same deflection of 10 division
then
(1) A is more sensitive than B
(2)
B is more sensitive than A
(3) A and B are equally sensitive
(4)
Sensitiveness of B is twice that of A
Sol. Answer (1)
The galvanometer which shows more deflection on small current is more sensitive.
49. The current sensitivity of a moving coil galvanometer can be increased by
(1) Increasing the magnetic field of the permanent magnet
(2) Increasing the area of the deflecting coil
(3) Increasing the number of turns in the coil
(4) All of these
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
17
Moving Charges and Magnetism
Sol. Answer (4)
 NAB

I
K
Current sensitivity
50. The restoring couple in the moving coil galvanometer is due to
(1) Current in the coil
(2)
Magnetic field of the magnet
(3) Material of the coil
(4)
Twist produced in the suspension wire
Sol. Answer (4)
Restoring torque is provided by twist produced in the suspension wire.
51. In an attempt to increases the current sensitivity of a moving coil galvanometer, it is found that its resistance
becomes double while the current sensitivity increases by 10%. The voltage sensitivity of the galvanometer
changes by
(1) –40%
(2)
–45%
(3)
55%
(4)
–55%
Sol. Answer (2)
Let the current sensitivity and voltage sensitivity are x and y
y=
x
R
1.1x
= 0.55y
2R
In second case x = 1.1x, R = 2R, y =
Change in voltage sensitivity
=
y – y
 100
y
= – 0.45 × 100
= – 45%
52. A magnetic field B  B0 jˆ exists in the region a < x < 2a and B  B0 jˆ , in the region 2a < x < 3a, where
B0 is a positive constant. A positive point charge moving with a velocity v  v 0iˆ , where v0 is a positive
constant, enters the magnetic field at x = a. The trajectory of the charge in this region can be like,
[IIT-JEE 2007]
B0
0
a
2a
3a
–B0
z
z
(1)
a
2a
3a
x
(2)
z
z
a
2a
3a
x
(3)
a
2a
3a
x
(4)
a
2a
3a
x
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
18
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (1)
Z
F
F
X
B0 ^j
–B0 j^
53. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise
direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into
the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
[IIT-JEE 2010]
(1) IBL
(2)
IBL

(3)
IBL
2
(4)
Sol. Answer (3)
From FBD of a small element on wire
d
T
⎛ d ⎞
2T sin ⎜ ⎟  idlB
⎝ 2⎠
 2T
idlB
d/2
IBL
4
d/2
T
d
 iRd B
2
 T = BiR
 T 
idlB
Bil
2
d2
d2
T
T
54. A long insulated copper wire is closely wound as a spiral of ‘N’ turns. The spiral has inner radius ‘a’ and outer
radius ‘b’. The spiral lies in the X-Y plane and a steady current ‘I’ flows through the wire. The Z-component
of the magnetic field at the center of the spiral is
[IIT-JEE 2011]
Y
I
(1)
0 NI
⎛b⎞
ln ⎜ ⎟
2(b  a ) ⎝ a ⎠
(2)
0 NI
⎛ba⎞
ln ⎜
2(b  a ) ⎝ b  a ⎟⎠
a
X
b
(3)
0 NI ⎛ b ⎞
ln
2b ⎜⎝ a ⎟⎠
(4)
0 NI ⎛ b  a ⎞
ln
2b ⎜⎝ b  a ⎟⎠
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
19
Moving Charges and Magnetism
Sol. Answer (1)
dB 
B
 NI dr
0dNI
 0
(b – a ) r
2r
 0NI ⎛ b ⎞
ln ⎜ ⎟
(b – a ) ⎝ a ⎠
55. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k̂ is coming out of the
plane of the paper. The magnetic moment of the current loop is
[IIT-JEE 2012]
y
I
a
x
a
(1) a2Ikˆ
(2)
⎛ ⎞ 2 ˆ
⎜⎝  1⎟⎠ a Ik
2
(3)
⎛ ⎞
 ⎜  1⎟ a 2Ikˆ
⎝2 ⎠
(4)
(2  1)a2Ikˆ
Sol. Answer (2)
M  IA
⎛ a⎞
⎜ ⎟
⎝ 2⎠
A  a2  4 
2
2
⎛ ⎞
 ⎜  1⎟ a2
⎝2 ⎠
56. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current
density along its length. The magnitude of the magnetic field, | B | as a function of the radial distance r from
the axis is best represented by
(1) | B |
(2)
R/2 R
r
[IIT-JEE 2012]
|B|
(3)
R/2 R
r
|B|
(4)
R/2 R
r
| B|
R/2 R
r
Sol. Answer (4)
R
⎧
⎪0, r  2
⎪
⎛ 2 R2 ⎞
⎪
⎪⎪  0 J ⎜⎝ r  4 ⎟⎠ R
B⎨
, r R
2r
2
⎪
2
⎪
⎛ 3R ⎞
⎪ 0J ⎜
⎝ 4 ⎟⎠
⎪
,r R
⎪⎩
2r
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
20
Moving Charges and Magnetism
Solutions of Assignment
57. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The
distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at
the center of the loop is
[JEE(Advanced)-2017]
I
4a
(1)
 0I ⎡
6 3 – 1⎤
⎦
4a ⎣
(2)
 0I ⎡
6 3  1⎤
⎦
4a ⎣
(3)
 0I ⎡
3 3 – 1⎤
⎦
4a ⎣
(4)
 0I ⎡
3 2 – 3⎤
⎦
4a ⎣
Sol. Answer (1)
Considering one section out of symmetric star shaped conducting wire loop.
From geometry :
O (Center of loop)
30
°
°
30
2a
0°
12
30°
I
a
Magnetic field at the center of the loop due to all 12 identical sections is additive in nature.
 Bnet = 12 ×
=
 0I
cos30  cos120
4a
 0I
 6  ⎡⎣ 3  1⎤⎦
4a
SECTION - B
Objective Type Questions (More than one option are correct)
1.
A thin wire carrying current i is bent to form a closed loop of one turn. The loop is placed in
y-z plane with centre at origin. If R is the radius of the loop, then
(1) At a point (x, 0) on x-axis Bx 
(3)
∫

–
B x dx   0 i
0 iR 2
2
2 3/2
2(R  x )

(2)
∫
0
(4)
∫
–
B x dx 

0i
2
B x dx  0
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
21
Sol. Answer (1, 2, 3)
Let us consider a length (dl) of the loop. The magnetic field at the point M due to this (dl) length is
dB =
 0 idl sin90
0 i
dl
=
2
2
2
4 R  x
4 R  x 2




From the symmetry of the figure it is clear that the component dB cos will cancel out
 Net field will be along the axis of the loop and is given by
BX =
=

0 i
2
4 R  x

2
0 i R
4 R 2  x 2

∫ Bx dx 
0
0 i R 2
2
∫ dl sin 


3
2
0
R
dx
2
 x2


0 i R
4 R 2  x 2
 2 R =

∫
=
3

0 i R 2
2 R2  x2

3

3
2
∫ dl
2
2
Let x = R tan 
Then dx = R sec2  d

 iR 2
∫0 Bx dx = 0 2

0 i
=
2

∫B
x
–
dx

2
∫

2
∫
0
R sec 2  d 
R 3 sec 3 
cos  d  =
0
=
0 i
2
=
0 i
 2 = 0 × i
2
0 i
2

⎡
⎤
0 i
⎢ sin 2 – sin0 ⎥ = 2
⎣
⎦
⎤
⎡ 
⎢ sin 2  sin 2 ⎥
⎣
⎦
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
22
2.
Moving Charges and Magnetism
Solutions of Assignment
The figure shows three long straight current carrying conductors. The straight parts are long and the circular
th
⎛3⎞
part in each case in ⎜ ⎟ of a complete circle. Let Ba, Bb and Bc represents the strength of field at the
⎝ 4⎠
centre O in the three cases, then
i
i
R
O
i
O
i
(a)
R
O
(b)
(c)
(1) Ba 
0i ⎛ 3 1 ⎞
⎜  ⎟
4R ⎝ 2  ⎠
(2)
Bb 
0i ⎛ 3 1 ⎞
⎜  ⎟
2R ⎝ 4  ⎠
(3) Ba 
0i ⎛ 3 1 ⎞
⎜  ⎟
4R ⎝ 2  ⎠
(4)
Bc 
3 0 i
8R
Sol. Answer (1, 2, 4)
Ba =
=
Bb =

⎡
⎤
⎢cos 2 – cos  ⎥
⎣
⎦
0 i ⎡
3 0 i 2 0 i
3

=
8R ⎢⎣
8R
8R
0 i
4R
0 i ⎡ 3 1 ⎤
2⎤

=
4R ⎢⎣ 2  ⎥⎦
 ⎥⎦
 o i 3  0 i
⎤
⎡
⎢cos0 – cos 2 ⎥ – 4R 2  4R
⎣
⎦
=
0 i
 i ⎛ 3 ⎞
 i
– 0 ⎜ ⎟  0
⎝
⎠
4 R 4R 2
4R
=
 i
2 0 i
 i ⎛ 3 ⎞
– o ⎜ ⎟ = 0
4R
4R 4 R ⎝ 2 ⎠
=
0 i ⎡ 1 3 ⎤
–
out of the page
2R ⎢⎣  4 ⎥⎦
=
0 i ⎡ 3 1 ⎤
–
into the page
2R ⎢⎣ 4  ⎥⎦
BC =
=
3.
 0 i 3
 i
 0
4R 2
4 R
0 i
4 R

⎡
⎤
⎢cos 2 – cos  ⎥
⎣
⎦
3 ⎞
⎛
⎜⎝ 2 –
⎟
2⎠
 ⎤  0 i ⎛ 3 ⎞  0 i
⎡
⎢cos– cos 2 ⎥  4R ⎜⎝ 2 ⎟⎠ – 4 R
⎣
⎦

⎡
⎤
⎢cos 2 – cos  ⎥
⎣
⎦
0 i
 i ⎛ 3 ⎞  i
3 0 i
 0 ⎜ ⎟– 0 =
4R 4 R ⎝ 2 ⎠ 4 R
8R
Two thick wires and two thin wires, all of same material and same length form a square in the three different
ways P, Q and R as shown. The current flowing through the arrangement is i. Let BP, BQ and BR represent
the magnetic field at the centre of squares in the three cases, then
i
P
(1) BP = 0
(2)
BQ = 0
Q
R
(3)
BR = 0
(4)
BP  0
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
Sol. Answer (1, 3)
P
BQ  0, because current in upper and lower section is not same.
Hence net field at Q will not be zero.
i
2
i
2
BR = 0 because current in upper and lower section is same.
4.
i
2
i
2
The field at P is zero. Because due to upper part the magnetic field
is into the page and due to lower part the field is out of the page.
23
A wire carrying current I, bent in the form of a quarter circle is held at rest on a smooth table with two threads
as shown. A uniform magnetic field exists in the region directed into the plane select the correct alternatives
I
R
R
O
B
(1) Net force on wire is zero
(2)
Net magnetic force on wire  BIR 2
(3) Tension in the threads is BIR
(4)
Tension in the threads is BIR 2
Sol. Answer (1, 2, 3)
Since the wire is held in rest
45°
 Net force on the wire is zero
Net magnetic force on the wire is F  i l B = i
F
2
F

F
2

2R B
R
T
2
From the figure it is clear that
T=
5.
F
2
T
= iRB
The figure shows two infinite parallel sheets of current, with current per unit length 1 and 2. If BP, BQ and
BR represent the magnetic field at the points P, Q and R respectively, then
P
1
y-axis
Q
R
x-axis
2
0
(along x-axis)
2
(2)
BQ 
⎛ 1   2 ⎞
⎟  0 (along x-axis)
(3) BR  ⎜
⎝ 2 ⎠
(4)
⎛   2 ⎞
BQ  ⎜ 1
⎟  0 (along –x-axis)
⎝ 2 ⎠
(1) BP  (1   2 )
(1   2 )
 0 (along x-axis)
2
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
24
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (2, 3)
0 
,  is current per unit length
2
x
 
 
BQ = 0 1  0 2
2
2
 0 1  0  2

BR =
= BP
2
2
A current carrying wire of length l is suspended horizontally by a spring of force constant K as shown. The
system is in equilibrium. A magnetic field B is switched on into the plane of paper suddenly. Which of the
following statements is correct?
B
6.
K
i
L
(1) The rod oscillates simple harmonically
(2) The rod goes down by a distance
(3) The rod goes up by a distance
BIL
and comes to rest
K
BIL
and comes to rest
K
(4) The maximum displacement of the wire from equilibrium is
2BIL
K
Sol. Answer (1, 4)
When the rod is in equilibrium and magnetic field is switch off then
Kx = mg
…(1)
When the magnetic field is switch on then at the new equilibrium position
Kx = mg iLB
i.e., the distance moved by the rod is 2(x – x) =
…(2)
mg iLB mg

–
K
K
K
2 iLB
option (2) is wrong and option (4) is correct.
K
When the rod is at a distance x0 from the 2nd equilibrium position then net force on the particle is
=
K(x + x0) – mg – iLB = – ma
 Kx + kx0 – mg – iLB = –ma
 Kx + Kx0 – Kx = –ma

a–
K
x
m 0
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
i.e. a  x0
Moving Charges and Magnetism
25
Hence the motion of the wire will be S.H.M. Option (1) is correct
The rod will never go up from the first equilibrium position.
7.
In a region of crossed fields as shown, the strength of electric field is E and that of magnetic field is B. Three
positively charged particles with speeds V1, V2 and V3 are projected and their paths are shown. From this it
implies that
×
×
×
+ + + + + + + + + + +
V1
V2 V
3
– – – – – – – – – – –
×
×
×
(1) V1 
E
B
(2)
V2 
E
B
(3)
V3 
E
B
(4)
All of these
Sol. Answer (1, 2, 3, 4)
When electric field force qE and magnetic field force qvB balance each other then the particle will move in
a straight line.
 qE = q V2B

V2 
E
B
If electric field force is not balanced by the magnetic field force then it may be either Eq > qvB or Eq < gvB
If Eq > qvB then V3 
8.
E
E
, option (3) correct and if Eq < vBq then, V1  .
B
B
Two charges q1 and q2 having same magnitude of charge are moving parallel to each other and they enter into
a region of uniform magnetic field as shown. If they have same mass and the time spent by them in the
magnetic field are t1 and t2 respectively, then
q1


q2
B
(1) For q1 = ± q2, t1 = t2
(2)
For q1 = q2, t1 = t2
(3) For q1 > 0, q2 < 0, t1 < t2
(4)
For q1 < 0, q2 > 0, t1 < t2
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
26
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (2, 4)
V
× × × × ×
 ×× × × × ×
× × × ×
× × ×
× × ×
×
× ×

× × ×
2
× × ×
× × × ×

× ×
× × ×
V
× × ×
× ×
× × × × × × ×
× × × × × ×
×
×
×
×
2
×

V
×
×
×
Figure (1)
Figure (2)
Distance described by the positively charged particle in the magnetic field is
S = 2 ( – ) R
= 2 ( – )
mv
qB
time taken by the positive charge particle in the magnetic field is
t=

2   –  m V
qB
V
t 
2   –  m
qB
…(1)
For the negative charge particle path is shown in figure (2)
t=
2 mv
2 R
=
V qB
V
 t=
2 m
qB
…(2)
For q1 = q2, t1 = t2 option (2) is correct
For q1 > 0, q2 < 0, t1 > t2 option (3) is wrong
For q1 < 0, q2>0, t1 < t2 option (4) is correct
9.
A charged particle goes undeflected in a region containing electric and magnetic fields. It is possible that
(1) E | | B, V | | E
(2)
E is not parallel to B
(3) V | | B, E is not parallel to B
(4)
E | | B, but V is not parallel to B
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
27
Sol. Answer (1, 2)
If E , B and V all are in the same direction then the charge particle will move on a straight line with
acceleration a 
Eq
m
In this case force exerted by the magnetic field is zero.
If E is not parallel to B but the force exerted by E is balanced by the force exerted by B , then the charge
particle will again move with constant velocity.
10. A charged particle moves in a gravity free space without change in velocity which of the following
is/are possible?
(1) E = 0, B = 0
(2)
E = 0, B  0
E  0, B = 0
(3)
(4)
E  0, B  0
Sol. Answer (1, 2, 4)
Force of a charge particle = q(E  v  B )
11. The figure shows a loop of wire carrying a current i as shown. There exists a uniform magnetic field along xaxis given by B  B iˆ . If the length of each side is a, then
0
z
i
y
x
2ˆ
(1) Magnetic moment of the loop is M  ia i
(2)
Magnetic moment of the loop is M  3ia 2 kˆ
(3) Torque experience by the loop is zero
(4)
Torque experienced by the loop is – 3ia 2B0 ˆj
Sol. Answer (1, 3)
z
1
2
y
3
x
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
28
Moving Charges and Magnetism
Solutions of Assignment
The magnetic moment of the loop will be the resultant of the magnetic moment of three loop shown in figure.
M1  Ia 2 – kˆ , M2  Ia 2 – iˆ and M3  Ia 2 kˆ
 
 

Net magnetic moment will be
M  M1  M2  M3 = Ia2 –iˆ
 
…(1)
Torque acting on the loop is given by
  M  B = Ia2 –iˆ × Bo = 0
 
12. A current carrying loop of radius R and mass ‘m’ is placed inside a uniform magnetic field an shown
B
I
(1) Acceleration of the loop is zero
(2)
Acceleration of the loop is towards right
(3) Torque on the loop is zero
(4)
The loop will turn if current exceeds
Sol. Answer (1, 4)
13. A particle of mass m and charge q, moving with velocity V enters Region II normal to the boundary as shown
in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of
the Region II is l. Choose the correct choice(s).
[IIT-JEE 2008]
Region I
q
V
Region II Region III
X X X X
X X X X
X X X X
X X X X
X X X X
l
(1) The particle enters Region III only if its velocity V 
qlB
m
(2) The particle enters Region III only if its velocity V 
qlB
m
(3) Path length of the particle in Region II is maximum when velocity V 
qlB
m
(4) Time spend in Region II is same for any velocity V as long as the particle returns to Region I
Sol. Answer (1, 3, 4)
When radius r > l, particle will move to region III.
mV
q Bl
>lV>
qB
m
q Bl
when V =
, the particle moves in the biggest semicircle possible in region II.
m
m
in region II, provided particle returns to region I.
Time spent t =
qB
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
29
14. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite
region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?
[IIT-JEE 2011]
(1) They will never come out of the magnetic field region
(2) They will come out travelling along parallel paths
(3) They will come out at the same time
(4) They will come out at different times
Sol. Answer (2, 4)
mv 2
 qvB
r
 r=
mv
qB
p
e
me < mp
 re < rp
 te < tp
15. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and
magnetic fields E  E0 jˆ and B  B0 jˆ. At time t = 0, this charge has velocity v in the x-y plane, making an
angle  with the x-axis. Which of the following option(s) is/are correct for time t > 0?
[IIT-JEE 2012]
(1) If  = 0°, the charge moves in a circular path in the x-z plane
(2) If  = 0°, the charge undergoes helical motion with constant pitch along the y-axis
(3) If  = 10°, the charge undergoes helical motion with its pitch increasing with time, along the
y-axis
(4) If  = 90°, the charge undergoes linear but accelerated motion along the y-axis
Sol. Answer (3, 4)
For  = 10°, path is helical.
Since there is an electric field, pitch is increasing.
For  = 90°, path is straight line as B will not exert any force.
16. A particle of mass M and positive charge Q, moving with a constant velocity u1  4iˆ ms1 , enters a region of
uniform static magnetic field, normal to the x-y plane. The region of the magnetic field extends from x = 0 to
x = L for all values of y. After passing through this region, the particle emerges on the other side after 10
milliseconds with a velocity u2  2( 3iˆ  jˆ)ms1 . The correct statement(s) is/(are)
[JEE(Advanced)-2013]
(1) The direction of the magnetic field is –z direction
(2) The direction of the magnetic field is +z direction
50M
units
3Q
100M
units
(4) The magnitude of the magnetic field is
3Q
(3) The magnitude of the magnetic field
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
30
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (1, 3)
t
 M

 qB
Clearly  = 30° =
C

6
F
100M 50M
M


B
6q
3Q
6q  10  10 3
B must be in –z direction.
v
2 ˆj

2 3iˆ

x = 0 4iˆ
x =L
17. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed
coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady
current I. Consider a point P at a distance r from the common axis. The correct statement(s) is/(are)
[JEE(Advanced)-2013]
(1) In the region 0 < r < R, the magnetic field is non-zero
(2) In the region R < r < 2R, the magnetic field is along the common axis
(3) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis
(4) In the region r > 2R, the magnetic field is non-zero
I
Sol. Answer (1, 4)
I
Assuming BC for cylinder and BS for solenoid
R
(1) R > r > 0
BS = 0ni > 0
(Correct)
(2) 2R > r > R
B
BS2  BC2 not along the axis of cylinder, hence, wrong.
(3) Wrong, not in the plane of circle.
(4) r > 2R, B = BC
18. A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform magnetic
field B . If F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s)
is(are)
[JEE(Advanced)-2015]
y
R
R
I
L
(1) If B is along ẑ , F  (L  R )
(3) If B is along yˆ, F  (L  R )
/6
R
/4
R
x
L
(2) If B is along xˆ, F  0
(4) If B is along zˆ, F  0
Sol. Answer (1, 2, 3)
For uniform field, F  i dl  B becomes F  I 2(L  R )iˆ  B
∫
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
31
3R
(region 2 in the figure) pointing
2
normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters
region 2 from region 1 at point P1 (y = –R). Which of the following option(s) is/are correct?
19. A uniform magnetic field B exists in the region between x = 0 and x 
[JEE(Advanced)-2017]
Region 1
y
O
+Q
P1
(y = –R)
×
×
×
×
×
×
×
×
×
Region 2
×
× B
×
×
×
×
×
×
×
Region 3
×
×
×
×
×
× P2
×
×
×
x
3R/2
(1) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the
change in its linear momentum between point P1 and the farthest point from y-axis is p / 2
(2) For B 
8 p
, the particle will enter region 3 through the point P2 on x-axis
13 QR
(3) For B 
2 p
, the particle will re-enter region 1
3 QR
(4) For a fixed B, particles of same charge Q and same velocity v, the distance between the point P1 and
the point of re-entry into region 1 is inversely proportional to the mass of the particle
Sol. Answer (2, 3)
The particle will follow circular trajectory inside the magnetic field region. The magnetic field cannot change
the magnitude of velocity and momentum.
For longest possible path, the radius of circular motion can be
3R
.
2
×
O
P2
P1
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
32
Moving Charges and Magnetism
Solutions of Assignment
At farthest point from y-axis, the momentum is directed upwards.
  p  2p
The radius and hence separation between p1 and re-entry point is proportional to m, if Q, v, B are same.
The particle will return to region only if it completes the half circle.
r
3R
2
mV 3R

B
2
p
3R

QB
2
r

2p
B
3QR
O
8p
p
13R
If B 
;r 

13QR
8
QB
R
r-r cos
It passes through point P2 if r – r cos = R
P2
P1

R
12
sin   2 
r
13
3
13R ⎛
5 ⎞
1–
R
8 ⎜⎝ 13 ⎟⎠
R=R
20. Two infinitely long straight wires lie in the xy-plane along the lines x = ±R. The wire located at
x = +R carries a constant current I 1 and the wire located at x = –R carries a constant current I 2 .
A circular loop of radius R is suspended with its centre at  0, 0, 3R  and in a plane parallel to the xy-plane.
This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in
the wire is taken to be positive
magnetic field B is(are) true?
(1) If I1 = I2, then B cannot be
(2) If I1 > 0 and I2 < 0, then B
(3) If I1 < 0 and I2 > 0, then B
if it is in the + ĵ direction. Which of the following statements regarding the
[JEE(Advanced)-2018]
equal to zero at the origin (0, 0, 0)
can be equal to zero at the origin (0, 0, 0)
can be equal to zero at the origin (0, 0, 0)
⎛  I⎞
(4) If I1 = I2, then the z-component of the magnetic field at the centre of the loop is ⎜ – 0 ⎟
⎝ 2R ⎠
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
33
Sol. Answer (1, 2, 4)
Magnetic field due to loop at origin

0I  R 2
2.8R
 kˆ   16 RI  kˆ 
0
3
Magnetic field at origin due to wires
 I ⎞
⎛ I
 ⎜ 0 1  0 2 ⎟ kˆ
⎝ 2R 2R ⎠
 I
(1) If I1 = I2 then B0  0 kˆ
16R
 
(2) It can be zero if I1 > 0, I2 < 0
B2
B1
 
(4) BLoop
I
z
SECTION - C
Linked Comprehension Type Questions
Comprehension-I
A moving coil galvanometer consists of a coil of N turns are area A suspended by a thin phosphor bronze strip in
radial magnetic field B. The moment of inertia of the coil about the axis of rotation is I and c is the torsional constant
of the phosphor bronze strip. When a current i is passed through the coil, it deffects through an angle .
1.
Magnitude of the torque experienced by the coil is independent of
(1) N
(2)
B
(3)
i
(4)
I
Sol. Answer (4)
 = iNAB = kQ
2.
The current sensitivity of the galvanometer is increases if
(1) N,A and B are increased and c is decreased
(2)
N and A are increased and B and c are decreased
(3) N,B and c are increased and A is decreased
(4)
N,A,B and c are all increased
Sol. Answer (1)
As kQ = NABi
⇒
Q NAB

i
k
Which is current sensitively hence (1)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
34
3.
Moving Charges and Magnetism
Solutions of Assignment
When a charge Q is almost instantly through the coil the angular speed  acquired by the coil is
(1)
NAB
QI
BAQ
NI
(2)
(3)
NABQ
I
(4)
NAQI
B
Sol. Answer (3)
If  is the angular speed acquired by the coil. When a charge Q is passed through it for a very short time
t, then

angular momentum I 

t
time interval
Or I = t = NABiQ = NABQ
⇒ 
4.
NABQ
I
In previous question, maximum angular deflection of the coil is
(1) max  
I
C
(2)
max 
1
I
C
(3)
max  I

C
(4)
max   IC
Sol. Answer (1)
By conservation of energy, we have
1 2 1 2
I   kQmax
2
2
Which gives Qmax = 
I
c
Comprehension-II
 0 i1i 2
. The force is attractive when the current
2r
is in same direction and repulsive, when they are in opposite directions. The force between the wires depends on
the distance between them. An arrangement of two parallel wires is shown. We can determine the equilibrium
position. Then we displace upper wire by a small distance, keeping lower wire fixed. If the wire returns to or tries
to return to its equilibrium position, its equilibrium is stable. We can thus show that upper wire can execute linear
simple harmonic motion or not. The length of wire AB is large as compared to separation between the wires.
The force per unit length between two parallel current carrying wires 
D (i2)
C
Free
A
1.
h
B (i1)
Fixed
If wire CD is in equilibrium position then which of the following may represent the directions of current in the
wires
(1) In AB, A  B and in CD, C  D
(2)
In AB, A  B and in CD, D  C
(3) In AB, B  A and in CD, D  C
(4)
Any of the above is suitable
Sol. Answer (2)
If the current in two parallel wires are in opposite direction then they repel each other.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
2.
Moving Charges and Magnetism
35
If  is mass per unit length of wire CD, then the equilibrium separation h is given by
(1) h 
 0 i1i 2
2g
(2)
h
2 0 i1i 2
2g
(3)
h
2g
 0 i1i 2
(4)
h
4g
 0 i1i 2
Sol. Answer (1)
 0 i1i 2
 g
2h
h
3.
0i1i 2
2g
If wire CD is displaced upward to increase the separation by dh, the magnitude of net force per unit length
acting on the wire CD becomes
 0 i1i 2
(1) 2 (h  dh )
(2)
 0 i1i 2
(3)
2 h 2
 0 i1i 2 dh
2
(4)
 gdh
h
Sol. Answer (4)
At equilibrium
 0 i1i 2
 g
2h
After displacing it dh upwards net force  g 
 g 
0 i1i 2
2(h  dh)
g
gdh

dh ⎞
h
⎛
⎜⎝ 1  h ⎟⎠
Comprehension-III
The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the
plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying
he same current I. The current in the loop is in the counterclockwise direction if seen from above.
[JEE(Advanced)-2014]
Q
Wire 1
P
1.
d
d
a
S
Wire 2
R
When d  a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop
is zero at a height h above the loop. In that case
(1) Current in wire 1 and wire 2 is the direction PQ and RS, respectively and h  a
(2) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and h  a
(3) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and h  1.2a
(4) Current in wire 1 and wire 2 is the direction PQ and RS, respectively and h  1.2a
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
36
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (3)
 0 ia 2
2(a 2 
a2
2

3
h2 )2
a h
2

2   0 ia
2 ( a 2  h 2 )
2a

a4
4a2
 2
2
a h

2
10a2  4a2  4h2
6a2  4h2
3a2
 h2
2
h  1.2a
2.
Consider d a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown
in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position
will be (assume that the net field due to the wires is constant over the loop)
(1)
 0I 2a 2
d
(2)
 0I 2a 2
2d
3 0 I 2 a 2
d
(3)
(4)
3 0 I 2 a 2
2d
Sol. Answer (2)
B
 0I
2
2d

 0I  2
1  I 2a 2
 I  a 2   0
2d
2
2d
Comprehension-IV
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure.
The length, width and thickness of the strip are l, w and d, respectively.
A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers
experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS
and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction
is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current
is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.
[JEE(Advanced)-2015]
l
y
K
I
S
w
d
P
R
I
x
z
M
Q
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
1.
Moving Charges and Magnetism
37
Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are
w1 and w2 and thicknesses are d1 and d2 respectively. Two points K and M are symmetrically located on the
opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M
in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength
B, the correct statement(s) is(are)
(1) If w1 = w2 and d1 = 2d2, then V2 = 2V1
(2) If w1 = w2 and d1 = 2d2, then V2 = V1
(3) If w1 = 2w2 and d1 = d2, then V2 = 2V1
(4) If w1 = 2w2 and d1 = d2, then V2 = V1
Sol. Answer (1, 4)
qE = qvd B
Also, E =
... (i)
V
V
⇒
 vd B
w
w
 V = (vd) B(w)
Also, I = ne Avd
I
I
 vd = neA  n e d  w
⎛ I ⎞
BI
⎟ Bw 
 V ⎜
ned
⎝ n e dw ⎠
V2 d1

V1 d 2
2.
Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with
carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed magnetic
field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K
and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct
option(s) is(are)
(1) If B1 = B2 and n1 = 2n2, then V2 = 2V1
(2)
If B1 = B2 and n1 = 2n2, then V2 = V1
(3) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
(4)
If B1 = 2B2 and n1 = n2, then V2 = V1
Sol. Answer (1, 3)
V2 B2 n1


V1 n2 B1
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
38
Moving Charges and Magnetism
Solutions of Assignment
SECTION - D
Assertion-Reason Type Questions
1.
STATEMENT-1 : A current carrying wire is placed parallel to magnetic field. The force on it due the magnetic
field is zero.
and
STATEMENT-2 : The net charge on current wire is zero.
Sol. Answer (2)
The force acting by the magnetic field on a current carrying wire is given by F  i l  B


Here wire is along the magnetic field
 l  B  lB sin0  0

F 0
Statement (1) is correct.
Current carrying wire remains always neutral. Statement (2) is correct.
2.
STATEMENT-1 : Path of a charged particle in a uniform and steady magnetic field cannot be parabolic.
and
STATEMENT-2 : Magnetic field cannot accelerate a charged particle.
Sol. Answer (3)
If only force is exerted by the magnetic field then the speed of the particle will not change. If the angle
between V and B is 0° or 180° then path will be straight line along B .

the path of the charge particle will be circular if the angle between V
If the angle between V and B is
2

and B is neither 0° nor 180° nor
then the path of the charge particle will be helical.
2
 Path will never be parabolic. Magnetic field can accelerates the charge particle as the particle moves on
curved path.
3.
STATEMENT-1 : The net magnetic flux through a spherical surface enclosing north pole of a bar magnet is
zero.
N
S
and
STATEMENT-2 : Magnetic field lines always form closed loops.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
39
Sol. Answer (1)
Since magnetic field forms a closed loop
S
N
 Net magnetic flux through the spherical surface will be zero, because for incoming line flux is (–ve) and
for outgoing lines the flux is (+ve) and total incoming lines is equal to total outgoing lines.
4.
STATEMENT-1 : A toroid produces uniform magnetic field.
and
STATEMENT-2 : A toroid is a simple solenoid bent into the shape of a hoop, so it is like an endless solenoid.
Sol. Answer (4)
The magnetic field inside the toroid may not has the same magnitude if its radius of cross-section is large.
Also magnetic field lines are closed loops. Hence statement (1) is wrong.
Toroid is also known as endless solenoid.
5.
STATEMENT-1 : When two particles having same charge and same de-Broglie wavelength enter in a region
of uniform transverse magnetic field, they follow circular paths of equal radius.
and
STATEMENT-2 : The radius of circular path depends on momentum and charge.
Sol. Answer (1)
The radius of the circular path in magnetic field is given by R =
as  

h

mV
R
h
qB 
mV =
mV
qB
h

since both the particle has same charge and wavelength 
 Radius will be same, Statement (1) is correct
As R  mv, Hence statement (2) is also correct.
6.
STATEMENT-1 : In the arrangement shown, the hoop carries a constant current. This hoop can remain stationary
under the effect of magnetic field of the bar magnet.
i
F = i (k × –j )
S
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
40
Moving Charges and Magnetism
Solutions of Assignment
and
STATEMENT-2 : When a magnetic dipole is placed in a non-uniform magnetic field, it experiences a force
opposite to direction of external magnetic field at its centre.
Sol. Answer (3)
The force acting on each small part of length (dl) of the loop by the magnetic field is dF  i dl  B .


Thus dF has two components one component is in the vertically upward direction which balances the weight
of the loop. The horizontal component of dF will cancel out.
Hence the loop can remain stationary. The force acting on a magnetic dipole is in different direction on its
two pole.
 Net force will be resultant of the forces on its pole and it may be in any direction.
 Statement (2) is wrong
7.
STATEMENT-1 : In a region of space, both electric and magnetic field act in same direction. When a charged
particle is projected parallel to fields, it moves undeviated.
and
STATEMENT-2 : Here, net force on the particle is not zero.
Sol. Answer (2)
If B , E and V are parallel then the deviation of charge particle is zero. But the acceleration of the charge
Eq
particle is given by a 
.
m
8.
STATEMENT-1 : Two parallel wires carrying current in same direction attract each other, whereas two proton
beams moving parallel to each other repel each other.
and
STATEMENT-2 : Wire carrying current is electrically neutral, therefore it is experiencing only magnetic attraction
while the electric force of repulsion between protons is more than the magnetic attraction.
Sol. Answer (1)
Between two wires carrying current in the same direction only magnetic force is the attractive force but
between two beams of proton, electrostatic repulsion is greater than the magnetic field force attraction.
 Statement (1) is correct
Statement (2) is also correct
9.
STATEMENT-1 : Electrostatic field lines are discontinuous at the surface of a conductor.
and
STATEMENT-2 : Magnetic field lines produced by permanent magnets are never discontinuous.
Sol. Answer (2)
Since the electric field inside the material of a metallic conductor is zero if it does not carry current.
 The electric field becomes discontinuous at the surface. But magnetic field lines always form a closed
loop. Therefore it can never be discontinuous.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
41
SECTION - E
Matrix-Match Type Questions
1.
Match the following
Column I
Column II
(A) Electric field
(p) Stationary charge
(B) Magnetic field
(q) Moving charge
(C) F  q E
(r)
Current carrying wire
(D) Magnetic force
Sol. Answer A(p, q), B(q, r), C(p, q), D(q, r)
(A) Electric field is produced by charge whether it is at rest or in motion.
(B) Magnetic field is produced by current; Charge in motion constitutes an electric current
 Magnetic field is also produced by charge in motion.
(C) F = qE is the force exerted by electric field on charge.
Electric field exerts force on charge whether it is at rest or in motion.
(D) Magnetic field exerts force on charge in motion and on current in a wire.
2.
Match the following
Column I
Column II
(A) Electric force of attraction
(p) Two stationary protons separated by a distance of 5 Å
(B) Magnetic force of attraction or repulsion
(q) Two protons moving parallel to each other
(C) Net force of attraction
(r)
An electron beam and proton beam moving parallel to
each other
(D) Net force of repulsion
Sol. Answer A(r), B(q, r), C(r), D(p, q)
(A) Electric force is repulsive between two protons and between two electrons whether it is at rest or in motion
but the electric force is attractive between an electron and a proton.
(B) Magnetic force of repulsion or attraction acts between two current and charge in motion constitutes an
electric current.
(C) Two stationary protons repel each-other when two protons move parallel to each-other then magnetic force
is attractive but the electric force is repulsive. Also electric force is stronger than the magnetic force.
Therefore net force will be repulsive.
(D) As discussed above.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
42
3.
Moving Charges and Magnetism
Solutions of Assignment
A charged particle of mass m enters in a uniform magnetic field B with speed v as shown. Angle of deflection
for the cases in column I is given in column II. Match the column I with column II.
v
m
d
Column II
Column I
(A) Charge of particle ‘q’ d 

mv
,  = 30°
qB
mv
,  = 60°
qB
(q) 30
mv
,  = 90°
2qB
(r) 60°
(B) Charge on particle ‘–q’ d 
(C) Charge on particle ‘q’ d 
(p) 0
(D) Charge on particle ‘–q’ d 
mv
,  = 90°
2qB
(s) 120°
(t) 180°
Sol. Answer A(r), B(r), C(q), D(q)
4.
Column I represents a charged particle being thrown in different combinations of magnetic and electric field.
Column II represents the path followed by the charged particle.
Column I
V
B
E
(A)
Column II
(p) Straight line
V
(B)
B
(q) Parabola
V
(C)
B
(r) Circle
V
(D)
(s) Helix
E
(t) Cycloid
Sol. Answer A(p), B(r), C(s), D(q)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
43
Answer Q. 5, Q. 6 and Q. 7 by appropriately matching the information given in the three columns
of the following table.
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity
v . A uniform electric field E and a uniform magnetic field B exist everywhere. The velocity v , electric field
E and magnetic field B are given in columns 1, 2 and 3, respectively. The quantities E0, B0 are positive in
magnitude.
Column 1
(I) Electron with v  2
(II) Electron with v 
E0 x
B0
E0 y
B0
(III) Proton with v  0
(IV) Proton with v  2
5.
E0 x
B0
(i)
Column 2
Column 3
E  E0 z
(P) B  B0 x
(ii) E  E0 y
(Q) B  B0 x
(iii) E  E0 x
(R) B  B0 y
(iv) E  E0 x
(S) B  B0 z
[JEE(Advanced)-2017]
In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along
y )?
(1) (IV) (ii) (S)
(2) (II) (iii) (Q)
(3) (III) (ii) (R)
(4) (III) (ii) (P)
Sol. Answer (3)
Proton will move in straight line along negative y-direction when
v  0 , E  E0 y and B  B0 y
6.
In which case will the particle move in a straight line with constant velocity?
(1) (II) (iii) (S)
(2) (III) (iii) (P)
(3) (IV) (i) (S)
(4) (III) (ii) (R)
Sol. Answer (1)
Electron will move in straight line with constant velocity if
E
v  0 y , E  E x , B  B z
0
0
B0
7.
In which case will the particle describe a helical path with axis along the positive z direction?
(1) (II) (ii) (R)
(2) (III) (iii) (P)
(3) (IV) (i) (S)
(4) (IV) (ii) (R)
Sol. Answer (3)
Proton will move in helical path with axis along positive z-direction
E0 x , E  E0 z and B  B0 z
If v  2
B0
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
44
Moving Charges and Magnetism
Solutions of Assignment
SECTION - F
Integer Answer Type Questions
1.
Figure shows sections of a long current carrying wires. At origin the current is divided in two equal parts. All
 0I x
 2  12 , then find out the value of x.
section lies in xy plane. If the net magnetic field at P is
8
y
I
I
45° ( 2, 0)
P
45°
x
2I
Sol. Answer (2)
Magnetic field due to section AB
B1 

0 I
 cos 45  cos0
4 1
0 2 ⎛ 1
⎞
1
4 ⎜⎝ 2 ⎟⎠
Due to section BC
B2 
0 ⎛ 1
⎞
1
4 ⎜⎝ 2 ⎟⎠
Due to section BD
B3 
0 1
 cos90  cos0
4 2
0 1
4 2
B = B1 + B2 + B3


0 ⎛
1
1 ⎞
 1
2 2
⎜
⎟
4 ⎝
2
2⎠

0
 2 2  2  1
4
B 
0
 2  12
4
B 
0 x
 2  12
8
x2
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
2.
Moving Charges and Magnetism
45
A spiral loop has inner radius 10 cm and outer radius 20 cm and carries current 1 A in each turn. The total
number of turn in loop is 100. A small rectangular loop of area 2 cm2 carries current 0.05 A is placed at the
centre of loop. If magnitude of potential energy of rectangular loop is
A=2cm²
x0 ln2
J, then find out the value of x.
400
I=1A
I=0.05A
10cm
20cm
Sol. Answer (2)
Magnetic field at centre of spiral loop
B


 0NI
b
ln
2  b  a a
 0  100  1
ln 2
2  10  10 2
A=2cm²
I=1 A
I=0.05A
0
 103  ln 2
2
10cm
N=100 time
20cm
 500 0 ln 2
Magnetic dipole moment
  IA  0.05  2  104  1  105 Am2


U   . B  0  103  ln2  10 5  0 ln2
2
200
x 0 ln 2
U
⇒x2
400
3.
A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x,
⎛ 0I ⎞
PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is k ⎜
⎟ , find the value
⎝ 48x ⎠
of k.
[IIT-JEE 2009]
Q
Sol. Answer (7)
BPR = 0
S
BPQ = 0
As 4x ×
PS 
12 x
5
3
 PS
5
3x
P
5x
37°
4x
R
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
46
Moving Charges and Magnetism

B

Solutions of Assignment
0
I

{sin53  sin37}
4 12 x / 5
 0 7I
⎧  I ⎫
0
5I ⎧ 4 3 ⎫
 7⎨ 0 ⎬

⎨  ⎬ 
x
4
12


4 12 x ⎩ 5 5 ⎭
⎩ 48x ⎭
 k = 7
4.
A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder
and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the
N
0 aJ, then the value of N is
[IIT-JEE 2012]
magnetic field at the point P is given by
12
P
a
O
2a
Sol. Answer (5)
B = B1 – B2
 0 Ja
(field due to complete cylinder)
2
Here, B1 is
2
⎛ a⎞
0J  ⎜ ⎟
⎝ 2⎠
 Ja
 0
B2 =
12
⎛ 3a ⎞
2 ⎜ ⎟
⎝ 2⎠
⇒ B=
5.
5 0aJ
12
A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990  resistance, it
2n
 resistance, it becomes an ammeter
can be converted into a voltmeter of range 0-30 V. If connected to a
249
of range 0-1.5 A. The value of n is
[JEE(Advanced)-2014]
Sol. Answer (5)
Ig = 0.006 A, R = 4990 
Rg
V = 30
R
V = Ig(Rg + R)
30 = 0.006(Rg + 4990)
30  1000
 Rg  4990
6
Rg = 10 
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Ig
I
II–
Moving Charges and Magnetism
47
Rg
g
⎛ 2n ⎞
⎜⎝
⎟
249 ⎠
0.006  10  (1.5  0.006) 
6.
2n 
0.06  249
 10
1.494

n5
2n
0.06
2n


249
1.5
249
Two parallel wires in the plane of the paper are distance X0 apart. A point charge is moving with speed u
between the wires in the same plane at a distance X1 from one of the wires. When the wires carry current
of magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast,
if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is
R2. If
R1
X0
is
 3 , the value of
R2
X1
[JEE(Advanced)-2014]
Sol. Answer (3)
B1 
 0I
 0I

2X1 2( X 0  X1)

⎤
0I ⎡ 1
1

⎢
⎥
2 ⎣ X 1 X 0  X 1 ⎦

 0I ⎡ X 0  X 1  X 1 ⎤
⎢
⎥
2 ⎣ X1( X 0  X1 ) ⎦
B1 
 0I ⎡ X 0  2 X 1 ⎤
⎢
⎥
2 ⎣ X1 ( X 0  X1 ) ⎦
B2 
⎤
 0I ⎡ 1
1

⎢
⎥
2 ⎣ X 1 X 0  X 1 ⎦

U
X1
X0 – X1
I
X0
⎤
 0I ⎡
X0
⎢
⎥
2 ⎣ X 1 ( X 0  X 1 ) ⎦
X0
X1
X0
R1 B2
3




3
32
R2 B1 X 0  2 X1 X 0
2
X1
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
48
7.
Moving Charges and Magnetism
Solutions of Assignment
A moving coil galvanometer has 50 turns and each turn has an area 2 × 10–4 m2 . The magnetic field
produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is
10–4 N m rad–1. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates
by 0.2 rad. The resistance of the coil of the galvanometer is 50 . This galvanometer is to be converted into
an ammeter capable of measuring current in the range 0 – 1.0 A. For this purpose, a shunt resistance is to
be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is _______.
[JEE(Advanced)-2018]
Sol. Answer (5.56)
 = BANim = K
im 
K
10 –4  0.2

BAN 0.02  2  10 –4  50

0.2
 0.1A
2
0.1 × 50 = 0.9 S
 S  50   5.56 
9
8.
In the xy-plane, the region y > 0 has a uniform magnetic field B1kˆ and the region y < 0 has another uniform
magnetic field B2 kˆ . A positively charged particle is projected from the origin along the positive y-axis with speed
v0 =  ms–1 at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when
the particle crosses the x-axis from below for the first time. If B2 = 4B1, the average speed of the particle,
in ms–1, along the x-axis in the time interval T is _________.
[JEE(Advanced)-2018]
y
v0 =  ms–1
B1
x
B2
Sol. Answer (2.00)
y
The particle will follow the path as shown
2mv 2mv

qB
4qB
 2.00 m/s
Average speed =
m m

qB 4qB
2R
2r
x
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
49
SECTION - G
Multiple True-False Type Questions
1.
STATEMENT-1 : Magnetic force is always non-central while the electric force may be central.
STATEMENT-2 : Magnetic field can accelerate a charged particle.
STATEMENT-3 : Consider a long, straight wire of radius R, carrying a current distributed uniformly over its
cross-section. The magnitude of the magnetic field is maximum at surface of the wire.
(1) F F F
(2)
FTF
(3)
TFT
(4)
TTT
Sol. Answer (4)
Statement -1
Electric force may or may not be central. The force acting on a charge kept at point on perpendicular
bisector of a dipole is non-central
Statement -3
B is maximum at surface.
B
r=R
2.
r
STATEMENT-1 : A charged particle when projected perpendicularly to a uniform magnetic field, its velocity
is constant through out its motion.
STATEMENT-2 : A charged particle its acceleration is zero.
STATEMENT-3 : A charged particle when projected perpendicularly to a uniform magnetic field, a magnetic
force acts on it but that does not change particle’s speed.
(1) T T F
(2)
FFT
(3)
FFF
(4)
TFT
Sol. Answer (2)
Speed of charged particle is constant but direction changes continuously so it has centripetal acceleration.
3.
STATEMENT-1 : Magnetic field inside an ideal solonoid is uniform.
STATEMENT-2 : Magnetic field outside an ideal solonoid is zero.
STATEMENT-3 : Magnetic field at centre of an ideal solonoid is twice the magnetic field at the ends.
(1) T T T
(2)
TTF
(3)
FFF
(4)
FFT
Sol. Answer (1)
As the solenoid approaches ideal behavior i.e. length () >> radius (r). Magnetic field inside becomes uniform
and outside it is zero.
4.
STATEMENT-1 : If a positive charge is thrown parallel to a current carrying wire it will be attracted by the
wire.
STATEMENT-2 : If a negative charge is thrown antiparallel to a current carrying wire it will be repelled by the
wire.
STATEMENT-3 : A current carrying wire can apply to a force on a charge particle placed near it.
(1) F T T
(2)
FTF
(3)
TFF
(4)
TTF
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
50
Moving Charges and Magnetism
Solutions of Assignment
Sol. Answer (3)
F  q(V  B )
×
B
F
+
V
V
B
F
In both case it is attracted.
SECTION - H
Aakash Challengers Questions
1.
A long straight copper wire, of circular cross section, contains n conduction electrons per unit volume, each
of charge q. Show that the current I in the wire is given by
I = nqva2,
where v is the drift velocity and a is the radius of the wire.
At a radial distance r from the axis of the wire, what is the direction of the magnetic field B due to the current I?
Assuming that the magnitude of the field is B = μ0I/2r(r ≥ a), obtain an expression for the Lorentz force F on an
electron moving with the drift velocity at the surface of the wire.
If I = 10 A and a = 0.5 mm, calculate the magnitude of (a) the drift velocity and (b) the force, given that for
copper, n = 8.5 × 1028 m–3.
Sol. The current I in the wire is defined as the rate at which charge crosses a plane perpendicular to the direction
of the current. Mathematically,
I=
dQ dQ dx

,
dt
dx dt
i.e., the current is given by the product of the charge per unit length of the wire with the mean velocity of
the charge carriers. Clearly, dQ/dx = nqA where n is the number of charge carriers per unit volume, q is their
charge and A is the cross-sectional area of the wire. Thus I = nqva2 as required.
x
r
I
Using Fleming’s right-hand rule, the direction of the field at X is out of the page if the conventional current
(flow of positive charge) is to right as shown. The electron drift is to the left since the electrons are negatively
charged. The Lorentz force F on a particle of charge q moving with velocity v in a magnetic field B is
F  qv  B ,
So the Lorentz force on an electron acts inwards. The magnitude of the magnetic force on an electron at
the surface of the wire is
F  Bqv 
0l
0l 2
l
q

2a nq a2 2 2na3
If I = 10 A, a = 0.5 mm and n = 8.5 × 1028 m–3, we find v = 9.4 × 10–4 ms–1 and F = 6.0 × 10–25 N.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
2.
Moving Charges and Magnetism
51
Two coaxial plane coils, each of n turns of radius a, are separated by a distance a. Calculate the magnetic
field on the axis at the point midway between them when a current I flows in the same sense through each
coil.
Electrons in a colour television tube are accelerated through a potential difference of 25 kV and then deflected
by 45° in the magnetic field between the two coils described above. If a is 100 mm and the maximum current
available for the coils is 2 A, estimate the number of turns which the coils must have.
Sol. B 
 0 l 2na
4 r 2
sin  
 0 nla 2
2 a2  x 2 
32
The field at x = a/2 (midway between the coils) is thus
0 nl
 4
2a 5
32
,
So the field due to both coils together is
⎛ 4⎞
⎜⎝ ⎟⎠
5
3/2
 0 xi
.
a
The constant (4/5)3/2 has a value of 0.716.
If we assume that the magnetic field strength between the deflecting coils has a constant value B, given by
the expression we have just calculated, everywhere between the coils and is zero outside this region, the
electrons will describe a circular path in the magnetic field, with an angular velocity  = eB/m.
When the electrons have travelled a distance x = a (measured parallel to their original direction, starting from
the point at which they enter the field), they have been deflected through an angle of 45° = /4 radians as
shown in figure. The relationship between a and the radius r of the circular path is thus a = r sin(/4) = r / 2,
So,
r = a/ 2
/4
r
a
x
coil
coil
path of electrons
/4
a
Now if the electrons are travelling at speed v (which is unchanged by the magnetic field), the relationship
between v, r and  is
v = r,
so we must have
2aeB
m
Putting
v=
V=
2eV
,
m
where V is the accelerating voltage, and
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
52
Moving Charges and Magnetism
B  0.716
Solutions of Assignment
 0 nl
,
a
and rearranging to make n the subject, gives
n=
1.397 mV
 0I
e
We can now substitute I = 2A and V = 25 kV (the radius of the coils does not matter) to obtain n = 210.
Since our assumption that the field is zero outside the coils is not a very accurate one, we can state the
answer as n  200.
3.
State the Biot-Savart law which gives the magnetic field B at a distance r from a current element. Hence obtain
an expression for the magnetic field BQ due to a point charge Q moving with constant velocity v (assumed
non-relativistic).
Point charges Q and Q are constrained to move along the x-and y-axes, respectively, with the same uniform
speed v. At time t = 0 both charges are at the origin. At time t calculate the Lorentz force F on Q′ due to
the magnetic field of Q.
Sol. The Biot-Savart law may be written as
 I dl  r
0
,
dB 
4 r 3
where dB is the contribution to the magnetic field due to a current I flowing in an element of a (notional) wire
of length dI located at position r. If we write I = dQ/dt and v = dI/dt, we derive the expression for the field B
due to a charge Q moving at velocity v:
 0Q v  r
B
4 r 3
Since both charges start at the origin at t = 0, we may write the position r1 of the charge Q as v1t , and
the position r2 of the charge Q as v 2t . The vector displacement r of Q with respect to Q is then
v 2 – v1 t, and the magnetic field at Q due to Q is thus
v1  v1
The Lorentz force on a charge Q moving at velocity v 2 in a magnetic field B is Q v 2  B, so the force F
acting on Q is
 QQ  v  [v  (v – v )]
1
2
1
F 0 2 2
.
4 t
| v 2 – v1 |3
Since v1  v1 = 0, this can be simplified to
 0QQ  v 2  (v1  v 2 )
F = 4 t 2 | v – v |3 .
2
1
Taking v1 = (v, 0, 0) and v2 = (0, v, 0) gives
v1  v 2 = (0, 0, v2),
v 2  (v1  v 2 )  (v 3 , 0, 0) and
|v1 – v2|3 = 23/2 v3.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
53
The Lorentz force is thus
0QQ 
8 2t 2
Parallel to the x-axis.
4.
In a helium dilution refrigerator 3He and 4He are mixed in a special chamber to obtain extremely low
temperature. A Bainbridge mass spectrometer is used to measure the ratio of the two isotopes.
(a) If the spectrometer were used with 100 V cm–1 between the plates and a magnetic field of 0.2 T, what
would be the speed of an ion that can pass through the velocity filter?
(b) If the velocity-filter exit slit were 1 mm wide, could this machine resolve the two isotopes?
Sol. Figure shows the arrangement of a Bainbridge mass spectrometer.
Positive ions
E
B into
page
v
r
Magnetic field B only (into page)
(a) Within the velocity selector (the region containing both E and B fields), a positive ion of charge q and
speed v will experience a force Eq to the left as a result of the electric field, and a force Bqv to the right
as a result to the magnetic field. The ions that travel in a straight line (and so emerge from the velocity
selector) must therefore have v = E/B. Taking E = 104 Vm–1 and B = 0.2 T gives v = 5 × 104 ms–1.
(b) Within the resin containing only a magnetic field, the ions will be deflected into a circular arc of radius
r. Equating the magnetic force Bqv to the centripetal force mv2/r gives the radius of the arc as
r=
mv
Bq
and clearly the sideways deflexion of an ion is given by 2r. The deflexion of a singly charged 3He ion is
thus
2  3  1.66  10 –27  5  104
m =15.6 mm
0.2  1.60  10 –19
and the deflexion of a single charged 4He ion will be 4/3 times as large, or 20.8 mm. The separation
between the ends of the two trajectories will thus be 5.2 mm. The exit slit width of 1mm will broaden
the two trajectories into bands of width 1 mm, so they will not overlap and the isotopes will be resolved.
5.
Two long concentric cylindrical conductors of radii a and b (b < a) are maintained at a potential difference V
and carry equal and opposite currents I. Show that an electron with a particular velocity u parallel to the axis
may travel undeviated in the evacuated region between the conductors, and calculate u when a = 50 mm,
b = 2.0 mm, V = 50 V and I = 100 A.
It is also possible for the electron to travel in a helical path. By regarding such a path as the combination of
a circular motion perpendicular to the axis with a steady velocity parallel to the axis, indicate without detailed
mathematics how this comes about.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
54
Moving Charges and Magnetism
Solutions of Assignment
Sol. We know that the electric field E in the region between the conductors varies with the distance r from the
axis as

E = 2 r ,
0
where  is the charge per unit length on the inner cylinder, and we also know that the potential difference V
between the conductors is given by
V=

⎛ a⎞
ln ⎜ ⎟ .
2 0 ⎝ b ⎠
Combining these expressions to eliminate  gives
E=
V
⎛ a⎞
r ln ⎜ ⎟
⎝ b⎠
We know that the magnetic field B varies with r as
B=
 0I
.
2 r
Let us consider a cross-section of the system in figure.
If the inner cylinder is at a positive potential with respect to the outer
cylinder, the electric field E is directed radially outwards. If we
assume that current flows into the page along the inner cylinder and
out of the page along the outer cylinder, the magnetic field will be
directed clockwise as shown.
Consider an electron travelling with velocity u into the page, at a
radial distance r from the axis of the cylinders. It experiences an
electric force radially inwards, and a magnetic force Beu radially
outwards. In order for the electron to be undeviated, these forces
must balance so that u = E/B. Substituting our expressions for E
and B gives
b
a
r
B E
V
⎛ a⎞
r ln ⎜ ⎟
⎝ b⎠
2 V

u=
.
 0I
⎛ a⎞
 0I ln ⎜ ⎟
⎝ b⎠
2 r
This expression is independent of r, so the electron’s position does not matter. Taking the values given in
the problem, the velocity of an undeflected electron is 7.8 × 105 ms–1.
If the electron is to describe a helical path about the axis, the net force acting upon it must be of constant
magnitude, and directed radially inwards (and this is clearly possible given the orientations of the fields). This
will provide the necessary centripetal acceleration for the circular component of the motion, while the fact that
the force has no component parallel to the axis will give the electron a steady component of velocity parallel
to the axis.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
6.
Moving Charges and Magnetism
55
Figure shows a long straight conductor carrying current i. A wooden disc of mass m and radius R is placed
near it such that conductor lies in the plane of the disc. The distance between the wire and centre of disc is
r ( r > R). A wire of length R and carrying current i0 is posted on the disc such that the wire is perpendicular
to the long straight conductor. The disc is hinged at the centre. Neglecting mass of conducting wire, find
i0
i
Hinge
r
R
(a) Angular acceleration of the disc at the instant.
(b) Reaction at the hinge at the instant
Sol. Let us consider a length (dx) of the wire at a distance x from the long straight conductor. Magnetic field at
this point is B 
0 i
2 x
Force acting on the length (dx) of the wire is dF 
0 i i0
dx
2 x
This force dF will be in the direction upward parallel to the current i.
 The disc will starts rotating in anticlockwise direction about its center in its plane.
Torque due to force dF is
d = (x – r)
0 i i0
dx
2 x
 0 i i0
Total initial torque = 2
r R
∫
 x – r  dx
r
x
 =
0 i i0
0 i i0 ⎡ r  R
r R
x r
– r ln x r ⎤ =

2
⎣
⎦
2
 =


I
 =
0 i i0 ⎡
r  R ⎤
R – r ln
⎥
2 ⎢
r ⎦
 mR ⎣
⎡
r  R ⎤
⎢R – r ln
⎥
r ⎦
⎣
⎡
0 i i0
r  R ⎤
⎢R – r ln
⎥
1
r ⎦
2  mR 2 ⎣
2
(b) Since the C.M of the (disc + wire) is in rest

Net force on (wire + disc) = 0

N–F=0

N=F
0 i i0
= 2
r R
∫
r
dx
0 i i0
r
ln x r
x = 2 
R
=
0 i i0 ⎛ R  r ⎞
ln ⎜
⎝ r ⎟⎠
2
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
56
7.
Moving Charges and Magnetism
Solutions of Assignment
A charged particle of mass m and charge + q is projected on a rough horizontal x-y plane, where both electric
and magnetic fields are acting in the region and given by E  E 0 kˆ and B  B0 kˆ . The particle enters into
the field at (a0, 0, 0) with velocity V  V0 ˆj . The particle starts moving into a circular path on the plane. If
coefficient of friction between the particle and the plane is , then calculate
(a) The time when the particle will come to rest.
(b) The time when the particle will hit the centre.
(c) The length of path travelled by the particle, when it comes to rest.
Sol. (a) This is the length of the path traveled by the particle
At any instant of time the tangential acceleration of the particle is
y

at = –  mg  Eq 
m

dv

= –  mg  Eq 
dt
m
0

∫
v0

dv  –  mg  Eq  dt
m
 V0 =

O
t
t 
∫
0
x
(a0, 0, 0)
V0
z

 mg  Eq  t
m
mV0
  mg  Eq 
This is the time in which the particle come in rest
(b) As the speed of the particle decreases the radius of the circular path decreases and when velocity
becomes zero, radius also becomes zero. i.e. the particle will be at the centre.
(c) Here work done by the electric field force, magnetic field force and gravitational force all are equal to zero.
Work is done only by the frictional force
 –fs s = 0 –
1
mV0 2 [Work energy theorem]
2
Let the particle come in rest after traveling a distance ‘S’ then
k N S =
1
mV0 2
2
  [Eq + mg] s =
1
mV0 2
2
mV0 2
S


2  Eq  mg 
8.
Each of two long parallel wires carries a current I along the same direction. The wires are separated by a
distance 2l. Calculate maximum magnitude of magnetic induction in the symmetry plane of this system located
between the wires.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Solutions of Assignment
Moving Charges and Magnetism
Sol. Let the magnetic induction at a height ‘y’ from the plane of
the two wires is B.
Then B 
2
l +
2
2 0 i
2 l  y
2
2
cos 
l
This net field is in the horizontal direction. The component
of the field perpendicular to the plane will cancel out.
 B =
2 0 i
2 l 2  y 2

For B to be maximum



l2  y2
2 0 i
y
=
2 l 2  y 2

y

y
I
l
I

dB
0
dy

2
2
2
2 0 i ⎡ l  y – 2y ⎤
⎢
⎥ 0
l2  y2
2 ⎢
⎥⎦
⎣
y  l
 Bmaximum =
9.
y
57
0 i
2 l
On an inclined plane, two conducting rails are fixed along the line of greatest slope. A conducting rod AB is
placed on the rails so that it is horizontal. A vertically upward uniform magnetic field is present in the region.
A current i flows through AB from B to A. The mass of AB is m and the coefficient of friction between the
rod and rails is . Find the minimum and maximum value of B, so that the rod can stay in equilibrium. Assume
that  < tan.
B
A
L
i
B

Sol. When magnetic field is maximum then the rod AB tends to slide up the inclined plane. For equilibrium of
the rod.
s
mg sin + flim = iLB cos
co
N
iLB
 mg sin + N = iLB cos

sin

n
B
i
s
 mg sin +  [iLB sin + mg cos] = iLB cos
iL

os
mg im
c
fl
Mg
mg  sin    cos  

B


iL  cos  –  sin  
This is the maximum value of the required magnetic field
For minimum Magnetic field B
mg sin = iLB cos + flim
= iLB cos + N
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
58
Moving Charges and Magnetism
Solutions of Assignment
 mg sin = iLB cos  +  [mg cos + iL B sin ]
B

mg  sin  –  cos  
iL  cos    sin  
N
s
co f
lim
iLB

sin
g
m

s
co
g
M

sin
B
iL
This is the minimum value of required magnetic field.
10. In a moving coil galvanometer, a rectangular coil of N turns, area of cross-section A and moment of
inertia I is suspended in a radial field B through a spring
(a) If a current i0 produces a deflection of

in the coil, find the torsional constant of the spring.
4
(b) Find the maximum deflection suffered by the coil, if a charge Q is passed through it in a short interval of
time.
Sol. Let the torsional constant = k
Then k
4N i0 AB

k =
Now,


 Ni0 AB
4
∫  dt  L
∫ NI AB dt  L
 NAB
dQ
∫ dt .dt  L
 NAB Q = L
From the energy conservation
1 2
1 L2
k max 
2
2 I
 max 
L2

k
I = moment of inertia, k is the torsional constant
Q NAB 
NAB Q
= 2
i0 I
4Ni0 AB.I

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456