Emi & Ac
Emi & Ac
Emi & Ac
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
1
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
7. A charged particle of mass m and charge q
enters a magnetic field B with a velocity v FORCE ACTING ON A CURRENT
at an angle θ with the direction of B. The CARRYING CONDUCTOR
radius of the resulting path is
mv sin θ mv mv mv tan θ 10. Two thin long parallel wires separated by a
(A) qB (B) q B sin θ (C) qB (D) qB distance b carry currents of equal magnitude
i . The magnitude of the force per unit length
8. A particle of mass m and charge q is
exerted by one wire on the other is
projected into a region having a
perpendicular uniform magnetic field B. Find μ 0i 2 μ 0i 2 μ i μ 0i
the angle of deviation of the particle as it (A) (B) (C) 0 (D) 2
b2 2πb 2 πb πb
comes out of the magnetic field if width d of 11. A conducting loop carrying a clockwise
m current ‘i’ is placed in a uniform magnetic
the region is field pointing into the plane of the paper as
2qB
shown. the loop will have a tendency to*
(A) contract (B) expand
(C) move towards positive x - axis
(D) move towards negative x - axis
12. Two long conducting rods suspended by
means of two insulating threads as shown in
figure, are connected to a charged capacitor
through a switch which is initially open. At
the other end, they are connected by a loose
wire.
(a) 300 (b) 600 (c) 450 (d) 900
The capacitor has a charge Q and mass per
9. Length of two parallel conducting plates is
0.17 m and separation between them is 0.1 unit length of the rod is . The effective
resistance of the circuit after closing the
m. A uniform electric field of E 10 3 N / C
switch is R.
is present in the space between the plates. What is the velocity of each rod when the
A charged particle having charge 106 C and capacitor is discharged after closing the
mass 10 10 Kg is fired from middle of the switch. Assume that the displacement of rods
during the discharging time is negligible.
plates at an angle 300 with plane of plates
as shown in the figure there exist a uniform
magnitic field perpendicular to the plane of
figure in the space outside the plates what
should be the magnitude of the magnitic field
so that the particle grazes the plates at P
and Q (neglect the gravity)
0Q 2 0Q
(A) (B)
4dRC 4dRC
0Q 3 0Q 4
(C) (D)
4dRC 4dRC
(a)
2
3
4
0 R k , R 4
3
(b)
2
3
4
0R2 k , R 4
3
(c)
2
3
4
0R3 k , R 4
3
Q tan h
2 2
Q tan h
2 2
(d)
2
3
4
0R4 k , R 4
3
(a) n (b) n 22. A conductor carries a constant current I
4 2
along the closed path abcdefgha involving 8
Q tan 2 h 2 of the 12 edges of length l . The magnetic
(c) n (d) Q tan 2 h 2 n
3 dipole moment of the closed path is
4amp O
(a) l 2 I j (b) l 3 I j
B D
(c) l 4 I j (d) l 5 I j
A E 23. A long current wire is bent in the shapes
(A) 8 / 3 10 T 5
(B) 8 / 3 10 T
4 shown in figure. The circular portion has
radius R. The magnetic induction at the
(C) 2 105 T (D) 2 10 4 T
center of the circular segment is
20. An infinitely long conductor PQR is bent to
form a right angle. A current I flows through
PQR. The magnetic field due to this current
at the point M is H1 . Now another infinitely
long straight conductor QS is connected at
Q so that the current is I/2 in QR as well as
in QS, the current in PQ remaining
unchanged. The magnetic field at M is now
H2. The ratio H1/H2 is given by:
(A) 1/2 (B) 1 (C) 2/3 (D)2
4
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
0 I I (A) rα rp rd (B) rα rd rp
(a) 2k i , 0 k 1 i ,
4R 4R
(C) rα rd rp (D) rp rd rα
0 I 3
k j i 27. For a positively charged particle moving in
4R 2 x–y plane initially along the x–axis, there is
0I I a sudden change in its path due to the
(b) 2 k i , 0 k 1 i , presence of electric and/or magnetic fields
4 R 4R
beyond P. The curved path is shown in the
0 I 3 x–y plane and is found to be non–circular.
4 R k j 2 i Which one of the following combinations is
possible ?
0 I I
(c) 2k i , 0 k 1 i ,
4R 4 R
0 I 3
4 R k j 2 i
0I I
(d) 2 k i , 0 k 1 i ,
4 R 4 R
(A) E 0; B biˆ ckˆ (B) E ai;
ˆ B ckˆ aiˆ
0 I 3
4 R k j 2 i
(C) E 0; B cjˆ bkˆ (D) E ai;
ˆ B ckˆ bjˆ
1 qR 2 dB
mg
l dt
(b)
K
1 2qR 2 dB
mg
l dt
(c)
K
2mK 3mK
1 qR 2 dB (a) ( b)
(d) mg
2l dt
qd qd
K
mK 2 mK
(c) (d)
qd qd
6
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
33. A particle of charge per unit mass is
FORCE ACTING ON A CURRENT
released from origin with velocity v 0 i CARRYING CONDUCTOR
in a magnetic field B B0 k for
35. Figure shows four wires placed in the same
3 0 uniform magnetic field B and carrying the
x same current in which case force acting on
2 B0
the wire is minimum
× × × ×
3 0 × × × ×
and B0 for x . The × × × I ×
2 B0 (A) × I I × (B) × ×
× × × ×
x coordinate of the particle at time b× ×
a b×
×a
O l/2 O l/3
t would be :
3B0
× × × ×
× × × ×
× I × × I ×
× × × ×
3 0 3 (C) (D) × ×
0 t a× ×
b× b×
(a) × ×a
2 B0 2 B0 O 1.5l O l
3 0 36. An alpha particle moving with a velocity
(b) 0 t
2 B0 3B0 v1 2iˆ m/s in a uniform magnetic field
experiences a force F1 4e( ˆj kˆ) N .
3 0 0 When the particle moves with a velocity
(c) t
2 B0 2 3B0
v2 ˆj m/s, then the force experienced by it
3 0 0 t is F2 2e(iˆ kˆ) N . The magnetic induction
(d)
2 B0 2
B at that point is :
34. Consider the uniform magnetic field shown (A) iˆ ˆj kˆ
^ ^
(B) i j k
^
diamter than the other, are suspended along
a2 1 2
2
their common diameter as shown in figure.
(c) a i 2 j a k
Initially the planes of the rings are mutually 2 2
perpendicular. When a steady current is set (d) none of these
up in each of them 41. A uniformly charged ring of radius R is
rotated about its axis with constant linear
speed v of each of its particles. The ratio of
electric field to magnetic field at a point P
on the axis of the ring distant x R from
centre of ring is (c is speed of light)
(A) The two rings rotate into a common plane
(B) The inner ring oscillates about its initial
position
(C) The inner ring stays stationary while the outer
one moves into the plane of the inner ring
(D) The outer ring stays stationary while the inner c2 v2 v c
(a) (b) (c) (d)
one moves into the plane of the outer ring v c c v
39. The square loop ABCD, carrying a current
I, is placed in a uniform magnetic field B, as MAGNETIC FIELD DUE TO CURRENT
shown. The loop can rotate about the axis CARRYING CONDUCTOR
XX’. The plane of the loop makes an angle
900 with the direction of B. Through 42. A non–planar loop of conducting wire
carrying a current I is placed as shown in
what angle will the loop rotate by itself
the figure. Each of the straight sections of
before the torque on it becomes zero ?
the loop is of length 2a. The magnetic filed
due to this loop at the point P (a, 0, a) points
in the direction.
z
6 3
45. A coaxial cable is made-up of two conductors. 47. The magnetic moment of the loop shown in
The inner conductor is a solid cylinder and the adjacent figure is :(length of each side
is of radius R1 and the outer conductor is is a )
hollow of inner radius R2 and outer radius
R3 . The space between the conductors is
filled with air. The inner and outer
conductors are carrying currents of equal
magnitudes and in opposite directions. Then,
the variation of magnetic field with distance (a) 3a 2 I k (b) 2a 2 I k
from the axis is best plotted as : (c) 2a 2 I j (d) 3a 2 I j
9
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
51. In a certain region uniform electric field E
MOTION OF CHARGE PARTICLE IN and magnetic field B are present in mutually
MAGNETIC FIELD oppsite directions. At the instant t = 0, a
particle of mass m carrying a charge q is
48. In a region of space uniform electric field is given veloicty v 0 at angle , with the y–axis,
in the yz plane. The time after which the
present as E E 0 ˆj and uniform magnetic speed of the particle would be mini mum
field is present as B B0 .k̂ . An electron is is equal to
x
released from rest at origin. Which of the
following best represents the path followed
by electron after release. ( E0 & B0 are z
E B
positive constants)
y v0
y
y
(A)
(B) mv 0 mv 0 sin mv 0 cos 2m
x x
(A) (B) (C) (D)
x
qE qE qE qB
(C) 1 / 2 2 (D) 1 / 2 1 / 2 a v
55. A current carrying loop is placed in a uniform
magnetic field in four different orientations,
A B
I, II, III and IV, arrange them in the
decreasing order of Potential Energy I
b
n B
I II
n W
n
III B IV B (A) A
n
(B) B
(A) I > III > II > IV (B) I > II> III > IV
(C) No emf will be induced in the rod
(C) I > IV > II > III (D) III > IV > I > II
56. A metal ring is placed in a magnetic field , (D) None of the above
with its plane to the field. If the magnitude 59. A non-conducting disk of radius R is rotating
of the field begins to change, the ring will about its own axis with constant angular
experience velocity in a perpendicular uniform
(A) a net force (B) a torque about its axis magnetic field as shown in figure. The emf
induced between centre and rim of disk is
(C) a torque about a diameter
(D) a tension along its length
11
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
62. A slender rod of mass m and length L is
x x x pivoted about a horizontal axis through one
x
end and released from rest at an angle of
x
x x x 300 above the horizontal. The force exerted
x by the pivot on the rod at the instant when
x x x the rod passes through a horizontal position
x x x x is
B R 2 B R 2 300
(A) (B) B R (C) Zero (D)
2
2 3
C3 O O
C2
D
C
(A) C2 , C1 , C3 (B) C3 , C1 , C2 (A) experiences no net force
(B) experiences no torque
(C) C1 , C2 , C3 (D) None of these (C) turns clockwise as seen by an observer
located at the dot (‘ ’)
(D) turns anti–clockwise as seen by an observer
MULTIPLE ANSWER QUESTIONS
located at the dot (‘ ’)
66. A long straight conductor, carrying a current 70. Two long thin, parallel conductors carrying
i, is bent to form an almost complete circular equal currents in the same direction are
loop of radius R on it. the magnetic field at fixed parallel to the x-axis, one passing
the centre of the loop
through y=a and the other through y=-a. The
resultant magnetic field due to the two
(A) is directed into the paper conductors at any point is B. Which of the
(B) is directed out of the paper following are correct?
0i 1
(C) has magnitude 1 Z
2R
0i 1
(D) has magnitude 1 –a
2R O a Y
i
i
X
13
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
(A) B = 0, for all points on the x-axis 75. A charged particle of unit mass and unit
charge moves with velocity of
(B) At all points on the y-axis, excluding the
origin, B has only a z-component.
v 8i 6 j m / s in a magnetic field of
(C) At all points on the z-axis, excluding the B 2kT . Choose the correct alternative(s)
origin, B has only a y-component. (A) The path of the particle may be
(D) B cannot have an x-component x 2 y 2 4 x 21 0
71. A long straight wire carries a current along (B) The path of the particle may be
the x-axis. Consider the points x 2 y 2 25
A 0,1,0 , B 0,1,1 ,C 1,0,1 and D 1,1,1 . (C) The path of the particle may be y 2 z 2 25
Which of the following pairs of points will (D) The time period of the particle will be 3.14s.
have magnetic fields of the same magnitude 76. Two circular coils of radius 5 cm and 10 cm
? carry equal current of 2A. The coils have
(A) A and B (B) A and C 50 and 100 turns respectively and are placed
(C) B and C (D) B and D in such a way that their planes as well as
72. A long straight wire AB carries a current of their centres coincide. Magnitude of
4A. A proton P travels at 4 106 ms 1 magnetic field at the common centre of
coil is
parallel to the wire, 0.2 m from it and in a
direction opposite to the current as shown (A) 8 10 4 T if current in the coil are in same
in Fig. 9.8 Calculate the force which the sense
magnetic field of current exerts on the (B) 4 10 4 T if current in the coil are in opposite
proton. Also specify the direction of force. sense.
A) 2.56 1018 N towards left (C) Zero if currents in the coils are in opposite
B) 2.56 1018 N towards right sense.
(D) 8 10 4 T if current in the coil are in
C) 5.12 1018 N towards left
opposite sense.
D) 5.12 1018 N towards right 77. A region has uniform electric and magnetic
73. Current flows through a straight cylindrical fields along the positive x-direction. An
conductor of radius r. The current is electron is fired from the origin at an angle
distributed uniformly over its cross-section.
The magnetic field at a distance x from the
900 with the x-axis. It will
axis of the conductor has magnitude B. (A) move along a helical path of increasing pitch
(A) B 0 at the axis (B) move along a helical path of decreasing pitch
(B) B x for 0 x r initially
1 (C) return to the yz plane at some time
(C) B for x r (D) come to rest momentarily at some position
x
(D) B is maximum for x = r 78. A charged particle goes undeflected in a
74. A semicircular wire of radius r, carrying a region containing electric and magnetic field.
current i, is placed in a magnetic field of It is possible that
magnitude B. The force acting on it
(A) can never be zero (A) E||B, v || E (B) E is not parallel to B
(B) can have the maximum magnitude 2Bir
(C) v||B but E is not parallel to B
(C) can have the maximum magnitude Bir
(D) can have the maximum magnitude Bir (D) E || B but v is not parallel to E .
14
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
79. A charged particle moves in a uniform 82. If upper wire is slightly displaced from its
magnetic field. The velocity of the particle mean position and released it will perform
at some instant makes an acute angle with simple harmonic motion. As wire moves then
the magnetic field. The path of the particle total mechanical energy of wire
will be (A) Remains constant
(A) a circle (B) Decreases
(B) a helix with uniform pitch (C) We can’t say anything about mechanical
(C) a helix with non uniform radius energy in magnetic field
(D) a helix with uniform radius (D) Increases
0 I1 I 2 0 I1I 2 0 I 0 I1 I 2
(A) (B) (C) (D)
4 g 2g g 6 g B
81. The upper wire can be in equilibrium if
(A) Direction of current in both wires is same
(B) Direction of current in both wires is
opposite (A) Parallel to
(C) Equilibrium does not depend upon the (B) Parallel to axis of ring
direction of currents (C) perpendicular to axis and magnitetic field
(D) Two wires attract each other (D) Coming out of plane of paper
15
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
85. If ring is having a radius R charge Q 87. Total thermal energy dissipated in one cycle
uniformally distributed over it. Ring is is
rotated with a constant angular velocity .
B 2 l 4 B 2 l 4
Torque acting on the ring due to magnetic (A) (B)
force is 16 R 4R
QR 2B R 2qB 3B 2 l 4 3B 2 l 4
(A) (B) (C) (D)
2 2 8R 16 R
88. Average power produced in wire frame is
q R 2 B
(C) (D) None of the above 3B 2 2l 4 B 2 2l 4
2 (A) (B)
Passage - 3 16 R 16 R
A wire frame in the form of a part of circle
(sector) of radius l and resistance R is free to B 2 2l 4 3B 2 2l 4
(C) (D)
rotate about on axis passing through O and 8R 8R
perpendicular to plane of paper as shown in the Passage - 4
A conductor of mass m and length l is sliding
figure. The angle of the sector is and it is
4 smoothly an two vertical conducting rails AB and
rotating with constant angular velocity a CD as shown in figure. The top ends of two
shown. Above line PQ uniform magnetic field of conducting rails are joined by a capacitor of
magnitude B exists in the direction perpendicular capacitance C. The conductor is released from
to plane of paper. In region I field is outward rest when it is very close to AC i.e., y 0 . A
while in region II, field is inward.
uniform magnetic field B0 perpendicular to plane
of figure existing. Neglect the resistance of rails
and connecting wires. Take acceleration due to
gravity to be g.
x
x x x x x
A C
P O Q x x x x x x
l x x x l x x x
4
x x x x x x
x x x x x x
B Y
D
Based on above information, answer the following
questions: Based on above information answer the following
86. The thermal energy dissipated in wire frame questions:
when it moves from region I to region II, is 89. Mark the correct statement about the
B l
2 4
B l
2 4 motion of conductor
(A) (B) (A)It is falling down with constant acceleration g
16 R 4R
(B) It is falling down with constant acceleration
3B 2l 4 3B 2 l 4 but not equal to g
(C) (D)
8R 16 R (C) It is fallig down with increasing acceleration
(D) It is falling down with decreasig acceleration
16
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
90. Charge on the capacitor as a function of y is 93. A circular current carrying loop of 100 turns
given by of radius 10cm is placed in xy plane. A
2mgy uniform magnetic field B i k Tesla
(A) CB0l
m CB02l 2 is present in the region. If the current in loop
is 5A.
m CB02l 2 Column I
(B) CB0l (A) Magnitude and direction of magnetic
2mgy
(B) Magnitude and direction of torque
mgy (C)Magnitude and direction of net force
m CB02l 2
(D) Magnitude and direction (Tesla) of
(C)
CB0le Column II
(p) Zero moment (in A -m) of loop
mgy (q) 5 (in N.m.) on loop
2mgy m CB02l 2 (r) along +Z axis (n N) on loop.
(D) CB0 l e
m CB02l 2 (s) along -Y-axis magnetic field at centre
(t) X 103
91. Current in the circuit is 94 Which of the field (s) given in column – II
(A) constant will be produced by a loop mentioned
(B) increasing with time in column – I
(C) decreasing with time Column I
(A) Lorentz-force equation
(D) First increases, then reaches a maximum (B) Velocity-selector work on the principle
value, and then starts decreasing to attain a (C) Force on a charge in magnetic field.
constant value finally (D)Magnetic moment due to moving charge
Column II
MATRIX MATCHING TYPE QUESTIONS
(p) F q(v B)
92. Column I (q) F q(E
v B)
(A) A charged particle is at rest in magnetic
(r) v, E and B are mutually perpendicular
(B) A charged particle is moving
(C)A charged particle is moving along the q
(s) times of angular momentum of a loop
(D) A charged particle is moving at an 2m
Column II
ASSERTION & REASON TYPE QUESTIONS
(P) The path of the particle is circular field
(Q) The path of the particle is perpendcular to Codes :
the magnetic field helical (A) Statement – 1 is True, Statement – 2 is True;
(R) Particle moves in a straight directional of Statement – 2 is a correct explanation for
magnetic field line Statement – 1.
(B) Statement – 1 is True, Statement – 2 is True;
(S) No any force acts on the angle with the
Statement – 2 is NOT a correct explanation
direction of magnetic particle field for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False.
(D) Statement – 1 is False, Statement – 2 is True.
17
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
95. STATEMENT – 1 101. STATEMENT -1
A direct current flows through a metallic rod. A current carrying closed loop remains in
It produces magnetic field only outside the equilibrium in a uniform and constant
rod. because magnetic field parallel to its axis. Consider
STATEMENT – 2 forces only due to this magnetic field.
STATEMENT - 2
The charge carriers flow through whole of
Torque on a current carrying closed loop due
the cross–section of rod.
to a magnetic field is maximum when the
96. STATEMENT – 1 plane of the coil if parallel to the direction
In electric circuits, wires carrying currents of the magnetic field.
in opposite directions are often twisted 102. STATEMENT -1
together. A rectangular current loop is in an arbitrary
because orientation in an external uniform magnetic
STATEMENT – 2 field. No work is required to rotate the loop
If the wires are not twisted together, the about an axis perpendicular to its plane.
combination of wires forms a current loop. because
The magnetic field generated by the loop STATEMENT -2
might affect adjacent circuits or components. All positions represent the same level of
97. STATEMENT – 1 energy.
Out of galvanometer, ammeter and 103. STATEMENT -1
voltmeter, resistance of ammeter is lowest In a conductor, free electrons keep on
and resistance of voltmeter is highest moving but no magnetic force acts on a
STATEMENT – 2 conductor in a magnetic field
An ammeter is connected in series and a STATEMENT -2
voltmeter is connected is parallel, in a circuit Force on free electron due to magnetic field
always acts perpendular to its
98. STATEMENT – 1
direction of motion
We can increase the range of an ammeter, 104. STATEMENT -1
but cannot decrease the range A wire carrying current differs from wire
STATEMENT – 2 which carries no current
Minimum range of an ammeter is fixed STATEMENT -2
99. STATEMENT – 1 A magnetic field develops around wire
The mathematical statement of ampere’s law carrying current. No such field develops
when current is zero
B . dl oi is true when variable electric
field is not present in the medium. INTEGER TYPE QUESTIONS
STATEMENT – 2
A variable electric field produces magnetic 105. A straight conductor of length 60 cm and
field. mass 10 gm is suspended horizontally by a
100. STATEMENT – 1 pair of flexible leads in a magnetic field of
0.4 T as shown in figure. A current at 0.41 A
A charged particle is moving in a circle with
constant speed in uniform magnetic field. If should be passed through the conductor to
we increase the speed of a particle to twice, just remove the tension in the supporting
its acceleration will become four times. cable. The total tension in the leads if the
STATEMENT – 2 direction of current is reversed is found
In Circular parth of radius R with constant 1.96 10 y N . Find y ?
v2
speed v, acceleration is given by .
R
18
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
110. A particle of charge q and mass m is
projected from the origin with velocity
× × × × × × × ×
v i in a non uniform magnetic field
× × × × × × × ×
× × × × × × × ×
× × × × × × × × o
× × × × × × × ×
× × × × × × × ×
B B0 xk . Here 0 and B0 are positive
× × × ×60cm
× × × ×
19
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
+
SUBJECTIVE TYPE QUESTIONS
× × ×
em
113. A particle of mass m having a charge q enters × × ×
into a circular region of radius R with
× × ×
velocity v directed towards the centre. The
l
strength of magnetic field is B. Find the
deviation in the path of the particle. 117. Two particles, each having a mass m are
R
placed at a separation d in a uniform
magnetic field B as shown in figure. They
v
have opposite charges of equal magnitude
114. A loop, carrying a current I, lying in the plane q. At time t = 0, the particles are projected
of the paper, is in the field of a long straight towards each other, each with a speed v.
wire with current 0 (inwards) as shown in Suppose the Coulomb force between the
the figure. Find the torque acting on the loop. charges is negligible in comparision to
magnetic force.
b
I (A) Find the maximum value vm of the
a
2 projection speed so that the two particles do
I0
not collide.
115. A long straight metal rod has a very long (B) What would be the minimum and
hole of radius a drilled parallel to rod axis maximum separation between the particles
at a distance c from the axis of the rod as if v = vm/2?
shown in the figure. If the rod carries a (C) At what instant will a collision occur
current i. Find the magnetic field (a) on the between the particles if v = 2vm?
axis of the rod. (b) on the axis of the hole.
(D) Suppose v = 2vm and the collision
a
between the particle is completely inelastic.
c Describe the motion after the collision.
b
118. An insulated square frame ABCD of side a
is able to rotate about one of its sides taken
116. There exists a uniform and constant
as z -axis. A magnetic field B is present in
magnetic field of strenght B in the space
the region given by B B j . A small block
between the plates of a charged parallel 0
plate capacitor. The charge density on the of mass m and charge q movable along side
CB is initially near C, when the frame lies
plate is and length of the plate is . An
in the xz-plane. Now the frame is given an
electron is projected in the space between
angular velocity about the z-axis. The
the plates along the length of the plate. It is whole system lies in gravity-free space. If
found that velocity of the electron does not
after time t the block reaches, B find B0 in
change. Find the time taken by the electron
terms of t.
to come out of the capacitor. The figure
describes the situation. Ignore the gravity.
20
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
119. A rectangular loop PQRS made from a 122. A charged particle carrying charge q 1C
uniform wire has length a, width b and mass moves in uniform magnetic field with velocity
m. It is free to rotate about the arm PQ, which
v1 106 m / s at angle 450 with x-axis in the
remains hinged along a horizontal line taken
as the y-axis. Take the vertically upward x-y plane and experiences a force
direction as the z-axis. A uniform magnetic F1 5 2 mN along the negative z-axis.
field B 3i 4k B0 exists in the region.
When the same particle moves with velocity
v2 106 m / s along the z-axis it experiences
The loop is now released and is found to stay
a force F2 in y-direction. Find (a) magnitude
in the horizontal position in equilibrium.
and direction of the magnetic field, (ii) the
(a) What is the direction of the current I in
magnitude of the force F2 .
PQ?
123. Deuterons in a cyclotron describes a circle
(b) Find the magnetic force on the arm RS.
of radius 32.0 cm. Just before emerging from
(c) Find the expression for I in terms of B0 , the D’s. The frequency of the applied
a, b and m. alternating voltage is 10 MHz. Find, (a) the
magnetic flux density (i.e., the magnetic
field). (b) the energy and speed of the
deuterons upon emergence.
21
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
frame due to the magnetic field ? (b) Find LEVEL - V
the angle by which the frame rotates under KEY
the action of this torque in a short interval SINGLE ANSWER QUESTIONS
of time t , and the axis about which this 1. (C) 2. (C) 3. (A) 4. (A) 5. (C)
rotation occurs ( t , is so short any variation 6. (A) 7. (A) 8.(A) 9. (A) 10. (B)
in the torque during this interval may be 11. (B) 12. (A) 13. (A) 14. (C) 15. (D)
neglected). Given : the moment of inertia of 16. (B) 17. (A) 18. (A) 19. (C) 20. (C)
21. (A) 22. (A) 23. (A) 24. (D) 25. (A)
the frame about an axis through its centre
26. (A) 27. (B) 28. (C) 29. (A) 30. (D)
per pendicular to its plane is 4/3 M L 2. 31. (D) 32. (A) 33. (C) 34. (D) 35. (B)
36. (D) 37. (A) 38. (A) 39.(C) 40. (B)
41. (A) 42. (D) 43. (D) 44. (A) 45. (C)
46. (A) 47. (A) 48. (C) 49. (D) 50. (D)
51. (B) 52. (A) 53. (B) 54.(C) 55. (C)
56. (D) 57. (C) 58. (c) 59. (c) 60. (B)
61. (B) 62 (C) 63.(A) 64. (A) 65. (B)
MULTIPLE ANSWER QUESTIONS
66. (A),(C) 67. (B),(C) 68. (B)
126. A ring of radius R having uniformly 69. (A),(C) 70. (A),(C),(D)
71. (B),(D) 72. (B)
distributed charge Q is mounted on a rod
73. (A),(B),(C),(D) 74. (B)
suspended by two identical strings. The
75. (B),(D) 76. (A), (C)
tension in strings in equilibrium is T0. Now a
77. (B), (C) 78. (A), (B) 79. (B), (D)
vertical magnetic field is switched on and
ring is rotated at constant angular velocity COMPREHENSION TYPE QUESTIONS
80. (B) 81. (B) 82. (A)
. Find the maximum with which the ring
83. (C) 84. (C)85. (A)
can be rotated if the strings can withstand a
86. (B) 87. (C) 88. (A)
maximum tension of 3T0/2.
89. (B) 90. (A) 91. (A)
127. An electron in the ground state of hydrogen
atom is revolving in anticlockwise direction MATRIX MATCHING TYPE QUESTIONS
92. (A) - s, (B) - p, (C) - r,s (D) - q
in a circular orbit of radius R (as shown in
93. (A)-(q, r), (B)-(q, s), (C)-(p), (D)-(r, t)
figure). (i) Obtain an expression for the
94. (A)-(q), (B)-(r), (C)-(p), (D)-(s)
orbital magnetic moment of the electron (ii)
ASSERTION & REASON TYPE QUESTIONS
The atom is placed in a uniform magnetic
95. (D) 96. (A) 97. (B) 98. (A) 99. (A)
induction B such that the plane normal of 100. (D) 101.(B) 102. (A) 103. (D)
the electron orbit makes an angle of 30° with 104. (A)
the magnetic induction. Find the torque INTEGER TYPE QUESTIONS
experienced by the orbiting electron. 105. (1) 106. (1) 107 (6) 108. (3 )
109 (5) 110 (2) 111 (2) 112 (6)
^
n
KEY FOR QUESTIONS 113 - 127 ARE
PRESENT IN HINTS
22
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
line (that of electric field) . Since the force due to
magnetic field is zero therefore the charged
LEVEL- V particle will move in a straight line (a) is the correct
HINTS & SOLUTIONS option.
7. The component v sin is perpendicular to B
0i
1. (c) B ;B i .
2 d
Hence if direction is same
0i1 0i2
B0 90 106........(1)
2 d 2 d
If direction is reversed in wire (1) mv1
0i1 0i2 Hence by R ; Where is v sin . So
B '0 30 10 6........(2) qB
2 d 2 d
mv sin
0I2 R
From eq (1) and (2) 120 106 ; qB
d
8. (a) The radius of the circular orbit is given by
0 I1 I m
60 10 6 2 2 r
d I1 qB
. So the angle of deviation for
o I m r d 1
2. B , B due to all three arc will be d is sin
4 r 2 2qB 2 r 2
0 I 1 1
mutually perpendicular Bnet 3 sin 2 6 30 .
0
8R
o I 0 I qE
3. 0 y x a
2 y 2 x 9. Since,
m
4. Magnetic field well inside the solenoid is constant.
As we move out of solenoid along axis field
10 10 10 ms
6 3
2
23
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
0.17 e
2g 20 2d RC
u 1.98 ms 1 Impulse due to this force
Speed of particle at Q is, 0 Q 2 2t / RC
e dt
0 2d R 2 C 2
u cos 1.98cos 300 1.7 ms 1
It will graze at P if, Hence
PQ 2 (radius of circular path in magnetic field) 0 Q 2 2 t / RC 0Q 2
0 2 d R 2 C 2
e dt
4 dRC
m
0.1 2
qB 0Q 2
or .
2 1010 1.7 4dRC
B 3.4 103 T 3.4 mT . 13. The magnetic force acting on the rod is given by
0.1106
10. The magnetic field due to current in wire 1 in the
0 2i
region of wire 2 will be B1
4 b
Since wire 2 having current i is placed in a
magnetic field B1 , it will experience a force given
by F i lB1 sin 90
0
fql 2 dq 1
;i ; i fdq {since f }
3 dt t
q 2Q
dq dl dl dl 2 x dx
l
2 h
16. Area vector of both the loops are in opposite
direction. Hence, the magnetic moment d associated
17. Applying K.V.L. I1 I 2 14 / 3 A with this current would be.
Q
d 2 x x tan dx n
2
h
Q tan x2
3
dxn
h2
( n unit vector along the axis)
1 I1 4 I 2 6 Q tan 2 h 3
x dx n
B 0
2 R 2 5 5 h 2
0
0 14
2
Q tan 2 h 2
n.
4R 3 5 4
4 10 7 14 3 / 2 0 I
100 19. B .
2 15 15 2 2R
8.21 10 6 T . 0 I
20. B
18. Surface area of the cone ph 2 tan sec 4 z
hence, charge per unit surface area 21. Charge on the differential circular strip is
Q
h tan sec
2
Q dx
2 x tan
h tan sec
2
cos
2Q
2 x dx
h
The current dl due to the rotation this charge is
given by, dq 2R sin R d
or dq 2R 2 sin d
25
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
dq The total magnetic moment is
Then dl R 2 sin d
2 or l 2 I j .
1 2 3
(a) Magnetic field at the centre is 23. (a) Figure-(d) shows magnetic field lines of a long
0 dI R 2 sin 2
wire. Note the direction of magnetic induction at
dB points P, Q, R and S. From right-hand thumb
2R3 rule the magnetic fields at points P, Q, R and S
0R sin 3 d
2
lie along the unit vectors j , k , j and
R 3 k respectively..
2 0
B 0 sin d
/2
0 R sin 3 d
0
2
or B 0 R k
3
(b) Magnetic moment due to elementary ring,
d dI r 2
dI r 2
1 2 4
Resultant magnetic field, rp : rd : r : : 1:= 2 :1 r rp rd
1 1 2
0 I 3 Hence (a) is the correct option
B R B1 B 2 B 3 4R k j 2 i . 27. (b) The velocity at P is in the X-direction (given).
24. (d) Magnetic force can’t change speed of a
Let V mi . After P, the positively charged
charged particle particle gets deflected in the x-y plane toward -
25. When only magnetic field will act path will be y direction and the path is non-circular.
circular, but if an electric force acts opposite of Now,
magnetic force then velocity of particle would
be in straight line. Hence FM qvB and
F q V B
q mi ck ai for
FE qE option (b)
By q mci k mai i mcq j
mv 2 Since in option (b) , electric field is also present
FM FE qvB qE .................(1)
R E ai , therefore it will also expert a force in
q v the +X direction. The net result iof the two forces
So by equation (1) will be a non -circular path. Only option (b) fits
m RB
fir the above logic . For other option , we get
q v2 some other results
Also ..................(2)
m RE
E 200
qvB qE ; v 500m / s
B 0.40
Now by equation (2)
27
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
mv
28. v v cos 60; find r ; r=0.1m
qB
2 m
Hence t 2 10 7 s . Hence answer is
qB
(c) and path is helix
29. Conceptual (i.e.) radius r of circular path is r d
Now for a particle of charge q, mass m and kinetic
dB R 2 dB
30. E 2l R 2 ; E
dt 2l dt 2K
energy K, where velocity the
qE mg Kx
m
qR 2 dB mg m m 2 K
x ; radius r is r d
K 2l dt K qB qB m
1 qR 2 dB m 2K 2mK
x mg Hence, B .
K 2l dt
. qd m qd
m0
q 33. r 0
31. , path of the particle will be a helix of B0 q B0
m
time period, x 3
sin
2m 2 r 2
T
B0 q B0 600
T
tOA
6 3B0
T
The given time t B 2
0
28
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
36. B xiˆ yjˆ zkˆ , q q
Hence, i and A R 2
T 2R
F1 2e 2iˆ xiˆ yjˆ zkˆ 4e ˆj kˆ
E 1 1 c2 1
c
0 0 .
F2 2e ˆj xiˆ yjˆ zkˆ 2e ˆi kˆ B 00
42. (d) Magnetic induction at (a,0,a) due to loop in
B ˆi ˆj kˆ xy plane is in +k direction. Due to loop in yz
37. Force exerted by air on the rod
plane, the magnetic field will be in i direction.
L / 2 2 R 2 LR2
Due to both the loops , the direction of B will
Balancing torque about O,
1
NI R 2 B LR2
3L be
2
i k
4
0 I1 2 0 I 2
32 L2 43. B1 ; B2
300lBR 2R 2 2 R 2
4
R R 2
1 L2 2 But I1 I2
1 0.01A . A A
400 BR
38. Ring will experience the torque I1 I 2 2 B1 B2
39. In the position shown, AB is outside and CD is
But direction of B1 and B2 are opposite so net
inside the plane of the paper. The Ampere force
on AB acts into the paper. The torque on the magnetic field at the centre is zero
loop will be clockwise, as seen from above. The I
loop must rotate through an angle 90
0
44.
2R
I
B 0 i 0 j 0 k
2R
I
2R
before the plane of the loop becomes normal to I
B 0 3.
the direction of B and the torque becomes zero 2R
40.
Ia2 k Ia2 j Ia2 cos450 j Ia2 sin450 j 45.
Ia 2 Ia 2
2
i
2
Ia 2 j Ia 2 k
a2 a 1 2
2
2
i
2
j a k .
2
41. At point P, From Ampere’s law, the field at the axis is zero.
1 qx From x 0 to R1 , he field increases linearly as
E
4 0 R x 2 3/ 2 ;
2 the charge enclosed increases.
From x R1 to R2 and from x R2 to R3 , the
2iA field decreases hyperbolically but with different
B 0
4 R x 2 3/ 2
2
slopes as the media are different.
29
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
46. At a distance x consider a small element of width 56. When the field changes, a current flows along
dx. Magnetic moment of the small elements is : the ring. each section of the ring now experiences
an ampere force acting either outward or inward.
The ring will tend to expand or contract, and hence
experience a tension along its length.
57. The ampere force on the conductor is F Bil .
q
dM IA dq f x dxf x 2
2 The impulse given by this force to the conductor
l
is J Fdt Bl idt BlQ
l/2
qf q f l 2
M
l l / 2
x 2 dx
12 .
58. As the rod is of plastic so no free electrons are
present and hence no emf will develop in the rod
59. As the ring is non-conducting, no motion of
47. u 3a 2 I charge carriers occur and hence there is no
48. Magnetic force FM will act in +x direction . And induced emf
60. Let at any time t, velocity of rod be v, then emf
FE will act in -y direction. Hence path followed
by electron (it will not be a complete circular developed across its ends is e B0vl . Due to
path) . Hence option is (c) this induced emf, a current will establish in circuit
E B0vl
49. eV B eE VB E B V given by, I .
R
Also 500 Volt E 2 10 m .
3
B0
d 3
50. sin 60
R 2 F
v f v cos i v sin j , v i vi
51. (b) v u at
V
qE mv sin I
O v.sin t; t 0
m qE
Electric force will declerate the component
v0 sin and will not effect to v0 cos dv B02l 2
For rod, m IB0l v
o I1 I 2 3 R dx I1 I 2 dt R
2 R x
52. ln 3
2 v
dv
t
B 2l 2 B 2l 2t
53. Coneptual 0 dt 0
v0
v 0
mR v v0e mR
54. 1 IA; 2 I 2 A 61. The situation is as shown in the figure below.
55. U M .B MB cos
B
In case I, 1800.U MB l
In case II, 900.U 0
In case III, acute.U ve(less then+MB)
In case IV, obtuse U ve ; Required induced emf,
I III II IV
30
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
Bl 2 3mg
e We have R1 mar , and
2 4
50 106 4 9 3mg mg
or e 2.83 mV mg R2 m at R2
2 4 4
62. The angular velocity of rod about pivot when it So reaction force by pivot on rod,
passes through horizontal position is given by 10mg
R R12 R22 at an angle of
4
R 1
ar tan 1 2 tan 1 with the
R1 3
mg at horizontal.
63. f mg sin
L mL2 2 3g For no sliding, f f L
0
mg sin 30
2 3 2 2L
N
Radial acceleration of centre of mass (as centre f f
mg sin
L mg sin
of mass is moving in a circle of radius ) is mg cos
2
L 3g
given by ar
2
2 4
Torque about pivot, in horizontal position is,
mg sin s mg cos
tan s
mgL
3g or mg sin f ma
L 22
mg I mL 2L
2 g sin 2 g sin
3 and a
K2 3
Tangential acceleration of centre of mass, 2
1
R
L 3g
at mg sin
2 4 f
3
Draw the FBD of rod, at an instant when it passes
through the horizontal position and use Newton’s For pure rolling, f f 2
2nd law equation
mg sin
s mg cos
R2 3
tan 3 s
R1
tan
so 3
ar tan
mg at
31
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
64. At P , E B0t , so force, F 9 B0t in anti- Force on proton,
clockwise (direction form Lenz’s law)
Fm qvB 1.61019 4106 4106 2.561018 N
65. Due to changing B, flux linked with cir5cuit By right-hand rule for cross product, the force
changes and hence a current induces in the on the proton acts parallel to horizontal towards
circuit. right.
The direction of this induced current is in anti-
clockwise direction. I
73. x r, B.2x 0 x 2
r 2
In C3 , R2 is shorted, so no current flows through
74. To find the Ampere force on a conductor of any
R2 and hence energy dissipated is zero. shape, replace the conductor by an imaginary
Under the assumption that induced emf is same sraight conductor joining the two ends of the given
conductor.
for both C1 and C2 , induced current in C1
mv 1 10 2m 21
would be less than that in C2 , as a result, thermal 75. r 5m, T
qB 1 2 qB 1 2
energy dissipated through R2 in C1 would be path of particle will be circle having radius equal
less than that in C2 . to 5 m
o N1 I 0 N 2 I
So, C 3 C1 C 2 2. 76. B1
2r1 2r2
I I 77. The electric field exerts a force opposite to the
66. B o 0
2 R 2R component of motion along the x-axis.
78. If B & E & v are in same direction then particle
67. N I A n unit vector perpendicular to A
will be undeflected cause FM 0
68. Using Ampere’s law magnetic field is zero inside
the pipe. (b) is also possible is not parallel to B
69. Force on circular arc will be zero but loop will
If FM qvB and FE qE then v is to E
experience toque due to the force acting on the
straight conductors. and v is to B E is not parallel to B
70. Magnitude of magnetic field on x-axis from both
the wire are same and opposite direction.So, B mv sin
79. Path will be helical , r
on axis is zero. qB
71. The magnitude of the magnetic field depends only
0 I1 I 2
on the distance from the x-axis. Points A and C 80. g
are at distances of 1 unit each from the x-axis. 2 d
Points B and D are at distances of 2 unit each 81. Wire will repel if they carry current in opposite
direction.
from the x-axis.
82. Wire will perform SHM i.e., total energy will be
72. Magnetic field at P due to current-carrying wire
constant.
AB is
83. MB sin , Rotating ring will behave as a
2 I 107 2 4 magnetic pole
B 0. 4 106 T
4 a 0.2 84. is perpendicular M and B
By right-hand grip rule, the magnetic field at P is
directed perpendicular to the plane of the paper q R 2
and in inward direction. 85. Torque = B MB
2
32
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
86.-88 Charge on the capacitor at this instant is,
Time taken by wire loop to enter into the magnetic q Ce B0Cl v
T the current in the circuit at this instant is,
field (Region I) above PQ is, t ,
4 8 dq dv
during this time induced emf and hence current I B0Cl B0Cl a
dt dt
will appear through wire frame.
where a is the acceleration of the conductor at
Bl 2 this instant.
e1 induced emf Writing Newton’s 2nd law equation for conductor,
2
mg IB0l ma
Thermal energy dissipated in time t1
mg
e2 B 2 l 4 a ,
H1 1 t1 m CB02l 2
R 16 R
Similarly, thermal energy developed when the mgt 2
mgt y
v
frame moves from region I to region II is, and
m CB02l 2 2 m CB02l 2
B 2 l 4
H2 After solving above equations, we could have q
4R as a function of y as
Thermal energy will also develop when the loop
comes out from magnetic field and is given by,
I B0Cl a
As a is constant, I is also constant.
B 2 l 4 92 MATRIX [1]
H3
16 R Magnetic field will not interact with charge particle
So, total thermal energy developed in one at rest. Path will be circular if velocity is
revolution is, perpendicular to magnetic field.
93. MATRIX [2] M = N I A, MB sin
3B 2 l 4
H H1 H 2 H 3
8R 94. MATRIX [4]
F q E v B
Average power produced, 105. In order to just remove the tension in the leads,
the weight of the conductor must be balanced
H 3B 2 2l 4
Pav by magnetic force Bil
T 16 R or Bil = mg
P89-91:
mg 10 10 3 9.8
Let us say that in time t the conductor falls down i
Bl 0.4 0.6
by y and acquires a velocity v, then at this instant,
induced emf is i 0.41 A
In second case mg and iBl both will act down
e B0vl [with Q at higher potential]
ward. Therefore tension in the cable is 2 mg =
1.96 101 N
A C
y
106 r 2 R cos
IB0l 2
mg
90
2
B D
33
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
Magnetic force on the particle at origin is along
0 I Rd sin
dB positive y-direction, so its trajectory will be
4 r concave upwards. As the magnetic field is
nonuniform, trajectory is not circular. Magnetic
I /2
Rd sin90 force is always perpendicular to velocity. So its
2
2 o magnitude remains constant. Let at point P (x, y)
4 0 2R cos / 2
its velocity vector make an angle with positive
o I
B
4 R
ln
2 1 n 1
x-axis. Then the magnetic force FB will be at an
angle with the positive y-direction. So,
d
107. m mg qE
dt F d y B0 x q0 cos
ay B cos (or)
m dt m
mv 2 mv
qvB r
r qB FB Bq0 sin 900 or
dv qB dr
d y dx B0 qx
dt m dt 0 cos where
dx dt m
t 0
1 qB
dt
o
mg qE dr , t 20sec
a
dx
x 0 cos
dt
n
o I o dr d y B q
2 r xo Thus 0 x
dx m
x 0 I0 ln
108. <B> =
x
= x At maximum x-displacement, velocity is along
dr
xo
0
x x positive y-direction
0 B q xmax
mo xo x or
0
d y 0
m 0
x dx
x0 x 2 mvo
q o I 0 ; o qIo =30m
ln x xoe B q x
2
2 or 0 0 max
m 2
2 br 2
r
109. I j 2 x dx 0 bx
2 xdx 2m0
2 or xmax
B0 q
a 2
2 br o br 1 111. The magnetic force on the conductor is
B 2 r o 2 2 n 25
110.
F B I L B ILk B0 a r B0 IL a
External force is
F ext B0 IL a
Where a is a unit vector in tangential direction.
Therefore work done required to turn the
conductor in one full revolution is :
2
W B0 IL a rd a 2rB0 IL
0
34
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
The negative sign shows that field does work. Torque about point O (perpendicular distace)
The fact that work is done around a closed path r sin ; d 0 r sin dF
shows that the force is non-conservative in this
case. 0i0 ii b
112. We can visualize the rod to consist of differential r sin idB ridr sin ; 0 0 0 sin dr;
2 r 2 a
elements dQ, which constitute a series of
concentric current loops. The charge per unit 0ii0 sin b a
length of the rod .
0 ;
2
Q Net Torque
L ii b a sin 0ii0 b a sin
So the charge on a differential element of length 2 0 0
2
dl,
dq dl 115. (a) On the axis of rod due to i1 B is zero on
The current dl due to rotation of this charge is i
j
given by axis i2 j a and b a2
2
2
dq
dl dq dl 0i
2 / 2 2 Due to infinite wire we know that B
2 d
The magnetic moment of this differential current
loop, Hence due to i2 at the axis of rod
2 2 i
d dI l 2 dl l l dl 0 a2
2 0i2 0 j a b a
2 2 2
2
To find total magnetic moment, we integrate B0
2 C 2 C 2 C
L 2 L 3
2 l dl
6 B0
0i2 a 2
B0
0i a 2
2 C b 2 a 2 2 C b 2 a 2
0
B B k 3i 4k B0 3B0 j
v a sin i cos j z k
If B is to be in negative direction, must point
Force on the block, downward. So the current in the coil must be
from P to Q.
F qv B a sin i cos j z k B0 j
(b) Force acting on arm
Force in z-direction
RS I l B
aB0 k sin
I b j 3i 4k B IB b 3k 4i
0 0
Acceleration of block,
(c) In equilibrium,
aB0 sin
az k gravity B 0
m
mga mg
d z aB0 Hence 3 abI B0 or I 6 B b .
or sin tk 2
dt m 0
0 I1I 2 sin
dF dy
0 II 0 L md 0 II 0 L 2 a 2 y 2
F
2x dx 2 x [ along positive z-direction, using right-hand
0 II 0 L dx thumb rule]
d
2m x y
But sin
II L
d D
dx a2 y2
0 d 20 m0 x
a 0 I1 I 2 ydy
dF
2 0 II 0 L 2 a 2 y 2
D
log e 1
2 2m d a
0 I1 I 2 3
ydy
D
F
2 a y2
2
0 II 0 L log e 1 3a
d.
ydy 1
m
Since, a log e a 2 y 2
121. The arrangement is shown here.
2
y 2
2
0 I1I 2 a
e
2 2
F log e log a y 3
4
3a
2 a2
II a
Please note that the direction of I1 and the F 0 1 2 log e 2 3 2
4 a 3a
z axis, both are outwards towards the reader..
Let us consider an infinitesimal element of length
dy at a distance y from x-axis. If B be the field 0 I1I 2 1 0 I1 I 2
due to the wire 1 at the element is F 4 log e 3 F log e 3 .
4
0 I1 0 I1 122. F2 is in y-direction when velocity is along z-axis.
B1
2r 2 a 2 y 2 Therefore, magnetic field should be along x-axis.
The force on this element due to the field of wire
So let, B B i 0
1 is
37
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
106 106 Hence, Bx B cos 450 B 2 and
i j
(a) Given v1
2 2 By B sin 450 B / 2
and F 1 5 2 103 k
B B
From the equation, F q v B ; we have So in vector form B i j
2
2
106 106 j B i And M I 0 k I 0 SkL2 k
5
2 103 k 106
2
i
2
0
B B
So, M B I 0 k i
2
j
B0 2 2
k
2 I 0 L0 B
B0 i.e ,
2
j i
5 2 103 or B 102 T
2 0
i.e torque has magnitude I 02 L2 B and is directed
Therefore, the magnetic field is, B 10 i T
2
along line QS from Q to S moment of inertia of
the frame about QS
(b) F2 B0 qv2 sin 900 126. When the ring is not rotating mg 2T0
As the angle between B and v in this case is When the ring is rotating its magnetic movement
900 . M
Q
r 2
F2 10 2 10 6 106 10 2 N . 2
123. (a) Frequency of the applied emf = Cyclotron Let the tension in the string be T1 and T2
Bq 2 mf QBr 2
frequency or f ; B T1 T2 mg , Torque acting on the ring
2m q 2
2 3.14 2 1.67 1027 10 106 T1
D D QBr 2
T2 T1
mg QBr 2
,
1.30 T
1.6 10 19 2 2 2 2 2D
(b) The speed of deuterons on the emergence maximum tension 3T0 / 2
from the cyclotron,
3T0 Qmax Br 2 DT0
BqR T0 max
v 2 fR 2 2D BQr 2
m
127. (i)Orbital magnetic dipole moment
2 3.14 10 106 32 102 2.01 107 m / s
e
M R 2
1 2 T
Energy of deuterons mv
2 1
M eR 2
1
2 1.67 10 27 2.01 107 J 4.22 MeV .
2 2
2 According to Borhr’s postulates
L mv LqB nh nhe
124. (a) sin But R sin mR2 M
R qB mv 2 4m
nhe nheB
(b)In this case, Width > Radius Now t (ii) MBsin Bsin 300
4m 8m
125. (a) As the magnetic field B is in x-y plane and
subtends an angle of 450 with the x axis
38
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
3. The figure shows a right angled isosceles
triangle wire frame. The wire frame starts
LEVEL- VI entering into the region of uniform magnetic
field with constant velocity v at t 0 . If I
SINGLE ANSWER QUESTIONS is the instantaneous current through the
frame, then choose the correct graph
BIOT-SAVART’S LAW & FORCE ON between I and t.
CONDUCTOR
a
1 A square frame carrying a current I is located
in the same plane as a long straight wire v
carrying a current I 0 . The frame side has a
length a. The axis of the frame passing the
midpoints of opposite sides is parallel to the
wire and is separated from it by the distance t0
which is 1.5 of times greater than the
side of the frame. Find the mechanical work l
to be performed in order to turn the frame l
39
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
7. A copper rod is bent into a semi-circle of
P
radius a and at ends straight parts are bent
along diameter of the semi-circle and are
passed through fixed, smooth and conducting
v0 ring O and O ' as shown in figure. A
capacitor having capacitance C is connected
Q
to the rings. The system is located in a
uniform magnetic field of induction B such
kx kx kx that axis of rotation OO ' is perpendicular
A. B. C. to the field direction. At initial moment of
m m B 2l 2 C m B 2l 2C
D. None of these time (t=0), plane of semi-circle was normal
5. A coil having N turns is wound tightly in to the field direction and the semi-circle is
the form of a spiral with inner and outer set in rotation with constant angular velocity
radii a and b respectively. When a current . Neglect the resistance and inductance
I passes through the coil, the magnetic of the circuit. The current flowing through
field at the centre is. the circuit as function of time is
μ 0 NI 2μ 0 NI
(A) (B)
b a O O1
μ 0 NI b 0 IN b
ln (D) b a ln a .
(C)
2b a a
X B
6. A rectangular loop with a sliding conductor
of length l is located in a uniform magnetic
field perpendicular to the plane of the loop. C
The magnetic induction is B. The conductor
has a resistance R. The sides AB and CD
have resistances R1 and R2 , respectively.. 1 2 2
A. a CB cos t
Find the current through the conductor during 4
its motion to the right with a constant velocity 1 2 2
v. B. a CB cos t
2
X X
X
X
1 2 2
X
A
B
C. a CB sin t
X X X 4
X X
R1 v
R2 X 1 2 2
X X X D. a CB sin t
X
2
x D X X X C X 8. A rectangular loop of wire with dimensions
X X X
X X shown in figure is coplanar with a long wire
carrying current I. The distance between the
wire and the left side of the loop is r. The
Blv R1 R2 Bl 2v
A. B. loop is pulled to the right as indicated. What
R1 R1 R2 R1 R1R2 are the directions of the induced current in
the loop and the magnetic forces on the left
Blv R1 R2 Bl 2v and the right sides of the loop when the loop
C.
R1R2 R R1 R2 D. R1R2 R R1 R2 is pulled?
40
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
10. PQ is an infinite current-carrying conductor.
r AB and CD are smooth conducting rods on
which a conductor EF moves with constant
velocity V as shown in figure. The force
needed to maintain constant speed of EF is
b R
I A C
P
E F
I V
a a
b
Induced current Force on left side Force on B D
Q
right side
A. Counterclockwise To the left To the left
B. Counterclockwise To the right To the left 2
1 0 IV b
C. Clockwise To the right To the left 1n
VR 2
A.
D. Clockwise To the left To the right a
9. There is a uniform magnetic field B in a 2
circular region of radius R as shown in figure 0 IV b 1
B. In
dB 2 a VR
whose magnitude changes at the rate of .
dt 2
The e.m.f. induced across the ends of a 0 IV b V
C. In
circular concentric conducting arc of radius 2 a R
R1 having an angle as shown 2
V 0 IV b
OAO ' is D.
R 2
In
a
11. AB is a resistanceless conducting rod which
forms a diameter of a conducting ring of
X
X X X
RX X X radius r rotating in a uniform magnetic field
X X X X
X X XXX X X
X X B as shown in figure. The resistors R1 and
X X XA X X X
X
X XX X R2 do not rotate. Then the current through
X XX X X
XX
X X
X R1
the resistor R1 is
xxxxxxxxxxxx
O'
xxxxxxxxxxxx R1
2 dB dB2 A B
A. R1 B. R
2 dt 2 dt
xxxxxxxxxxxx
R2
2 dB xxxxxxxxxxxx
C. R D. None of these
2 dt
41
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
B r 2 B r 2 B
A. B.
2 R1 2 R2
a b
B r 2 B r 2
C. 1 2 D. 2 R R
R R
2 R1R2 2 2 (A) qb B/m
(C) qa B/m
(B) q (b–a)B/m
(D) q(b+a) B/2m
12. A conducting wire of length l and mass m 14. A particle of mass m, carrying a charge q, is
is placed on two inclined rails as shown in lying at the origin in a uniform magnetic field
figure. A current l is flowing in the wire in directed along + X axis. At the instant t = 0
the direction shown. When no magnetic field it is given a velocity v0 at an angle with
is present in the region, the wire is just on the y–axis, in the xy plane. The coordinates
the verge of sliding. When a vertically of the particle after one revolution will be
upward magnetic field is switched on, the y
Rails z
43
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
A. For r R , the induced electric field is 24. A charged particle is fired at an angle to
proportional to r a uniform magnetic field directed along the
B. For r R , the induced electric field is x-axis. During its motion along a helical path,
if the pitch of the helical path is equal to the
1 maximum distance of the particle from the
proportional to
r x-axis
C. For r R , the induced electric field is 1 1
maximum (A) cos (B) sin
D. If a coaxial “non-conducting” ring of radius 1
(C) tan (D) tan
R
is placed in the magnetic field region, 25. A chared particle enters into a region
2
which offers a resistance its motion and a
R 2 uniform magnetic field exists in the region.
then emf induced in the ring is
4 The particle traces a spiral as shown in the
fig. Which of the following correct ?
22. Two long, thin, parallel conductors are kept
very close to each other , without touching.
One carries a current i, and the other has
charge per unit length. An electron Q
P
moving parallel to the conductors in
undeflected. Let C = velocity of light.
b) at point Q is anticlockwise
(B) the radius of curvature at 5 3,5 is 10 m
c) at point Q is clockwise (C) the speed of the particle is 2 m/s
d) at point R is zero (D) the particle moves in a helical path
44
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
COMPREHENSION [2]
COMPREHENSION TYPE QUESTIONS A charge particle of mass m and charge q is
projected on a rough horizontal XY plane. Both
electric and magnetic fields ar given by
COMPREHENSION [1]
E 10kN / C and magnetic field B 5k
There is region of space where uniform magnetic
tesla are present in the region. The particle enters
field of induction B exists. The field exists at all
into the magnetic field at (4,0,0) m with a velocity
points for which x-coordinates are positive. The
direction of field is along negative z axis 50 j m/sec. The particle starts into a curved parth
Now a certain charged particle of mass m and 1
charge q having a certain speed enters in this on the plane. If coefficient of friction
3
region. A magnetic field at a point whose between particle and plane, then
coordinates are x = 0; y = -d and z = 0. Magnetic
force will start acting on the particle and particle qE 2mg , g 10m / s .
2
45
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
34. Speed of the particle at a general point P (x, 37. In the graphs below, the resistance R of a
y) is given by superconductor is shown as a function of its
2qE0 x 2 y 2 2qE0 y temperature T for two different magnetic
(A) (B) fields B1(Solid line) and B2(dashed line)if
m m
B2is larger than B1which of the following
2qE0 x E0
(C) (D) B graphs shows the correct variation of R with
m 0 T in these fields?
35. The y - coordinate at the highest point of
trajectory is
4m 2m 2mE0 4mE0
(A) qB 2 (B) qB 2 (C) qB 2 (D) qB
0 0 0 0
B) B 5T ,75K Tc B 100K
P1-2 Electrical resistance of certain materials,
known as super conductors, changes abruptly C) B 10T , 75 K Tc 100 K
from a non-zero value to zero as their temperature D) B=10T, Tc=70K
46
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
Column II
A
q R2
(p) Magnetic moment
4
q R2
(q) Magnetic moment
8
R E S
q R2
(r) Magnetic moment
2 B
(s) Ratio of magnetic moment and angular
q
momentum
2m
X R1 Y
q R2
(t) Magnetic moment angular Column I
10
momentum of a loop. (A) Switch S is closed
40. A square loop of conducting wire is placed [exclude small switching time]
near a long straight current carrying wire (B) Switch S is closed for long time, then
as shown. Match the entries of Column I it is opened for this transition time
with the entries of Column II (C) If coil A is brought nearer to B, switch R1
(D) The battery of constant emf is replaced by a
varying emf, switch S is closed
Column II
(P) Current in R1 is from X to Y
I
(Q) Current in R1 is from Y to X
(R) No current is flowing through S is closed
(S) current in R1 can be from X to Y or from
Y to X
Column I
(A) If loop is moved away from the wire be INTEGER ANSWER TYPE QUESTIONS
clockwise
(B) If loop is moved towards the wire 42. An electron gun G emits electrons of energy
(C) If current I is increased 2 KeV travelling in the positive x-direction.
(D) If current I is decreased The electrons are required to hit spot S
Column II where GS = 0.1 m, and the line GS makes
(P) Induced current in the loop will an angle of 600 with the x-axis as shown in
(Q) Induced current in the loop will be
fig. A uniform magnetic field B parallel to
anticlockwise GS exists in the region outside the electron
(R) Wire will attract the loop
gun. Find the minimum value of B needed to
(S) Wire will repel the loop
make the electrons hit S is found
41. In the circuit shown in the figure two coils
are arranged as shown. In Column I, some 4.737 10 n T . Find n.
operation which is carried out in circuit 1 is S
mentioned and in Column II, are given the
effects of the operations its effect. Match
B
the entries of Column I with these of Column B
II. Take heating effect of current also into 600
account. X
G
47
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
43. A coil carrying a current of i = 10 mA is
at the centre P and given a velocity v along
placed in uniform magnetic field so that its NP find its instantaneous acceleration. (b)
axis consists of only one turn and is made of
If an external uniform magnetic induction
copper. The diameter of the wire is 0.1 mm,
the radius of coil is R = 3 cm. An approximate B Bi is applied, find the force and torque
external field B will rupture the coil is found acting on the loop.
y 103 . Find y. Breaking stress = 3 × 108 N/
m2.
44. A particle of mass m and charge q enters a
region of electric field E as shown in the
figure with some velocity at point P. At the
moment the particle collides elastically with
smooth surface at N, the electric field E is 47. A constant current I flows through a metal
switched off and a magnetic field B rod of length L and mass m that slides on
perpendicular to the plane of paper frictionless rails as shown in figure. If the
automatically switched on. If the particle hits
initial speed of the rod is v0 and a magnetic
mE field B acts vertically up, find the total
the surface at point O, then if B x .
qd distance moved by the rod before coming to
What is the value of x1 a stop.
B
I
L
48
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
(b) time when the particle will hit the centre.
i0
(c) distance travelled by the particle when it
R comes to rest.
52. A circular loop of radius R is bent along a
diameter and given a shape as shown in
i figure (a). One of the semicircles (KNM) lies
in the xz-plane and the other one (KLM) in
6. Kinetic energy of each electron in a beam the yz-plane with their centres at the origin.
of television picture tube is 12.0 keV. The Current I is flowing through each of the
electron are emitted horizontally from semicircles as shown in figure.
geomagnetic south to geomagnetic north. (a) A particle of charge q is released at the
The vertical component of earths magnetic
origin with a velocity v v 0 i . Find the
field points down and has a magnitude
instantaneous force F on the particle. Assume
55.0 J . (a) In which direction will the
that space is gravity free.
elecrtrons deflect? (b) How far will the beam
deflect in moving 20.00 cm through the (b) If an external uniform magnetic field B j
television tube?
is applied, determine the forces F 1 and F 2
50. Two long concentric cylindrical conductors on the semicircles KLM and KNM due to
this field and the net force F on the loop.
of radii a and b b a are maintained at a
potential difference V and carry equal
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.
51. A charged particle of mass m and charge q
is projectd on a rough horizontal xy-plane
surface with z-axis in the vertically upward
direction. Both electric and magnetic fields 53. A conductor carries a current I parallel to a
are acting in the region and given by current strip of current per unit width j and
width w, as shown in figure. Find an
E E k and B B k respectively. The
0 0 expression for the force per unit length on
particle enters into the field at (a,0,0) with the conductor. Discuss the result when the
width w approaches infinity.
velocity v j . The particle states moving
0
49
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
54. A long cylindrical conductor of radius a has 56. Three infinitely long thin wires, each carrying
two cylindrical cavities of diameter a through current i in the same direction are in the x-y
its entire length as shown in cross-section
plane of a gravity free space. the central wire
in figure. A current I is directed out of the
page and is uniform throughout the cross- is along the y-axis while the other two are
section of the conductor. Find the magnitude along x = ±d. (i) Find the locus of the points
and direction of the magnetic field in terms for which the magnetic field B is zero. (ii) If
of 0 , I, r and a. the central wire is displaced along the z-
(a) at point P1 and (b) at point P2 direction by a small amount and released,
show that it will execute simple harmonic
motion. If the linear density of the wires is
, find the frequency of oscillations.
LEVEL- VI
KEY W U f Ui
51
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
Thus in this case, we have r e and so we
e 2 Bv 2t
I , where R is the resistance of
get tan 1 and . Therefore, the angle R R
4
3 loop. The above expression is valid only for
between d1 and r is . Also
4 a
t
dr 2v
d1 2dr
4. Let the velocity of rod be v when it has been
sin
4 displaced by x . Due to motion of rod an emf
From Biot-Savart’s law, we know that there is
no contribution from the straight portion of the will be induced in rod given by e Bvl , due to
this induced emf, charging of the capacitor takes
wire since d1 r 0 . For the field of the
spiral, we have place as a current, flows in the circuit [for very
dB
0 I d 1 r
small time] as a result of this current, the rod
experiences a magnetic force given by IBl .
4 r 2
2 d 1 sin r
0 I P
4 0
B
r2
2
0 I 3 1
4
0
2dr sin 2
4 r
2
2
0 I 0 I 1
B 4 r 2 dr
4
r
v0
0 0
Substitute re , we get
Q
2
0 I
B
4
e
0
52
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
Small amount of magnetic field produced at Induced emf in the circuit,
O due to thickness dx of the wire
d 1
0 NI dx NI 0 e a 2 B sin t
dB 2 b a x 2 b a
dt 2
Since resistance of the circuit is negligible,
On integrating we get therefore, potential difference across the
b
0 NI dx 0 NI capacitor is equal to induced emf in the circuit.
B Charge on the capacitor at time t is q=Ce
a
2 ba x 2 b a
1
a 2CB sin t
dx 0 NI
b
log e x a ; 2
b
a
x 2 b a
dq 1 2 2
But current I a CB cos t
0 NI b dt 2
B log e
2 b a a 8.
Hence correct option is (c)
r
6. Induced emf Blv . R is internal resistance of A B
seat of emf, i.e., of rod
R1 R2 I V
Total resistance of circuit = R
R1 R2
D C
Blv Blv R1 R2
I As the flux decreases, to maintain flux, current in
R
R1R2 R1R2 R R1 R2 the loop is clockwise. Force on DA due to the
R1 R2 long wire is towards left while on BC is towards
right.
7. When the copper rod is rotated, flux linked with 9. Required emf
the circuit varies with time dq 1 2 2
Therefore, anemf is induced in the circuit. I a CB cos t
dt 2
At time t, plane of semi-circle makes angle t
b b
with the plane of rectangular part of the circuit. I
Hence, component of the magnetic induction 10.
Induced emf BVdx 20 x BVdx
a a
normal to plane of semi-circle is equal to
B cos t . 0 IV b
Flux linked with semicircular part is Induced e.m.f. = 2 In a
1
1 a 2 B cos t E2
2 Power dissipated
Let area of rectangular part of the circuit be A. R
Flux linked withthis part is E2
2 BA Also, power = FV F
VR
Total flux linked with the circuit is
2
1 1 0 IV b
a 2 B cos t BA F In
2 VR 2 a
53
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
11. The equivalent diagram is 14. Pitch = (T) (Horizontal Velocity)
2 m
R1 V0 sin
qB
So (x,y,z) co-ordinates are
R2 2 mv0 sin
, 0, 0
qB
e
15. Magnetic moment for one loop of radius r is
The induced emf across the centre and any point M = IA = I ( r2)
on the circumference is N
Total number of turns per unit length is
ba
1 B r 2
e Bl
2
N
2 2 dM = (dN)(IA) dM = dr
ba
B r 2
Current through R1 2 R ( r2I) =
NI 2
r dr
1 ba
12. The front view of the arrangement is shown in b
NI r 3
b
NI 2
figure M = ba r dr
b a 3 a
a
IBl sin
N IBl cos
NI b3 a 3
IBl = 3 ba
f mg cos
mg sin
mg 16. Stationary charges produces only electric
field. Moving charges produces both electric
From initial condition, mg sin mg cos and magnetic fields.
tan 0I
17. B due to a long conductor
ma IBl cos mg sin N 2r
18. A is stationary and observes the current I. B
N mg cos IBl sin observes the free electrons to be at rest, but
the unbalanced positive charges in the
IBl IBl sin 2
a cos q 2 g sin conductor will appear to move in the direction
m m cos opposite to that of v. Thus , A and B observe
the same current and hence the same
IBl cos 2
2 g sin magnetic field.
m cos
Now,
1 2 1 IBl cos 2
s at 2 g sin t 2 .
2 2 m cos
19. Treat the gas as a thick conductor carrying
mv 2 a uniform current. Apply Ampere’s law to find
13. qvB where radius R is (b-a); So the magnetic field. Then apply the left- hand
R
rule to find the direction of the Ampere force.
qB
mv qB b a 20. At any time t, the situation would be as shown
So vmin
b a m in the figure below:
54
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
2x 1 c 2
or v .
2 0 x 0i i 0 0 i
23. Concuptual
600
R/2 V m
Pitch 2 v cos
QB
24.
Also, mv sin QBr for motion
perpendicular to the magnetic field
2 mv
R 2 or r sin
PO R vt QB
2 Maximum distance of the particle from the
Emf induced, e Bv 2 PO x-axis = 2 r
e
i m mv
2 R 2 v cos 2. sin or tan
As flux is decreasing with time, induced emf will QB QB
try to oppose the decreasing flux and hence 25. If particle has velocity component parallel to
induced current would be in clockwise direction. magnetic field path will be helical. Radius of
path is decreasing so it enters at Q.
dB
21. For r R, 2 r r2 mv sin
dt 26. r 5
qB 5 10
r
E
2
x 5 3 y x tan 1800
27. Radius of circular path will be equal to
dB
For r R, E 2 r R2 d
mv
v
qBd
dt qB m
R2 28. Velocity will be along x axis
E
2r qB qB m
29. t , , .t t
m m qB
R
For r , the emf induced, mv m 50 m 10
2 r 10 5m
30.
2
qB q5 2mg
dB R R 2
e dv mg qE 1 mg 2mg
dt 2 4 31. 10 ;
dt m 3 m
22. At P, electric field E
20 x (to the right), dv 10dt t 5sec
32. Work done by electric force will be zero as
0i force is perpedicular to velocity
and magnetic field B
2x mg qE dv mg qE
( into the paper) 33. a ; v g
m ds m
E
For no deflection , E vB or v vdv 50 50
B ds g
s
2 10
125m
55
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
to be in clockwise direction. The magnetic frield
1 2 2qE 0 y
34. mv qE0 y v at the location of left side of loop is more than
2 m that of right side, so force direction is decided
by directioin of force acting on left side of loop,
E0 qB
35. Vy sin 0 t , in this case, F1 is towards the wire i.e., attractive.
B0 m
y t
E0 qB
dy
0 0
B0
sin 0 t dt
m
mE 0 qB0 mE 0
y
qB02 1 cos m t ; y max qB2
0
57
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
m N 2mEd 1 v0 E B v B
R --------------(1) i , j And
k
qB q B
But because here,
v0 E B v0 B
d d
From figure R 2 v0 qBt E qE v0 B qB
2
So v0 v0 cos t
2 v0 sin t
Putting in equation (1), v0 m E m v0B m
d 1 2mEd mE 46. The magnetic field at the centre P due to
or B 2 .
2 B q qd current in wire NM is
45. Because the forces to parallel electric and
magnetic fields on a charged particle moving
perpendicular to the fields will be at right
angles to each other (electric force being
along the direction of E while magnetic force
perpendicular to the plane containing v and
B ) so magnetic force will not affect the
motion of charged particle in the direction of 0 I
electric field and vice-versa. So the problem B1 sin 600 sin 600 ;
is equivalent to superposition of two 4 r
independent motion as shown in the figure.
0 I 3 3 2I 3
So for motion of the particle under electric B1 ; B1 0
field alone 4 a / 2 2 2 4 a
qE dv y qE Directed away from the reader perpendicular
ay i.e.,
m dt m r a
to the plance of paper sin 30 r
0
y t t a 2
qE qE
or dvy dvy m
dt i.e.vy t...............(1)
m MS 3a
0 0 0
cos 300 MS MN 3a
While at the same instant, the charged a 2
particle under the action of magnetic field will The magnetic field at the centre P due to
describe a circle in the x-z plane current in are MN is
mv0 v qB 0 2 I 0 2 I 2 /3 0 2 I
With r i.e., 0 B2
qB r m 2 a 2 4 a 2 4 3a
So angular position of the particle at time t in Directed towards the reader perpendicular to
the x-y plane the plance of paper. The net magnetic field
qB 0 2 3I 0 2I 0 2I 2I
will be given by t t and therefore B B1 B2
4a a 4 3a 4a 3 0 .68
3 4 a
m
in accordance with figure (Directed away from the reader perpendicular
and to the plane of paper). The force acting on
the charged particle Q when it has a velocity
qB v and is instantaneously at the centre is
vz v0 sin v0 sin t v0 sin t..........(3)
m F QvB sin QvB sin 900 QvB
So in the light of equations (1), (2) and (3) ,
The acceleration produced
we get v ivx jv y kvz
F QvB Qv 0 2I 0.110IQv
A 0.68; A
qB ˆ qE ˆ qB ˆ M m m 4 a ma
v v 0 cos ti j v 0 sin k
m m m
58
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
rotate to occupy equilibrium position and teh
energy converts into kinetic energy and kinetic
energy of the system is maximum when
stored energy is completely released.
Magnetic induction, at centres due to current
0 I
in larger coil is B
2R
Magnetic dipole moment of smaller coil is
The direction of acceleration is given by the
i r 2 .
vector product v B or by applying Initially planes of two coils are mutually
Fleming’s left hand rule RPN 900 and perpendicular, therfore is 90 or energy of
the system is
MPN 1200 MPR 120 90 300
0 Ii r 2
Since, MPQ 600 RPQ 300 U (i r 2 ) B(1 cos90) U
2R
i.e. the acceleration vector makes an angle
When coils are release, both the coils start
of 300 with the negative x-axis to rotate about their common diameter and
(b) The Torque acting on the loop in the their kinetic energies are maximum when they
become coplanar.
magnetic field is given by M B Moment of inertia of larger coil about axis of
Where M=IA where A = (Area of PMQNP)-
(area of triangle PMN) 1
rotation is I1 mR 2
2
1 1 a2 1 a 3
a2 MNPS 3a a2 1 2
3 2 3 2 2 3 4 and that of smaller coil is I 2 mr
2
Since, two coils rotate due to their mutual
3
2 3
A a2 k Ia k iB ; interaction only, therefore, if one coil rotates
3 4 3 4 clockwise then the other rotates
anticlockwise.
3 Let angular velocities of larger and smaller
BIa 2 j 0.614 Bia J
2
coils be numerically equal to 1 and 2
3 4
respectively when they become coplaner,
The force acting on the loop is zero u V0 According to law of conservation of angular
momentum,
ILB
a I11 I 2 2
m and according to law of conservation of
V 0 energy,
Using
1 1
2 I112 I 2 22 U
ILB mV 2 2
V 2 u 2 2as V02 2 SS 0
59
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
49. (a) Figure shows the electron travelling north. The On solving the above quadratic equation, we get
magnetic field points inward. From right hand rule
d R R2 l 2
v B vector is toward west. Since electron is The plus sign corresponds to angle 1800 and
negative, the force on it is toward east.
minus sign corresponds to our case.
1 l2
R l2
1/ 2
For l R ,
2
R
2R
1 l 2 1 0.200
2
Hence d 0.00298 m .
2R 2 6.72
50. The elecric field E in the region between the
conductors varies with the distance r from the
(b) The electron speed can be determined from
kinetic energy.
axis as E 2 r ,
0
1 2K
K m 2 ; Where is the charge per unit length on the
2 m
inner cylinder, and we also know that the
2 12.0 103 1.60 1019 potential difference V between the conductors
is given by
9.1110 31
a
6.49 107 m / s V ln .
20 b
The magnetic force on the electron is
Combining these exppressions to eliminate
FB eB sin where is angle between the
gives
electron velocity and the magnetic field. In this
case, velocity andmagnetic field B are V
E
perpendicular, 900 . The magnetic force a
r ln
provides centripetal force for circular motion, b
therefore The magnetic field B varies with r as
m2 0 I
eB B
R 2r
If the inner cylinder is at a positive potential w.r.t
m 9.111031 6.49 107 the outer cylinder, the electic field E is directed
R 6.72 m
eB 1.6 10 19 55 106 radially outwards. If we assume that current flows
Let the arc traced by the electron beam subtend into the page along the inner cylinder and out of
the page along the outer cylinder, the magnetic
an angle at the centre, l be the length of the
field will be directed clockwise as shown.
tube, and d the deflection.
From figure, l R sin
d R R cos or R cos R d
On squaring and adding these two equations, we
have
R 2 R d l 2 or d 2 2 Rd l 2 0
2
60
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
Consider an electron travelling with velocity u m0
into the page, a a radial distance r from the axis Here Ri
B0 q
of the cylinders. It experiences an electric force
Ee radially inwards, and a magnetic force Beu m 0
radially outwards. In order for the electron to be Thus t
mg qE0
undeviated, these forces must balance so that
u E / B . Substituting our expressions for E and d
(c) m mg qE0 or,
B gives dl
0 t
V m d mg qE0 dt or
r ln a / b
0 0
2V
u
0 I 0 I ln a / b m02
l
2r 2 mg qE0 .
This expression in independent of r, so the 52. (a) The magnetic field of semicircular arc KLM,
electrons’s position does not matter. I
51. (a) N mg qE0 ----------------(1) at its centre B1
0
4R
i
m2 The magnetic field of semicircular arc KNM at
qB0 ------------------(2) I
R
its centre B2
0
4R
j
d
and m N ----------------(3) I
dt
The resultant magnetic field BR
0
4R
j i
m
From equantion (2), R
B0 q ---------(4)
From equations (1) and (3),
d
m mg qE0 (or)
dt
0 t
m d mg qE0 dt
0 0
0 mg qE0 t
Ri
dR
qB0 dt
0
qB0 Ri
or t
mg qE0
61
ELECTROMAGNETISM & EMI JEE ADVANCED - VOL-VII
(b) We take a differential element on arc KLM 0 Ij dx h
as shown in figure (c); length dl R
2 h 2 x 2 h2 x2
In vector form, d1 dl cos j dl sin k Integrating over half width of strip,
The magnetic force on a differential element d1 F 0 Ijh w / 2 dx 0 Ij 1 w
0 h 2 x 2
tan
d F Id1 B L 2h
I dl cos j dl sin k B j When w ,
F
L
Ij
0 , the force is
2
Idl sin B i IRB sin d i attractive as expected for parallel currents.
0 I 2r 2 a 2
54. (a)
4r 4r 2 a 2 to the left
Resultnat force F KLM IRB 0 sin d i
dF I dl cos i dl sin k B j
I dl cos k dl sin i B 1 2I
J
a 2
2
I Rd cos k sin i B Current density a
a 2 2
2
FKNM IRB cos d k sin d i
Let us consider both the cavities are carrying
0 0 equal and opposite currents with current density
J.
IRB 0 2i 2IRBi Let B1 , B2 and B3 be magnetic fields due to
Thus resultant force, complete cylinder, upper and lower cavity
respectively.
F F KLM F KNM 4 IRBi .
(a) At point P1
53. We consider a differential segment of thickness
dx, carrying current j dx. The attractive force per 2I 2 J a 2 I
B1 0 1 i 0 i 0 i
unit length is 4 r 4 r r
dF 0 I j dx a
2
2J
L 2r 2I 2 i
B 2 0 2 i 0
A symmetrical segment at x exerts same 4 r a 4 a
r
magnitude of force, the x components cancel 2 2
and the resultant force is
0 I i
dF 0 I jdx
cos a
4 r
L 2r 2
62
JEE ADVANCED
MAINS - CW--VOL-VII
IIT ADVANCED VOL - I ELECTROMAGNETISM
NAME OF THE & EMI
TOPIC
2I 0 I 2 4r
B 3 0 3 i i
B 0 2 j
4 r a a 2 r 4r a 2
4 r
2 2
0 I 2r 2 a 2
j
B B1 B 2 B 3 4r 4r 2 a 2
0 I 2r 2 a 2
I
0
4
1
1
i
B
4r 4r 2 a 2
, upwards.
4 r a a
r r 55. For finding the magnetic field produced by this
2 2
circuit at the centre we can consider. It to contain
I 2r 2 a 2
B 0 2 i of two semicircles of radius r1 0.08m and
4r 4r a 2 r2 0.12m . Since current is flowing in the same
I 2r 2
a 2 direction the magentic field created by circular
B 4r 4r
0
arcs will be in the same direction and hence added
a 2
2 , towards left.
i i
(b) At point P2 B1 0 and B2 0
4r1 4r2
0i 1 1
B (Direction outwards)
4 r1 r2
B 6.54 10 5 T (Right hand thumb rule)
(b) Force acting on a curent carrying conductor
2 I placed in a magnetic field is given by
I
B1 0 1 j 0 j F I l B IlB sin
4 r r
2I 2 In this case 0 F 0
B2 0 sin i cos j
4 2 (ii) On arc AC due to current at the centre B at
a
r2
4 0 I
AC will be B 2 r . The direction of this
0 I
sin i cos j
1
64