Solution

KLP40

JEE main - Physics

1. (a)

Explanation:
i. When the ring is just entering the magnetic field, emf is induced and the current flows in one direction.
ii. When the whole ring is surrounding the flux inside, there is no change in the flux. Induced emf = 0, current is zero.
iii. When the ring just goes out of the region of flux, a pulse of current is produced which is opposite to that of (i) given above.
Hence,

is the only figure that describes this.

2.
(b) QV
Explanation:
∮E.dl=−dϕ/dt=Einduced

Given, Einduced=V ...(i)

The force on the charge Q, due to the induced electric field, is,

F=QE⇒E=F/Q ...(ii)

From equations (i) and (ii),

∮F/Q.dl=V

⇒∮F.dl=QV

∴Work done=QV

3.
(d) 2 Blv
Explanation:
RATE OF CHANGE OF FLUX = 2Blv

4.
(d) 8
Explanation:
8

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 5.
(b) −παr 2

Explanation:
Given that, the magnetic field vary with time as,
B = B0 + at ...(i)
(Here, B0 and α are positive constants.)
dϕ
We know that, the induced emf, e = − dt
...(ii)
and the magnetic flux is given as:
ϕ = BA cos θ

[Here θ is the angle between the magnetic field B and area of ring A.]
So θ = 0°
or ϕ = BA cos 0°
or ϕ = BA ...(iii)
From eqns. (ii) and (iii),
e = − [BA]
d

dt

On putting the value of B from eqn. (i)

So e = − [(B + αt) ⋅ A]
d

dt
0

or e = −A d

dt
[B0 + αt]

or e = -A [0 + α ]
or e = -Aα
But A is the area of ring, here r is the radius of ring
So, e = −πr α 2

or e = −παr 2

6.
(d) tan δ = 2 tan λ
Explanation:
mjjdhjkl

7.
(c) None of These
Explanation:
4 × 103 = 40/10-2 x i

8.
(d) cot2δ = cot2δ′ + cot2δ′′
Explanation:
basic understanding

9.
(d) only ii
Explanation:
2
μ μ πr

M12 = M21 =
0 r 1

2r2

where πr is the overlapping area, i.e., that of the smaller coil. This formula is valid under ideal conditions if the coupling is
2
1

maximum. This depends on μ , the intrinsic magnetic property of the core.

r

10.

(b)

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 Explanation:

11.
(c) 4 V/m
Explanation:
After every T/2 the field will change from B to –B
Thus
e = 2BAf

12. (a) 2 In 2
Explanation:
Given circuit is a series L-R circuit
In an L-R circuit, current increases as
−R
E t
i = (1 − e L )
R

Now, energy stored in inductor is

1 2
UL = Li
2

where, L = self inductance of the coil and energy dissipated by resistor is

2
UR = i R

Given, rate of energy stored in inductor is equal to the rate of energy dissipation in resistor. So, after differentiating, we get
di 2 di R
iL = i R⇒ = i
dt dt L

R R
E R − t R E − t
⇒ ⋅ e L = ⋅ (1 − e L )
R L L R

R
− t
⇒ 2e L = 1
R
− t 1
⇒ e L =
2

Taking log on both sides, we have

−R 1 R
⇒ t = ln( )⇒ t = ln 2
L 2 L
L 20
⇒ t = ln 2 = ln 2 ⇒ t = 2 ln 2
R 10

13.
(b) 0.016 s
Explanation:
In given circuit,
Inductance of circuit is
L = 10 mH = 10 × 10-3 H
Resistance of circuit is
R = (Rs + r) = 0.1 + 0.9 = 1 Ω

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 Now, from
R
− ⋅t
i = i0 (1 − e L ) .......(i)

Given, i = 80% of i0
Substituting the value of i in Eq.(i), we get
R R R
− t − t t
0.8 = 1 − e L ⇒ e L = 0.2 ⇒ e L = 5
R
t R
⇒ ln(e) L = ln 5 ⇒ t = ln 5
L
−3
L 10×10
⇒ t = ⋅ ln(5) = × ln(5)
R 1

= 10 × 10 −3
× 1.6

= 1.6 × 10 −2
s = 0.016s

14. (a) only i

Explanation:
At the poles, θ = 90°
e = vBl sin θ = vBl sin 90° = vBl
So, when the car moves on a plane road at poles, the induced emf produced across the axle is maximum.
15.

(b)

Explanation:
Magnetic flux associated with the outer coil is
−αt
ϕ = μ0 πN R ⋅ I = μ0 N πR (kte )
outer
−αt
= Cte
where, C = μ 0 N πRk = constant
Induced emf,
−dϕ
outer −αt −αt
e= = Ce + (−αC te )
dt

= C e (1 − αt)
−αt

∴ Induced current, I =
e

Resistance
⇒ At t = 0, I = -ve

16.
(d) perpendicular to the plane of paper and downwards
Explanation:
According to fleming right hand rule, the direction of B will be perpendicular to the plane of paper and act downward.

17.
mgR sin θ
(c) 2 2
B l

Explanation:
From Faraday’s law of electomagnetic induction,
dϕ d(BA) d(Bll)
e= dt
=
lt
=
dt
=
Bdl×l

dt
= BVl

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 (l2B) =
2 2

Also, F = ilB = ( BV

R
)
B l V

At equilibrium
2 mgR sin θ
mg sin θ = B lV

R
⇒ V= 2 2
B l

18. (a)

Explanation:
Inside the sphere, the field varies linearly i.e., E ∝ r with distance and outside varies according to E ∝ 1

2
.
r

19.
2 2 2 2 2

(b) π a b B ω

2R

Explanation:
As we know, emf ε = NABω cos ω t, Here N = 1
Average power,
2 2 2 2 2 2 2 2
ε A B ω cos ωt A B ω 1
P= R
= R
= R
(
2
)

Therefore average power loss in the loop due to Joule heating

2 2 2 2

P= π a b B

2R
(ω )
2

20.
(d)

Explanation:
Case (1): When bar magnet is entering with constant speed, flux (ϕ) will change and an e.m.f. is induced, so galvanometer will
deflect in a positive direction
Case (2): When bar magnet is entering with constant speed, flux (ϕ) will not change, so the galvanometer will show null
deflection.
Case (3): When bar magnet is making on exit, again flux (ϕ) will change and an e.m.f. is induced in the opposite direction so
the galvanometer will deflect in a negative direction i.e. reverse direction.

21.
(d) 5000
Explanation:

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 Force on the strip when it is at stretched position x from mean position is

F = -kx - iIB = -kx - BIv

R
× IB
2 2

F = -kx - B I

R
× v
Above expression shows that it is case of damped oscillation, so its amplitude can be given by
bt

⇒ A=A 0e
2m

bt
A0 A0
⇒
e
= A0 e 2m [as per question A = e
]
−3
2×50× 10 ×10
⇒ t =
2m
= 0.01×0.01
B2 I 2
( )
R

Given, m = 50 × 10-3 kg
B = 0.1 T
l = 0.1 m
R = 10 Ω
k = 0.5 N
−−
Time period, T = 2π√ m

k
≃ 2s
so, required number of oscillations,
N= = 5000
10000

22. (a) πμV

Explanation:
Here,
Magnetic field, B = 0.025 T
Radius of the loop, r = 2 cm = 2 × 10-2 m
Constant rate at which radius of the loop shrinks,
,= 1 × 10-3 ms-1
dr

dt

Magnetic flux linked with the loop is

ϕ = BA cos θ = B(π r2) cos 0° = Bπ r2
The magnitude of the induced emf is
dϕ d dr
2
|ε| = = (Bπr ) = Bπ2r
dt dt dt

= 0.025 × π × 2 × 2 × 10-2 × 1 × 10-3

=π × 10-6 V = πμV
23.
(c) 1.61 × 10-4 V
Explanation:
Here, H = B = 0.4 × 10-4. l = 0.8 m
v = 120 rpm = 2rps
emf induced across the ends of each spoke
e = Bω l = B(2πv)l [∵ ω = 2πv ]
1

2
2 1

2
2

= Bπ vl2
= 0.4 × 10-4 × π × 2 × (0.8)2
= 1.61 × 10-4 V
Note: Remember that number of spokes is not relevant because the emfs across spokes are in parallel.

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24.

(b)

Explanation:

25.
Δϕ
(c) Q = R

Explanation:
We have, e = ∣∣ ΔΦ
∣
∣
Δt

∴ Current, I = e

R
=
ΔΦ

RΔt
Δϕ
So, charge Q = I ⋅ Δt = ΔΦ
⋅ Δt = R
RΔt

26.
μ
(c) 0

2π
Ivl

Rr

Explanation:
Magnetic field at a distance r from the wire
μ I
B= 2πr
0

Magnetic flux for small displacement dr,

ϕ = B⋅ A = Bldr [∵ A = l dr and B.A = BA cos 0o]

μ I
0
⇒ ϕ = ldr
2πr
dϕ μ0 Il
Emf, e = dt
=
2πr
⋅
dr

dt
μ0 Ivl
⇒ e= 2π
⋅
r

e μ0 Ivl
Induce current in the loop, i = R
=
2π
⋅
Rr

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27.
2
qBa
(b) 2
2mb

Explanation:
Suppose E is the electric field generated around the charged ring of radius r, then
dϕ
|e| = dt
→ 2

or ∮ ⃗
E ⋅ dl =
Bπr

Δt
2
Bπa
or E(2πb) = Δt
2

or Eb = Ba

2(Δt)
2
qBa
∴ Torque acting on the ring, τ = b × force = bqE = 2(Δt)

If ΔL is the change in angular momentum of the charged ring, then torque,

ΔL L2 − L1
τ = =
Δt Δt
2 2
qBa Δt qBa
or (L2 - L1) = τ Δt = =
2
2Δt

As initial angular momentum L1 = 0

2
qBa
Hence, L2 = 2
= I ω = mb ω
2

2
qBa
∴ ω =
2
2mb

28.
(c) A is true but R is false.
Explanation:
Faraday's laws of electromagnetic induction concern with the conversion of mechanical energy into electric energy in
accordance with the law of conservation of energy. But in a purely resisistive a.c. circuit, the emf is in phase with the current.

29. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
When bar magnet is dropped through a metallic cylindrical pipe, magnetic flux linked with cylindrical pipe increases and eddy
currents are produced. This oppose the motion of magnet through a metallic cylindrical pipe. As a result it takes more time to
come down as compared to a non-magnetic bar with same geometry and mass.
30.
(b) 1.0 T, 50 A/m and 1.5 T
Explanation:

31.
(c) 1 mA
Explanation:
Given Number of turns,
n = 1000 turns/cm = 1000 × 100 turns/m
Coercivity of ferromagnet, H = 100 A/m
Current to demagnetise the ferromagnet, I = ?
Using, H = nI
or, 100 = 105 × 1
100
∴ I = = 1 mA 5
10

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32. (a) is zero, otherwise there would be a field falling as ( 1
) at large distances outside the toroid.
r3

Explanation:
In a toroid, magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and
outside the toroid magnetic field is zero.
33.
(c) from stronger to weaker parts
Explanation:
We know that diamagnetic materials are magnetized in the opposite direction of the magnetising field. Therefore, diamagnetic
material moves from stronger to weaker parts of the magnetic field.

34. (a) VB = 45 mV; VW = 120 mV with left side of pilot at higher voltage
Explanation:
√5
sin θ = 2

3
so, cos θ = 3

Wings will cut vertical component of magnetic field so,

Vw = BV × lwing × V = 5 × 10-5 × 2

3
× 15 × 240 = 120 mV
whereas, vertical section of plane will cut horizontal component of earth magnetic field.
VB = BH × 15 × 240 = 5 × 10-5 ×
√5

3
× 15 × 240 ≃ 45 mV

35. (a) 2.56 × 10 Wb/m −4 2

Explanation:
Bnet = B1 + B2 + BH

μ ( M1 + M2 )
0
B = + BH
net 4π r
3

−7
10 (1.2+1)
= 3
+ 3.6 × 10 −5

(0.1)

= 2.56 × 10-4 Wb/m2

36.
(d) Mu metal
Explanation:
Mu metal is a range of nickel-iron alloys with low hysteresis loss.

37.
(c) 1500A turns/m, 0.42 T, 7.5 × 106 Am-1
Explanation:
H = nl = 3 00 × 5 = 1500A turn/m
B = μH = 5000μ0 H

= 5000 × 4π × 10
−7
× 1500 =0.42 Tesla
B = μ0 (H + M ) or μH = μ 0 (H + M)

5000μ0 H = μ0 (H + M )

or 5000 H = H + M
or M = 4999 × H
or = 4999 × 1500

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38.
(d) 2 A-m
Explanation:
2 A-m

39.
(d) χ = 1

Explanation:
Paramagnetic substances obey Curie’s law, χ ∝ 1

40.
(b) sink of H-vector
Explanation:
sink of H-vector

41.
(b) is zero, otherwise, there would be a field falling as 1

3
at large distances outside the toroid.
r

Explanation:
The magnetic field is only confined inside the body of a toroid in the form of concentric magnetic lines of force. For any point
inside the empty space surrounded by toroid and outside the toroid, the magnetic field B is zero because the net current
enclosed in these spaces is zero. Thus, the magnetic moment of the toroid is zero.

42.
(d) diamagnetic
Explanation:
Hydrogen molecule behaves as diamagnetic as no net magnetic moment is associated with it.

43.
(c) 6 J
Explanation:
Given that: The initial direction of magnet = parallel to the field = 0o
The final direction of the magnet = 60o
Magnetic field, B = 6 × 10−4T
Magnetic Moment, M = 2 × 104 JT−1
Work done in rotating a magnet in the magnetic field is given by;
W = B. T (cos θ1 − cos θ2 )

−4 4
W = 6 × 10 × 2 × 10 (1 − 0.5)

W = 6J
Thus, It requires a 6 J of work to rotate the magnet from 0o to 60o in the magnetic field.

44. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
The discrepancy between the orientation of a compass and true north-south direction is known as magnetic declination. Angle
between magnetic axis and geographical axis is about 9.69o.
45.
(c) A is true but R is false.
Explanation:
Diamagnetism is non-cooperative behaviour of orbiting electrons when exposed to an applied magnetic field. Diamagnetic
substance are composed of atom which have no net magnetic moment (i.e., all the orbital shells are filled and there are no
unpaired electrons). When exposed to a field, a negative magnetization is produced and thus the susceptibility is negative.

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 Behaviour of diamagnetic material is that the susceptibility is temperature independent.

46.
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
Explanation:
Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).

47.
(c) A is true but R is false.
Explanation:
A tangent galvanometer is certainly used for measuring very small current, but it is not direct reading current is calculated from
I= 2a

N μ0
BH tan θ
Where a is radius and N is turns of a coil, BH is horizontal component of earth's magnetic field, θ is deflection of the needle.

48. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Both A and R are true and R is the correct explanation of A.
49.
(d) A is false but R is true.
Explanation:
If horizontal component of earth's magnetic (H) field is known, a deflection magnetometer can be used for measuring magnetic
movement. If H is not known, then to measure M, we have to use both the deflection and oscillation magnetometer.

50. 243.0
Explanation:
Time period of oscillation.
−−−−
I 1
T = 2 π√ M
⇒ T∝
BH √BH

But, BH = B cosθ
Here, B = total magnetic field due to earth, θ = angle of dip
−−−−−−
T1 B2 cos θ2

T2
=√ B1 cos θ1

60 −−−−−−−
B2 ∘

⇒
20

60
=√ B1
cos 60
∘
cos 30
30
−−−−
3 B2 9 B2
⇒
2
=√ ⇒
4
=
√3B1 √3B1

B1
⇒
B2
= 4
= 4

9√3 √243

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