JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Electromagnetic Induction

A thin conducting rod MN of mass 20 gm, length 25 cm and resistance

10Ω

is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field

B_{0} = 4 T

directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time

t = 0

and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.
[Given: The acceleration due to gravity

g = 10 ms^{- 2}

and

e^{- 1} = 0.4

]

List-I		List-II
(P)	The magnitude of the induced emf (in Volt) at time t = 0.2 sec	(1)	0.07
(Q)	The magnitude of the magnetic force (in Newton) on the rod at time t = 0.2 sec	(2)	0.14
(R)	The power (in Watt) dissipated as heat at time t = 0.2 sec	(3)	1.20
(S)	The magnitude of the terminal velocity (in m/s)	(4)	0.12
		(5)	2.00

(A) $P \to 5, Q \to 2, R \to 3, S \to 1$
(B) $P \to 3, Q \to 1, R \to 4, S \to 5$
(C) $P \to 4, Q \to 3, R \to 1, S \to 2$
(D) $P \to 3, Q \to 4, R \to 2, S \to 5$ ✓ Correct

Answer: (D)

Solution

Equation of motion:

m g - B i ℓ = m \frac{d v}{d t}

where

i = \frac{B ℓ v}{R}

m g - \frac{B ^{2} ℓ ^{2} v}{R} = m \frac{d v}{d t}

Terminal velocity:

v_{T} = \frac{m g R}{B ^{2} ℓ ^{2}} = \frac{0.02 \times 10 \times 10}{16 \times 0.0625} = 2

m/s → (S) matches 5

Solving the differential equation:

v = v_{T} (1 - e^{- t / τ})

where

τ = \frac{m R}{B ^{2} ℓ ^{2}} = 0.2

s

At t = 0.2 s:

v = 2 (1 - e^{- 1}) = 2 (1 - 0.4) = 1.2

m/s

(P) Induced emf:

ε = B ℓ v = 4 \times 0.25 \times 1.2 = 1.2

V → 3

(Q) Magnetic force:

F = B i ℓ = B \times \frac{B ℓ v}{R} \times ℓ = \frac{B ^{2} ℓ ^{2} v}{R} = \frac{16 \times 0.0625 \times 1.2}{10} = 0.12

N → 4

(R) Power dissipated:

P = \frac{ε ^{2}}{R} = \frac{1. 2 ^{2}}{10} = 0.144

W ≈ 0.14 W → 2

Answer: (D)

P \to 3, Q \to 4, R \to 2, S \to 5

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Electromagnetic Induction

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