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JEE Advanced Physics Electromagnetic Induction 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Physics Question (2023) — Solution

Question

A thin conducting rod M N of mass 20   gm , length 25   cm and resistance 10 Ω  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 At t = 0 . 2   s , the magnitude of the induced emf in Volt 1 0 . 07 Q At t = 0 . 2   s , the magnitude of the magnetic force in Newton 2 0 . 14 R At t = 0 . 2   s , the power dissipated as heat in Watt 3 1 . 20 S The magnitude of terminal velocity of the rod in  m   s − 1 4 0 . 12     5 2 . 00

Options

  1. A. P → 5 ,   Q → 2 ,   R → 3 ,   S → 1
  2. B. P → 3 ,   Q → 1 ,   R → 4 ,   S → 5
  3. C. P → 4 ,   Q → 3 ,   R → 1 ,   S → 2
  4. D. P → 3 ,   Q → 4 ,   R → 2 ,   S → 5

Answer

D. P → 3 ,   Q → 4 ,   R → 2 ,   S → 5

Step-by-step solution

Induced emf ε = B l v ⇒  Induced current  i = ε R = B l v R Direction of induced current would be from  N   to   M . Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write ⇒   m g - i l B = m a  [Applying 2 nd  law] ⇒   m g - B 2 l 2 v R = m d v d t ⇒ ∫ 0 v   d v m g - B 2 l 2 v R = ∫ 0 t d t m   ⇒   ln m g - B 2 l 2 v R 0 v - B 2 l 2 R = t m ⇒   m g - B 2 l 2 v R m g = e - B 2 l 2 m R t Now,  m g = 20 × 10 - 3 × 10 = 0 . 2 ,  B 2 l 2 R = 4 2 × 0 . 25 2 10 = 0 . 1  and  B 2 l 2 m R = 4 2 × 0 . 25 2 20 × 10 - 3 × 10 = 5 Therefore, ⇒   0 . 2 - 0 . 1 v 0 . 2 = e - 5 t ⇒   v = 2 [ 1 - e – 5 t ] ⇒  At t = 0 . 2   s ,   v = 2 1 - 1 e = 2 × 0 . 6 = 1 . 2   m   s - 1 ⇒   ε = B l v = 4 × 0 . 25 × 1 . 2 = 1 . 2   V  volts Then, current at  t = 0 . 2   s  will be  i = 4 × 0 . 25 × 1 . 2 10 = 0 . 12   A and magnetic force = i l B = 0 . 12 × 0 . 25 × 4 = 0 . 12   N and power dissipated = F v = 0 . 12 × 1 . 2 = 0 . 144   W also for maximum velocity(terminal velocity)  t → ∞ ,   e - 5 t → 0  and hence  v = 2 1 - 0 = 2   m   s - 1 ⇒  Correct match is (D)

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