Quantrex Academy · Free JEE Advanced PYQ solutions
JEE Advanced Physics Electromagnetic Induction 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Physics Question (2023) — Solution

Question

A rectangular conducting loop of length 4   cm and width 2   cm is in the x y -plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction  3 2 x ∧ + 1 2 y ∧  with a constant speed v . The wire is carrying a steady current I = 10   A in the positive x -direction. A current of  10   μ A  flows through the loop when it is at a distance d = 4   cm from the wire. If the resistance of the loop is  0 . 1   Ω , then the value of v is _____  m   s - 1 . [Given: The permeability of free space  μ 0 = 4 π × 10 - 7   N   A - 2 ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

The two sides of the rectangular loop which are perpendicular to the wire would not contribute in induced emf. Angle of velocity vector with x-axis is  tan α = 1 2 3 2 = 1 3 . Therefore,  α = 30 ° . For parallel sides: E → = B → × v → = μ 0 I 2 π x × v ⇒  Net emf  = E 1 cos 60 o - E 2 cos 60 o × width = 1 2 × 2 100 × μ 0 I v 2 π 1 4 100 - 1 8 100 = 1 100 × 2 × 10 - 7 × 10 × v × 100 8 = 2 . 5 v × 10 - 7 Now, net EMF  = i × R ⇒   v = 10 × 10 - 6 × 0 . 1 2 . 5 × 10 - 7 = 4   m   s - 1

Practice more on Quantrex App →

Related: Physics — Electromagnetic Induction · All PYQ Banks