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JEE Advanced Physics Magnetic Effects of Current 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Physics Question (2022) — Solution

Question

A small circular loop of area A  and resistance R  is fixed on a horizontal x y -plane with the center of the loop always on the axis n ^ of a long solenoid. The solenoid has m  turns per unit length and carries current I  counter clockwise as shown in the figure. The magnetic field due to the solenoid is in n ^ direction. List-I gives time dependences of n ^ in terms of a constant angular frequency ω . List-II gives the torques experienced by the circular loop at time t = π 6 ω , Let α = A 2 μ 0 2 m 2 I 2 ω 2 R .   List-I   List-II (i) 1 2 sin ω t j ^ + cos ω t k ^ (p) 0 (ii) 1 2 sin ω t i ^ + cos ω t j ^ (q) - α 4 i ^ (iii) 1 2 sin ω t i ^ + cos ω t k ^ (r) 3 α 4 i ^ (iv) 1 2 cos ω t j ^ + sin ω t k ^ (s) α 4 j ^     (t) - 3 α 4 i ^ Which one of the following options is correct?

Options

  1. A. i → q , ii → p , iii → s , iv → t
  2. B. i → s , ii → t , iii → q , iv → p
  3. C. i → q , ii → p , iii → s , iv → r
  4. D. i → t , ii → q , iii → p , iv → r

Answer

C. i → q , ii → p , iii → s , iv → r

Step-by-step solution

Note: This question was given bonus by JEE council. The magnetic field in a solenoid is given by,  B → = μ 0 m I n ^ . Therefore, the flux through the loop will be, ϕ = A k ^ · μ 0 m I n ^ For case I: ϕ = B A 2 cos ω t The emf will be, ε = B A ω 2 sin ω t Therefore, the current in the loop will be, i = B A ω 2 R sin ω t The magnetic moment of the loop will be, m → = i A k ^ = B A 2 ω 2 R sin ω t k ^ Therefore, the torque τ → = m → × B → = B 2 A 2 ω 2 R sin ω t k ^ × n ^ ⇒ τ → = - B 2 A 2 ω 2 R i ^ sin 2 ω t ⇒ τ → = - B 2 A 2 ω 2 R sin 2 π 6 = - α 4 i ^ Hence,  I → q For case II ϕ = 0 Therefore, we can directly write  τ = 0 Hence  II → p Following the same process as in case I for case III ϕ = B A 2 cos ω t i = B A ω 2 R sin ω t m → = B A 2 ω 2 R sin ω t k ^ τ → = m → × B → = B 2 A 2 ω 2 × 2 R sin ω t k ^ × sin ω t i ^ + cos ω t k ^ τ → = B 2 A 2 ω sin ω t 2 R sin ω t j ^ τ → = B 2 A 2 ω 2 R sin 2 ω t j ^ τ → = α 4 j ^ Hence,  III → s Following the same process as in case I for case IV ϕ = B A 2 sin ω t i = - B A ω 2 R cos ω t m → = - B A 2 ω 2 R cos ω t k ^ τ → = m → × B → = - B 2 A 2 ω 2 R k ^ × j ^ cos 2 ω t τ → = + B 2 A 2 ω 2 R i ^ · cos 2 π 6 τ → = + 3 4 α i ^ Hence,  IV → r

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