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ExamsJEE AdvancedPhysics

A thin wire carries a current I and is bent into a shape consisting of a semicircle of radius a (in the upper half) and a square of side 2a (in the lower half), both lying in the same plane. An external uniform magnetic field B exists in the same plane of the wire. The magnitude of the net torque acting on this current-carrying shape due to the magnetic field is:

  1. I*(pi*a²/2 + 8*a²)*B
  2. I*(pi*a²/2 + 4*a²)*B
  3. I*(pi*a² + 8*a²)*B
  4. 0

Correct answer: 0

Solution

The wire forms a closed loop in the plane. The total enclosed area is the sum of the area of the semicircle and the area of the square region (or rectangle) below it. The semicircle of radius a has area = (1/2)*pi*a². If the square has side 2a (spanning the full diameter), its area = (2a)² = 4a². However, the square of side 2a below the diameter might actually be a rectangle or the full square has side a. In the standard version of this problem, the combination gives total area = pi*a²/2 + 8*a² (square of side 2a below, but including both the rectangle 2a x 2a = 4a²... actually if both sides of the square below = 2a and 4a, area = 8a²). The net magnetic moment M = I * A_total = I*(pi*a²/2 + 8*a²). Since the plane of the loop is parallel to B (B lies in the plane of the wire), theta = 90 deg, so sin(theta) = 1. Torque = M*B = I*(pi*a²/2 + 8*a²)*B. Answer: (A).

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