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JEE Main2014PhysicsMagnetic Effects of CurrentActual

A conductor lies along the z-axis at - 1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in - a ^ z direction (see figure). For a field B → = 3.0 × 1 0 - 4 e - 0.2x a ^ y T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 1 0 - 3 s . Assume parallel motion along the x-axis.

Options

  1. A1.57 W
  2. B2.97 W
  3. C14.85 W
  4. D29.7 W

Correct answer

B. 2.97 W

Step-by-step solution

Power P = W Δ t Force on conductor F → = I ( l → × b → ) F → = -10 ( 3 a ^ z ) × ( 3 × 10 − 4 e − 0.2 x a ^ y ) ⇒ F → = 90 × 10 - 4 ( e - 0.2 x ) along with x-axis work done on the conductor in moving along x-axis. W = ∫ x = 0 2 F → . d x → W = ∫ 0 2 Bi ℓ dx = ∫ 0 2 3 × 10 - 4 e - 0.2x × 3 × 10dx = 90 × 10 - 4 e - 0.2 x - 0.2 0 2 = 450 × 10 - 4 [ 1 - e - 0.4 ] = 180 × 10 - 4 ( Using e - x = 1 - x ) P = W Δ t = 180 × 10 - 4 5 × 10 - 3 = 3.6 W. Hence, closest option is 2.97 W. (if we used exact value of exponential t

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