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JEE Main 2014 (19 Apr, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2014 (19 Apr, Shift 1)

Q53JEE Main 2014 · 19 APR · SHIFT 1Difficult

Motional EMF

Figure

A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 ms−1{}^{-1} (see figure). The e.m.f induced at the time the left arm of the frame is at x = 10 cm from the wire is:

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Answer: (b)

  1. Left arm at x1=0.1 mx_1 = 0.1\text{ m}, right arm at x2=0.2 mx_2 = 0.2\text{ m}; their emfs oppose.
  2. e=μ0I2π(1x1−1x2)lve = \dfrac{\mu_0 I}{2\pi}\left(\dfrac{1}{x_1} - \dfrac{1}{x_2}\right) l v
  3. e=2×10−7×1×(10−5)×0.1×10=1×10−6 Ve = 2 \times 10^{-7} \times 1 \times (10 - 5) \times 0.1 \times 10 = 1 \times 10^{-6}\text{ V}
  4. So the answer is (b) 1 μ\muV.
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