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 $10 m s^{- 1}$ (see figure). The e.m.f induced at the time the left arm of the frame is at $x = 10 c m$ from the wire is:

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$0.5 μ V$

$2 μ V$

$1 μ V$

$0.75 μ V$

answer is B.

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Detailed Solution

Time in which left arm reach at $x = 10 c m$

$t = \frac{0.1}{10} = \frac{1}{100} = 0.01 \sec, r_{1} = 10 \times 0.01 = 0.1 m$

$e = l V (B_{1} - B_{2}) = 0.1 \times 10 (\frac{μ_{0} i}{2 π r_{1}} - \frac{μ_{0} i}{2 π r_{2}}) = 0.1 \times 10 \times 2 \times 10^{- 7} \times 1 (\frac{1}{0.1} - \frac{1}{2 \times 0.1})$

$= 2 \times 10^{- 7} (\frac{2 - 1}{2 \times 0.1}) = 10^{- 6} V$

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