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Q.

A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.4 sec. If magnetic flux between the pole pieces is known to be 8 × 10⁴Wb, then induced emf in the wire, is

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a

2 × 10^–3V

b

6 × 10^–3V

c

8 × 10^–3V

d

4 × 10^–3 V

answer is C.

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Detailed Solution

$Emf (e) = \frac{d ϕ}{dt} = \frac{8 \times 10^{- 4}}{0.4} = 2 \times 10^{- 3} V$

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