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If in a coil rate of change of area is $\frac{5 m^{2}}{m i l l i s e c o n d}$ and current become 1 amp from 2 amp in 2 × 10^–3 sec. If magnetic field is 1 T then self inductance of the coil is

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2 H

20 H

5 H

10 H

answer is D.

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Detailed Solution

$e = B.\frac{{dA}}{{dt}} = L\frac{{di}}{{dt}} \Rightarrow 1 \times \frac{5}{{{{10}^{ - 3}}}} = L \times \frac{{(2 - 1)}}{{2 \times {{10}^{ - 3}}}} \Rightarrow L = {\text{ }}10H$

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