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A circular coil of radius 10 cm, 500 turns and resistance 2 $Ω$ is placed with its plane, perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 90° in 0.25 s. The induced e.m.f. in the coil is $(T a k e H_{E} = 3.0 x 10^{- 5} T) :$

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$1.9 x 10^{- 3} V$

$2.6 x 10^{- 2} V$

$3.8 x 10^{- 3} V$

$1.4 x 10^{- 2} V$

answer is D.

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Detailed Solution

$\begin{array}{l} ε = - \frac{d ϕ_{B}}{d t} \\ = \frac{N B A}{t} \\ = \frac{500 \times 3 \times 10^{- 5} \times 3.14 \times 0.1 \times 0.1}{0.25} \\ = 18.84 \times 10^{- 4} \\ = 1.9 \times 10^{- 3} V \end{array}$

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