Magnetic field changed at the rate of $0.4 T s^{- 1}$ in a square coil of side $4 c m$ . Kept perpendicular to the field. If the resistance of the coil is $2 \times 10^{-} 3 Ω$ , then induced current in coil is:

0.16 A

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0.32 A

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3.2 A

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1.6 A

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Solution

The correct option is B $0.32 A$
According to Faraday's Law the relation is as follows

$i R = - A \frac{d B}{d t}$

$i = \frac{- A}{R} \frac{d B}{d t}$

Using all the given values

$\frac{d B}{d t} = 0.4 T / sec$

$R = 2 \times 10^{- 3} Ω$

$i = \frac{16 \times 10^{- 4} \times 0.4}{2 \times 10^{- 3}}$

$i = 0.32 A$

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