A square shaped coil of 10 cm side, having 100 turns, is placed perpendicular to a magnetic field which is increasing at the rate of 1.0 T/s. The induced emf in the coil would be

1.0 V

The correct answer is Option (2) → 1.0 V

Induced emf is given by Faraday’s law:

$\mathcal{E} = -N \frac{d\Phi}{dt}$

Magnetic flux: $\Phi = BA$

So, $\frac{d\Phi}{dt} = A \frac{dB}{dt}$

Here, $N = 100$, side = $0.1 \ \text{m}$ → $A = (0.1)^2 = 0.01 \ \text{m}^2$

$\frac{dB}{dt} = 1.0 \ \text{T/s}$

$\mathcal{E} = 100 \times 0.01 \times 1 = 1.0 \ \text{V}$

Induced emf = 1 V