A circular coil of area $2 \sqrt{2} \mathrm{~cm}^2$ and resistance $2 \boldsymbol{\Omega}$ is arranged vertically in the east -west direction. A uniform magnetic field 0.2 T is set up across the plane in the north to east direction. Now the magnetic field is removed at a steady rate in 0.4 s What is the current developed in the coil?

The area of the coil is $A = 2 \sqrt{2} \text{ cm}^2 = 2 \sqrt{2} \times 10^{-4} \text{ m}^2$. The magnetic field $B = 0.2 \text{ T}$ is in the north-east direction, while the coil is in the east-west plane. The normal to the coil points in the north-south direction.

The angle $\theta$ between the magnetic field (north-east) and the normal to the coil (north or south) is $45^{\circ}$.

The initial magnetic flux $\Phi_i$ through the coil is given by $\Phi_i = B A \cos(45^{\circ}) = 0.2 \times (2 \sqrt{2} \times 10^{-4}) \times \dfrac{1}{\sqrt{2}} = 0.4 \times 10^{-4} \text{ Wb}$.

The final magnetic flux $\Phi_f$ is $0$ since the magnetic field is removed. The change in flux is $\Delta \Phi = |\Phi_f - \Phi_i| = 0.4 \times 10^{-4} \text{ Wb}$.

The induced electromotive force (emf) is $\varepsilon = \dfrac{\Delta \Phi}{\Delta t} = \dfrac{0.4 \times 10^{-4}}{0.4} = 10^{-4} \text{ V}$.

The induced current $I$ is given by $I = \dfrac{\varepsilon}{R} = \dfrac{10^{-4}}{2} = 0.5 \times 10^{-4} \text{ A}$.

Answer: $0.5 \times 10^{-4} A$