A square loop of side $10\ \text{cm}$ and resistance $0.7\ \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.20\ \text{T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $1\ \text{s}$ at a steady rate. Then, the magnitude of induced emf is $\sqrt x\times10^{-3}\ \text{V}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
The loop lies in the east-west vertical plane, so its normal points north-south. The field (north-east) makes $45^\circ$ with the normal:
$$\Phi = BA\cos45^\circ = 0.20\times0.01\times\frac{1}{\sqrt2} = \sqrt2\times10^{-3}\ \text{Wb}$$
$$\varepsilon = \frac{\Delta\Phi}{\Delta t} = \sqrt2\times10^{-3}\ \text{V} \;\Rightarrow\; x = 2$$
Solution by Sreeraj P, M.Sc Physics