Question 11 of 153

A square loop of side $10cm$ and resistance $0.7\Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.20T$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $1s$ at a steady rate. Then, the magnitude of induced emf is $\sqrt{x}\times {10}^{-3}V$. The value of $x$ is _______.

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Answer: 2

The area vector for the square loop is given by
$\vec{A}=(0.1{)}^{2}\hat{j}{m}^{2}$
According to the above diagram, the magnetic field is given by
$\vec{B}=\dfrac{0.2}{\sqrt{2}}\hat{i}+\dfrac{0.2}{\sqrt{2}}\hat{j}$
Hence, the magnitude of induced emf can be calculated as follows
$\begin{matrix}e & = & \dfrac{\Delta \phi }{\Delta t} \\ & = & \dfrac{\vec{B}\cdot \vec{A}-0}{1} \\ & = & \dfrac{\left(\dfrac{0.2}{\sqrt{2}}\hat{i}+\dfrac{0.2}{\sqrt{2}}\hat{j}\right)\cdot \left((0.1{)}^{2}\hat{j}\right)-0}{1}V \\ & = & \sqrt{2}\times {10}^{-3}V\end{matrix}$
Hence, $x=2$.