Question 38 of 276

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii ${R}_{1}=2\pi m$ and ${R}_{2}=4\pi m$ carrying current $I=4A$ as per figure given below is $\alpha \times {10}^{-7}T$. The value of $\alpha$ is _____. (Centre $O$ is common for all segments)

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Answer: 3

From the figure, it is evident that the magnetic field at $O$ due to the straight portion of the loop is zero, only the contributions from the circular parts survive.
Also, the direction of the magnetic field due to the circular loops can be determined by the right-hand corkscrew rule.
Hence, the net magnetic field at $O$ can be calculated as follows:
$\begin{matrix}B & = & \dfrac{{\mu }_{0}i}{2{R}_{2}}\left(\dfrac{\pi }{2\pi }\right)⊗+\dfrac{{\mu }_{0}i}{2{R}_{1}}\left(\dfrac{\pi }{2\pi }\right)⊗ \\ & = & \left(\dfrac{{\mu }_{0}i}{4{R}_{2}}+\dfrac{{\mu }_{0}i}{4{R}_{1}}\right)⊗ \\ & = & \dfrac{4\pi \times {10}^{-7}\times 4}{4\times 4\pi }+\dfrac{4\pi \times {10}^{-7}\times 4}{4\times 2\pi }⊗T \\ & = & 3\times {10}^{-7}⊗T\end{matrix}$
Hence, $\alpha =3$.