Question

A wire loop $\text{PQRSP}$ have two semicircular coils of radii $R_1=0.5\text{m}$ and $R_2=1\text{m}$ as shown in the figure. Current $I=1\text{A}$ is flowing in the loop. The net magnetic induction at point $\text{C}$ is

• A. $\pi \times {10}^{-6}\text{T}$
• B. $2\pi \times {10}^{-7}\text{T}$
• C. $\pi \times {10}^{-7}\text{T}$
• D. $5\pi \times {10}^{-6}\text{T}$

Answer 1

The correct option is C $\pi \times {10}^{-7}\text{T}$

Magnetic field due to a circular arc at its centre is given by,

$B=\frac{{\mu }_{0}I}{2R}\left(\frac{\theta }{2\pi }\right)$

Where,

$\theta$ is the angle subtended by the circular arc at its center.

From the given diagram, we can see that the magnetic field produced by both the arcs will be opposite to each other due to the opposite direction of the current through each arc.

Magnetic field due to the wires $\text{RS}$ and $\text{PQ}$ will be zero, because point $\text{C}$ lies on their linear extension.

Now,

$B_1=\frac{{\mu }_{0}I}{2R_1}\left[\frac{\pi }{2\pi }\right]\Rightarrow B_1=\frac{{\mu }_{0}I}{4R_1}(\odot )$

$B_2=\frac{{\mu }_{0}I}{2R_2}\left[\frac{\pi }{2\pi }\right]\Rightarrow B_2=\frac{{\mu }_{0}I}{4R_2}(\otimes )$

$\therefore B_{\text{net}}=B_1-B_2=\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}-\frac{1}{R_2})$

Substituting the values,

$B_{net}=\frac{4\pi \times {10}^{-7}\times 1}{4}(\frac{1}{0.5}-\frac{1}{1})$

$=\pi \times {10}^{-7}\text{T}$

Hence, $\left(C\right)$ is the correct answer.