Question

A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of $10\text{A}$. The magnetic field at point $O$ will be close to:

• A. $1.0\times {10}^{-5}\text{T}$
• B. $1.0\times {10}^{-7}\text{T}$
• C. $1.5\times {10}^{-7}\text{T}$
• D. $1.5\times {10}^{-5}\text{T}$

Answer 1

The correct option is A $1.0\times {10}^{-5}\text{T}$


There will be no magnetic field at $O$ due to wire $PQ$ and $RS$ as the point $O$ lies along $PQ$ and $RS$

The Magnetic field at $O$ due to arc $QR$,

${\overrightarrow{B}}_1=\frac{{\mu }_0}{4\pi }\frac{\left(\frac{\pi }{4}\right)i}{r_1}\odot$

The Magnetic field at $O$ due to arc $PS$

${\overrightarrow{B}}_2=\frac{{\mu }_0}{4\pi }\frac{\left(\frac{\pi }{4}\right)i}{r_2}\otimes$

$\therefore$ Net magnetic field at $O$

$|\overrightarrow{B}|=B_1-B_2$

$\Rightarrow |\overrightarrow{B}|=\frac{{\mu }_0}{4\pi }\left(\frac{\pi }{4}\right)\times 10[\frac{1}{(3\times {10}^{-2})}-\frac{1}{(5\times {10}^{-2})}]$

$\Rightarrow |\overrightarrow{B}|={10}^{-7}\times \left(\frac{\pi }{4}\right)\times 10\times \frac{2\times {10}^{-2}}{15\times {10}^{-4}}$

$\Rightarrow |\overrightarrow{B}|=\frac{\pi }{3}\times {10}^{-5}\text{T}\approx 1\times {10}^{-5}\text{T}$

Hence, option $\left(D\right)$ is correct.