Question

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

• A. $2.0\times {10}^{-7}T$
• B. $1.5\times {10}^{-5}T$
• C. $1.0\times {10}^{-7}T$
• D. $1.0\times {10}^{-5}T$

Answer 1

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


Magnetic field due to wires $PQ$ and $RS$ will be zero at point $O$ as $O$ lies on the axis of both of these wires.

Magnetic field at centre due to a circular arc forming angle $\theta$ at centre is given by

$\underset{B}{\to }=\frac{{\mu }_{0}i}{4\pi r}$ $\theta$

where $\theta$ is in radian, $r=$ radius of arc

Field by arc $QR$ at centre $O$,

${\underset{B}{\to }}_{QR}=\frac{{\mu }_{0}i}{4\pi r_1}\left(\frac{\pi }{4}\right)$ $⨀$ (out of the plane)

$=\frac{{\mu }_{0}i}{4\pi \left(0.03\right)}\left(\frac{\pi }{4}\right)$ $⨀$

${\underset{B}{\to }}_{PS}=\frac{{\mu }_{0}i\left(\frac{\pi }{4}\right)}{4\pi \left(0.05\right)}$ $⨂$ (into the plane)

At centre $O$,

${\underset{B}{\to }}_{net}=$ ${\underset{B}{\to }}_{QR}-$ ${\underset{B}{\to }}_{PS}$

(taking out of the plane as positive)

${\underset{B}{\to }}_{net}=\frac{4\pi \times {10}^{-7}}{16}\times 10[\frac{1}{0.03}-\frac{1}{0.05}]$

${\underset{B}{\to }}_{net}=\frac{\pi }{3}\times {10}^{-5}$ $T$

$\approx 1\times {10}^{-5}$ $T$

Hence, option $\left(a\right)$ is correct