Question

A current loop having two circular arcs joined by two radial line as shown in figure. It carries a current of $10\text{A}$. The magnetic field at point $\text{O}$ 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 $1.0\times {10}^{-5}\text{T}$

The field due to segments $\text{PQ, RS}$ at point $\text{O}$ is zero because for both wires segment, point $\text{O}$ is lying on the axis.

The magnetic field due to the arc $\text{PS}$ at $\text{O}$ is given by,

$B_{\text{PS}}=\frac{{\mu }_{0}I}{2R}\frac{\theta }{{360}^{\circ }}$

Here,$\theta ={45}^{\circ };R=3\text{cm}=3\times {10}^{-2}\text{m}$
$I=10\text{A}$

$B_{\text{PS}}=\frac{4\pi \times {10}^{-7}\times 10}{2(3\times {10}^{-2})}\times \frac{{45}^{\circ }}{{360}^{\circ }}$

$B_{\text{PS}}=\frac{\pi }{12}\times {10}^{-4}\text{T}⨀$

Similarly, magnetic field due to the arc $\text{QR}$ at $\text{O}$ is given by,

$B_{\text{QR}}=\frac{4\pi \times {10}^{-7}\times 10}{2(5\times {10}^{-2})}\times \frac{{45}^{\circ }}{{360}^{\circ }}$

$B_{\text{QR}}=\frac{\pi }{20}\times {10}^{-4}\text{T}⨂$

So, net magnetic field at point $\text{O}$ will be

$B_{net}=B_{\text{PS}}-B_{\text{QR}}$

(considering outward direction to the plane of paper is positive)

$B_{net}=[\frac{\pi }{12}\times {10}^{-4}-\frac{\pi }{20}\times {10}^{-4}]$

$\therefore B_{net}=1.04\times {10}^{-5}\text{T}⨀$

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