Question

As shown in the figure, a single conducting wire is bent to form a loop in the form of a circle of radius 'r' concentrically inside a square of side 'a', where a: r = 8 : $\pi$. A battery B drives a current through the wire. If the battery B and the gap G are of negligible sizes, determine the strength of magnetic field at the common centre O.

• A. $\frac{{\mu }_{0}I}{2\pi a}\sqrt{2}(\sqrt{2}-1)$
• B. $\frac{{\mu }_{0}I}{2\pi a}(\sqrt{2}+1)$
• C. $\frac{{\mu }_{0}I}{\pi a}2\sqrt{2}(\sqrt{2}+1)$
• D. $\frac{{\mu }_{0}I}{\pi a}2\sqrt{2}(\sqrt{2}-1)$

Answer 1

The correct option is D $\frac{{\mu }_{0}I}{\pi a}2\sqrt{2}(\sqrt{2}-1)$

For square wire
$B=\frac{{\mu }_0}{2\pi }\frac{I}{r}(sin\alpha +sin\beta )$

$B=\frac{{\mu }_0}{4\pi }\frac{I}{\frac{a}{2}}(sin{45}^{\circ }+sin{45}^{\circ })\text{for one wire}B=\frac{{\mu }_{0}I\sqrt{2}}{2\pi a}\text{for four wires}{B}_1=\frac{{\mu }_{0}I2\sqrt{2}}{\pi a}$(Out of the plane)

For circular wire
$B_2=\frac{{\mu }_{0}I}{2r}\Rightarrow \frac{{\mu }_{0}I\times 8}{2\pi a}\left(\text{Into the loop}\right)as\frac{a}{r}=\frac{8}{\pi }\Rightarrow r=\frac{\pi a}{8}NetB=B_2-B_1=\frac{{\mu }_{0}I}{\pi a}2\sqrt{2}(\sqrt{2}-1)$(Into the plane)