Question

The magnetic field at the point of intersection of diagonals of a square wire loop side $L$ carrying a current $I$ is

• A. $\frac{{\mu }_{0}I}{\pi L}$
• B. $\frac{{2\mu }_{0}I}{\pi L}$
• C. $\frac{\sqrt{2}{\mu }_{0}I}{\pi L}$
• D. $\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$

Answer 1

The correct option is D $\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$

The side length of the square loop, $AB=L$

$\therefore$ $BC=\frac{L}{2}$

From geometry, we get $\angle BOC={45}^o$

$\therefore$ In $△OCB$ $\Rightarrow \frac{BC}{x}=tan{45}^o$

$\Rightarrow \frac{\frac{L}{2}}{x}=1$

$\Rightarrow$ $x=\frac{L}{2}$

Magnetic field at O due to AB,

$I_o^{\prime }=\frac{{\mu }_o}{4\pi }\frac{I}{x}[sin{\theta }_1+sin{\theta }_2]$

where, ${\theta }_1={\theta }_2={45}^o$

$\Rightarrow I_o^{\prime }=\frac{{\mu }_o}{4\pi }\frac{I}{L/2}[sin{45}^o+sin{45}^o]$

$\Rightarrow I_o^{\prime }=\frac{{\mu }_o}{4\pi }\frac{2I}{L}[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}]$

$\Rightarrow I_o^{\prime }=\frac{{\mu }_{o}I}{\pi L\sqrt{2}}$

$\Rightarrow$ Net magnetic field at point O due to four such wires, $I_o=4I_o^{\prime }=\frac{{\mu }_{o}2\sqrt{2}I}{\pi L}$