Question

The magnetic field at the point of intersection of diagonals of a square loop of 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

Answer: (D)

$\frac{2 \sqrt{2} \mu _{0} I}{\pi L}$

Answer 2

From Biot-Savart's law, magnetic field due to current carrying conductor is $B= \, \frac{\left(\mu \right)_{0}}{4 \pi }\frac{i}{r} \, \left(sin \left(\phi\right)_{1} + sin ? \left(\phi\right)_{2}\right)$ Where, $\phi_{1}=45^\circ ,\phi_{2}=45^\circ$ $B=\frac{\left(\mu \right)_{0}}{4 \pi } \, .\frac{i}{L/2}\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)=\frac{\left(\mu \right)_{0} i \sqrt{2}}{2 \pi L}$ Since there are four sides, the total magnetic field is $B_{net}=4B=\frac{2 \sqrt{2} \mu _{0} i}{\pi L}$