Q. The magnetic field at the point of intersection of diagonals of a square wire 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) $\frac{2\sqrt{2}\mu_{0}\ I}{\pi L}$
Explanation
(d) $B = 4\left\lbrack \frac{\mu_{0}}{4\pi}\frac{I}{\left( \frac{1}{2} \right)}\ \left( \sin{45{^\circ}} + \sin{45{^\circ}} \right) \right\rbrack$
$= \frac{\mu_{0}}{4\pi}\frac{2I}{L}\ .\frac{2}{\sqrt{2}}\ $
$B = \frac{\mu_{0}}{\pi}\frac{2\sqrt{2}I}{L}$
[image:4_image108.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.