Question

A square wire of each side $l$ carries a current $I$ .The magnetic field at the mid point of the square is :

• A. $4\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$
• B. $8\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$
• C. $16\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$
• D. $32\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$

Answer 1

The correct option is B $8\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$

Magnetic field due to a conductor subtending angles. ${\theta }_{1}and{\theta }_2$at centre is
$B=\frac{{\mu }_0}{4A}\frac{i}{R}(\sin {\theta }_2-\sin {\theta }_{1)}$
field due to one side of square is $B=\frac{{\mu }_0}{4\pi }\frac{i}{(\frac{L}{2})}(\frac{1}{\sqrt{2}}-\left(\frac{-1}{\sqrt{2}}\right))$
as shortest distance $R=\frac{L}{2}and{\theta }_2={45}^o{\theta }_1=-{45}^o$
$B=2\sqrt{2}\frac{{\mu }_{o}i}{4\pi i}$
$\therefore B$ due to four sides $B_1=4B=8\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{l}$