Question

Find the magnetic field $B$ at the center of square loop of side $a$ carrying a current $I$.

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

Answer 1

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

Since $\text{ABCD}$ is a square, so

Length of $\text{PE}=\frac{a}{2}$

Now, calculating magnetic field due to $\text{AB}$ part at centre $\text{E}$,

$B_{AB}=\frac{{\mu }_0}{4\pi }\frac{I}{\text{PE}}[\sin {\theta }_1+\sin {\theta }_2]$

$B_{AB}=\frac{{\mu }_0}{4\pi }\frac{I}{\left(a/2\right)}[\sin {45}^{\circ }+\sin {45}^{\circ }]$

$B_{AB}=\frac{{\mu }_0}{4\pi }\frac{2I}{a}\frac{2}{\sqrt{2}}$

Since all sides of square create same magnetic field at the centre (direction will also same), so total magnetic field $B$ will be

$B=4\times B_{AB}=4\times \frac{{\mu }_0}{4\pi }\frac{2I}{a}\frac{2}{\sqrt{2}}$

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{I}{a}\left(8\sqrt{2}\right)$

Hence, option $\left(b\right)$ is correct.