Question

An equilateral triangular loop of wire of side $2l$ carries a current $I$. The magnetic field produced at the centre of the loop is

• A. $\frac{{\mu }_0}{4\pi }\frac{3\sqrt{3}I}{l}$
• B. $\frac{{\mu }_0}{4\pi }\frac{18I}{l}$
• C. $\frac{{\mu }_0}{4\pi }\frac{6I}{l}$
• D. $\frac{{\mu }_0}{4\pi }\frac{9I}{l}$

Answer 1

Answer: (D)

The magnetic field at the centre $O$ due to current $I$ through one side $BC$ of the triangle is
$B=\frac{{\mu }_{0}I}{4\pi r}\left[sin{\varphi }_1+sin{\varphi }_2\right]$
Here,$r=OD=\frac{BD}{tan60^{\circ }}$
$\frac{l}{tan60^{\circ }}=\frac{l}{\sqrt{3}}$
${\varphi }_1={\varphi }_2=60^{\circ }$
$\therefore B=\frac{{\mu }_0}{4\pi }\frac{I}{\left(l\sqrt{3}\right)}\left(sin60^{\circ }+sin60^{\circ }\right)$
$=\frac{{\mu }_0}{4\pi }\frac{3I}{l}$
Since the direction of magnetic field at $O$ due to the current through all the three sides is same in magnitude and direction, hence total magnetic field at $O$ due to current through the triangle is
$B_O=3B$
$=\frac{{\mu }_0}{4\pi }\frac{9I}{l}$