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)

$\frac{\mu_{0}}{4\pi} \frac{9I}{l}$

Answer 2

The magnetic field at the centre $O$ due to current $I$ through one side $BC$ of the triangle is $B = \frac{\mu_{0}I}{4\pi r} \left[ sin\phi_{1} + sin \phi_{2}\right]$ Here,$r = OD = \frac{BD}{tan \,60^{\circ}}$ $\frac{l}{tan \,60^{\circ}} = \frac{l}{\sqrt{3}}$ $\phi_{1}=\phi_{2}= 60^{\circ}$ $\therefore B= \frac{\mu_{0}}{4\pi} \frac{I}{\left(l \sqrt{3}\right)} \left(sin\, 60^{\circ} +sin \,60^{\circ}\right)$ $= \frac{\mu_{0}}{4\pi} \frac{3I}{l}$ 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 $B_O = 3B$ $= \frac{\mu_{0}}{4\pi} \frac{9I}{l}$