Question

A length $L$ of wire carries a current $i$. If the wire is formed into a circular coil, then the maximum torque in a given magnetic field is developed when the coil has one turn only. Find the magnitude of maximum torque.

• A. $\frac{iLB}{2\pi }$
• B. $\frac{i{L}^{2}B}{\pi }$
• C. $\frac{i{L}^{2}B}{4{\pi }^2}$
• D. $\frac{i{L}^{2}B}{2\pi }$

Answer 1

The correct option is C $\frac{i{L}^{2}B}{4{\pi }^2}$

Given, Length of the wire is $L$.

Since, it is made into a circular coil, $L=2\pi R$

$\Rightarrow R=\frac{L}{2\pi }$

Magnetic moment of a circular current carrying loop is given by $\overrightarrow{\mu }=i\left(\pi R^2\right)\hat{n}$

$\therefore \overrightarrow{\mu }=i\left(\frac{L^2}{4{\pi }^2}\right)$

Torque acting on the circular loop is given by $$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$For maximum torque, $\theta ={90}^{\circ }$

Thus, $\tau =\frac{i{L}^{2}B}{4{\pi }^2}\sin {90}^{\circ }\Rightarrow \tau =\frac{i{L}^{2}B}{4{\pi }^2}$

Hence, option (c) is the correct answer.