Question

A wire of length $l=2\text{m}$ is bent to form a circular coil having some turns. A current $i=10\text{A}$ is established in the coil, and it is placed in a uniform magnetic field $B=10\pi \text{mT}$. The maximum torque that acts on the coil is

• A. $0.4\text{Nm}$
• B. $0.1\text{Nm}$
• C. $0.2\text{Nm}$
• D. Zero

Answer 1

The correct option is B $0.1\text{Nm}$

Let the number of turns be $n$, then,

$l=n\left(2\pi r\right)$

$\Rightarrow r=\frac{l}{2\pi n}$

Using, $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$,

$\tau =\mu B\sin \theta$

$=niAB\sin \theta$

$=ni\left(\pi r^2\right)B\sin \theta =ni\times (\pi \times \frac{l^2}{4{\pi }^2{n}^2})\times B\sin \theta =\frac{i{l}^{2}B\sin \theta }{4\pi n}$

For maximum torque, $\sin \theta =1$ and $n=1$

So, ${\tau }_{max}=\frac{iB{l}^2}{4\pi }$

Substituting the data given in the question,

${\tau }_{max}=\frac{10\times 10\pi \times {10}^{-3}\times (2{)}^2}{4\pi }$

$\Rightarrow {\tau }_{max}=0.1\text{Nm}$

Hence, option (b) is the correct answer.