Question

A rectangular coil of wire of area $400c{m}^2$ contains $500$ turns. It is placed in a magnetic field of induction $4\times {10}^{-3}T$ and it makes an angle ${60}^o$ with the field. A current of $0.2A$ is passed through it. The torque on the coil is :

• A. $8\sqrt{3}\times {10}^{-3}N-m$
• B. $8\times {10}^{-3}N-m$
• C. $8\sqrt{3}\times {10}^{-4}N-m$
• D. $8\times {10}^{-4}N-m$

Answer 1

The correct option is B $8\times {10}^{-3}N-m$

$A=400\times {10}^{-4}{m}^2$
$n=500$
$B=4\times {10}^{-3}$
Area vector makes complementary angle with field therefor, angle between magnetic field and normal of coil is $\theta ={90}^o-{60}^o={30}^o$
$i=0.2$

$\tau =$$\overrightarrow{M}\times \overrightarrow{B}$
$=niA$$B\sin \theta$
$=500\times 0.2\times 400\times {10}^{-4}\times 4\times {10}^{-3}\times \frac{1}{2}$
$=8\times {10}^{-3}Nm$