Answer to Question #310288 in Electricity and Magnetism for Manisha

Question

A square coil RSTU of sides 5.0 cm carrying a current, I of 2.5 A is placed in a uniform magnetic field, B of 2.4 T.Calculate the torque on the coil.

Answer 1

We know that

Magnetic moment

$M=NIA$

$M=NIa^2$

$$M=1\times2.5\times0.05^2=6.25\times10^{-3}Am^2$$

Torque

$\tau=M\times B$

$\tau=MBsin\theta$

$\tau=MBsin90°$

$$\tau=MB=6.25\times10^{-3}\times2.4=0.015Nm$$