Question

If the plane of coil and uniform magnetic field $B(=4\text{T})$ is in the same plane, then torque on the current carrying coil is

• A. $4\pi R^{2}i$
• B. $2\pi R^{2}i$
• C. $\pi R^{2}i$
• D. Zero

Answer 1

The correct option is A $4\pi R^{2}i$

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

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

$=NiAB\sin \theta$

Here, the magnetic field and the coil are on the same plane. So, the angle $\theta$ will be ${90}^{\circ }$ as the magnetic moment vector in this case is always perpendicular to the magnetic field.

Note: The angle ${30}^{\circ }$ is given to confuse you.

$\Rightarrow \tau =1\times i\times \pi R^2\times 4\times \sin {90}^{\circ }=4\pi R^{2}i$