Question

A square coil $20\text{cm}\times 20\text{cm}$ has $100$ turns and carries a current of $1\text{A}$. It is placed in a uniform magnetic field $B=0.5\text{T}$ with the direction of magnetic field parallel to the plane of the coil. The magnitude of the external torque required to hold this coil in this position is

• A. Zero
• B. $200\text{Nm}$
• C. $2\text{Nm}$
• D. $10\text{Nm}$

Answer 1

The correct option is C $2\text{Nm}$


$\overrightarrow{B}$ is parallel to plane of coil.

$\therefore$ Angle between magnetic field $\overrightarrow{B}$ and magnetic moment $\overrightarrow{\mu }={90}^{\circ }$

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

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

$=NiAB\sin \theta$

$\Rightarrow \tau =100\times 1\times (0.2\times 0.2)\times 0.5\times \sin {90}^{\circ }=2\text{Nm}$

External torque of $2\text{Nm}$ in opposite sense will be required to hold this coil in this position.