Question

When a bar magnet is placed in a uniform magnetic field at an angle ${30}^{\circ }$ to the direction of field, it experiences a torque of $x\text{Nm}$. The same magnet experiences a torque of $\sqrt{3}\text{Nm}$ when it makes an angle of ${45}^{\circ }$ to the field. Find the value of $x\left(\text{in Nm}\right)$.

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}$

Let $M$ and $B$ be the magnetic moment and magnetic field strength, respectively.

Torque acting on the bar magnet in a uniform magnetic field is given by

$\tau =MB\sin \theta$

For same $M$ and $B$,

$\tau \propto \sin \theta$

$\frac{{\tau }_1}{{\tau }_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}.........\left(1\right)$

Given:
${\theta }_1={30}^{\circ };{\tau }_1=x\text{Nm}$
${\theta }_2={45}^{\circ };{\tau }_2=\sqrt{3}\text{Nm}$

So, equation $\left(1\right)$ becomes,

$\frac{x}{\sqrt{3}}=\frac{\sin {30}^{\circ }}{\sin {45}^{\circ }}$

$x=\sqrt{3}\times \frac{1/2}{1/\sqrt{2}}$

$\therefore x=\sqrt{\frac{3}{2}}$

Hence, option $\left(C\right)$ is correct.