Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is

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

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between the work done in rotating a magnet in a magnetic field and the torque exerted to maintain that position.

The work done ($W$) in turning the magnet from its equilibrium position by an angle $\theta$ in a magnetic field is given by the change in potential energy formula:

W = MB (1 - \cos(\theta))

where $M$ is the magnetic moment of the bar magnet, $B$ is the magnetic field, and $60^{\circ}$ is given as $\theta$.

To calculate the torque required to maintain this angular displacement, we use the formula for torque ($\tau$) in a magnetic field:

\tau = MB \sin(\theta)

We have:

\frac{W}{MB} = 1 - \cos(60^{\circ}) = 1 - \frac{1}{2} = \frac{1}{2}

Thus:

W = \frac{1}{2} MB

Now, substitute MB from above into the torque equation:

\tau = MB \sin(60^{\circ}) = MB \cdot \frac{\sqrt{3}}{2}

Substitute MB = \frac{2W}{1} from W = \frac{1}{2} MB, we have:

\tau = \frac{2W}{1} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} W

Thus, the torque required to maintain the magnet at a $60^{\circ}$ angle in the magnetic field is $\sqrt{3} W$.

This matches with the given correct answer: $\sqrt{3} W$.