JEE Physics 2025 Question with Solution

Given: Torque of the magnetic dipole is $\tau = 80\sqrt{3} \, \text{N m}$ and the angle with the magnetic field is $\theta = 60^\circ$.

Find: The potential energy $U$ of the dipole.

For a magnetic dipole in a uniform magnetic field,

$$\tau = MB \sin\theta$$

Substituting the given values,

$$80\sqrt{3} = MB \sin 60^\circ$$

Using $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

$$80\sqrt{3} = MB \left(\frac{\sqrt{3}}{2}\right)$$

So,

$$MB = \frac{80\sqrt{3} \times 2}{\sqrt{3}} = 160$$

Now the potential energy is

$$U = -MB \cos\theta$$

Substituting $MB = 160$ and $\theta = 60^\circ$,

$$U = -160 \cos 60^\circ$$

Using $\cos 60^\circ = \frac{1}{2}$,

$$U = -160 \left(\frac{1}{2}\right) = -80 \, \text{J}$$

Therefore, the potential energy of the dipole is $-80 \, \text{J}$. The correct option is D.

Given: $\tau = 80\sqrt{3} \, \text{N m}$ and $\theta = 60^\circ$.

Find: Potential energy $U$.

The torque relation is

$$\tau = mB\sin\theta$$

Hence,

$$80\sqrt{3} = mB\sin 60^\circ = mB \cdot \frac{\sqrt{3}}{2}$$

Therefore,

$$mB = \frac{80\sqrt{3}}{\sqrt{3}/2} = 160$$

The potential energy of a magnetic dipole is

$$U = -mB\cos\theta$$

So,

$$U = -160\cos 60^\circ$$ $$U = -160 \times \frac{1}{2} = -80 \, \text{J}$$

Thus, the required potential energy is $-80 \, \text{J}$.