Animated Solution for Physics - Magnetic Effects of Current: A small bar magnet placed with its axis at $30^{\circ }$ with an external field of $0.06\text{ T}$ experiences a torque of $0.018\text{ Nm}$. The minimum work required to rotate it from its stable to unstable equilibrium position is

Imagine you are holding a small bar magnet inside a region where an invisible, uniform magnetic field is flowing steadily from left to right. If you hold the magnet at an angle to this field, you will immediately feel it trying to twist out of your hand. This twisting force is what physicists call torque.

In our specific problem, the bar magnet is placed such that its central axis makes an angle of $\theta =30^{\circ }$ with the external magnetic field, which has a strength of $B=0.06\text{ T}$. At this exact orientation, the magnet experiences a torque of $\tau =0.018\text{ Nm}$.

Our ultimate goal is to find the work required to rotate this magnet from its most comfortable position (stable equilibrium) to its most uncomfortable position (unstable equilibrium). But before we can calculate the work, we are missing a crucial piece of information: the intrinsic magnetic strength of the bar magnet itself, known as its magnetic dipole moment ($M$).

The torque experienced by a magnetic dipole in a uniform magnetic field is governed by the cross product of the magnetic moment and the magnetic field. The magnitude of this torque is given by the master equation:

Why is there a sine function? Because only the component of the magnetic field that is perpendicular to the magnet's axis actually contributes to the twisting motion. The parallel component just tries to stretch or compress the magnet, which doesn't cause rotation.

Let's substitute the values provided in the problem to uncover the hidden magnetic moment $M$:

We know that $\sin 30^{\circ }=\frac{1}{2}$. Plugging this in, we get:

Solving for $M$, we find:

Now we have the complete profile of our bar magnet. Its magnetic dipole moment is $0.6{\text{ A-m}}^2$.

Stable vs Unstable Equilibrium

When a magnet is placed in a magnetic field, it possesses potential energy based on its orientation. The formula for the potential energy $U$ of a magnetic dipole is:

Let's understand the two extreme states mentioned in the problem:

1. Stable Equilibrium: This occurs when the magnet aligns perfectly with the magnetic field. The angle ${\theta }_1=0^{\circ }$. Here, the potential energy is $U=-MB\cos 0^{\circ }=-MB$. This is the minimum possible energy state. It's like a ball resting at the bottom of a valley; if you nudge it, it will naturally oscillate and settle back down.

2. Unstable Equilibrium: This occurs when the magnet is forced to point exactly opposite to the magnetic field. The angle ${\theta }_2=180^{\circ }$. Here, the potential energy is $U=-MB\cos 180^{\circ }=+MB$. This is the maximum possible energy state. It's like balancing a ball perfectly on the peak of a hill; the slightest disturbance will cause it to violently flip around to the stable position.

The work done ($W$) by an external agent to rotate the magnet slowly from an initial angle ${\theta }_1$ to a final angle ${\theta }_2$ is exactly equal to the change in its potential energy:

We need to rotate the magnet from stable (${\theta }_1=0^{\circ }$) to unstable (${\theta }_2=180^{\circ }$) equilibrium. Let's substitute our known values into the work equation:

Be very careful with the minus signs here. We know $\cos 0^{\circ }=1$ and $\cos 180^{\circ }=-1$:

To match the format of the given options, we convert this into scientific notation:

This is the minimum energy you must expend to force the magnet to point completely against the magnetic field. The correct option is (b).