Question

A bar magnet is kept such that it is making an angle of 30$^\circ$ with the magnetic field. The torque acting on the magnet is 0.016 N-m. Find the amount of work done by external agent in rotating the magnet from most stable position to most unstable position.

• A. 0.064 J
• B. 0.020 J
• C. 0.034 J
• D. 0.055 J

Hint:

Stable equilibrium is always at $\theta = 0^\circ$ and unstable at $\theta = 180^\circ$. The work done to flip a dipole end-to-end is always $2MB$. If you have the torque at $30^\circ$, which is $\tau = MB/2$, you can directly see that $MB = 2\tau$, and the work done is $2MB = 4\tau = 4 \times 0.016 = 0.064$ J without writing down the formulas.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A bar magnet experiences a torque when placed in a uniform magnetic field at a given angle. We need to use this information to find the magnetic moment or the product $\mu B$ (or MB). Then, we must calculate the work done to rotate the magnet from its most stable equilibrium to its most unstable equilibrium.
Step 2: Key Formula or Approach:
1. Torque on a magnetic dipole: $\tau = M B \sin\theta$, where $M$ is the magnetic moment.
2. Potential energy of a magnetic dipole: $U = -M B \cos\theta$.
3. Most stable position is at $\theta_{1} = 0^\circ$ (Minimum PE).
4. Most unstable position is at $\theta_{2} = 180^\circ$ (Maximum PE).
5. Work done by external agent: $W_{ext} = \Delta U = U_{f} - U_{i}$.
Step 3: Detailed Explanation:
Given:
Initial angle, $\theta = 30^\circ$
Torque, $\tau = 0.016 \text{ N-m}$
Using the torque formula to find $MB$:
$\tau = M B \sin 30^\circ$
$0.016 = M B \times \frac{1}{2}$
$MB = 0.032 \text{ J}$
Now, calculate the work done to rotate the magnet from $\theta_{1} = 0^\circ$ (most stable) to $\theta_{2} = 180^\circ$ (most unstable).
$W = U_{f} - U_{i} = (-M B \cos 180^\circ) - (-M B \cos 0^\circ)$
Since $\cos 180^\circ = -1$ and $\cos 0^\circ = 1$:
$W = (-M B \times -1) - (-M B \times 1)$
$W = M B + M B = 2MB$
Substitute the value of MB:
$W = 2 \times 0.032 \text{ J}$
$W = 0.064 \text{ J}$
Step 4: Final Answer:
The amount of work done by the external agent is 0.064 J.