Question

A short bar magnet placed with its axis at $30^\circ$ with an external magnetic field of $800$ Gauss experiences a torque of $0.016\,\text{N m}$. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3}\,\text{J}$. The value of $\alpha$ is ___.

Hint:

Work done in rotating a magnetic dipole from stable to unstable position is always $2MB$.

Answer 1

Correct Answer: 64

Step 1: Converting magnetic field into SI units.
\[ B = 800\,\text{Gauss} = 800 \times 10^{-4} = 0.08\,\text{T} \] Step 2: Using torque formula for a magnetic dipole.
\[ \tau = MB\sin\theta \] \[ 0.016 = M \times 0.08 \times \sin 30^\circ \] \[ 0.016 = M \times 0.08 \times \frac{1}{2} \] \[ M = 0.4\,\text{A m}^2 \] Step 3: Work done in rotating the dipole.
Work done in rotating a magnetic dipole from stable ($0^\circ$) to unstable ($180^\circ$) position is:
\[ W = 2MB \] Step 4: Substituting values.
\[ W = 2 \times 0.4 \times 0.08 = 0.064\,\text{J} \] Step 5: Writing in required form.
\[ 0.064 = 64 \times 10^{-3}\,\text{J} \] Step 6: Final conclusion.
The value of $\alpha$ is $64$.