This problem involves calculating the work done on a bar magnet in an external magnetic field, using the given torque and magnetic field strength.

Understanding Torque and Magnetic Field

The torque ($\tau$) experienced by a magnetic dipole (like a bar magnet) in an external magnetic field ($B$) is given by the formula:

$ \tau = MB \sin \theta $

Where:

• $M$ is the magnetic dipole moment of the magnet.
• $B$ is the strength of the external magnetic field.
• $\theta$ is the angle between the axis of the magnet and the magnetic field.

We are given:

• $B = 800 \text{ Gauss}$
• $\theta = 30^\circ$
• $\tau = 0.016 \text{ N.m}$

First, convert the magnetic field from Gauss to Tesla (T):

$ 1 \text{ Gauss} = 10^{-4} \text{ T} $

$ B = 800 \times 10^{-4} \text{ T} = 0.08 \text{ T} $

Calculating Magnetic Moment (M)

Using the torque formula, we can find the magnetic dipole moment $M$:

$ 0.016 \text{ N.m} = M \times (0.08 \text{ T}) \times \sin(30^\circ) $

Since $\sin(30^\circ) = 0.5$:

$ 0.016 = M \times 0.08 \times 0.5 $

$ 0.016 = M \times 0.04 $

$ M = \frac{0.016}{0.04} = \frac{16}{40} = 0.4 \text{ A.m}^2 $

Calculating Work Done (W)

The work done ($W$) in moving a magnetic dipole from an angle $\theta_1$ to $\theta_2$ in a magnetic field is:

$ W = MB (\cos \theta_1 - \cos \theta_2) $

The most stable position corresponds to $\theta_1 = 0^\circ$ (parallel alignment).

The most unstable position corresponds to $\theta_2 = 180^\circ$ (anti-parallel alignment).

Substitute the values of $M$, $B$, $\theta_1$, and $\theta_2$:

$ W = (0.4 \text{ A.m}^2) \times (0.08 \text{ T}) \times (\cos 0^\circ - \cos 180^\circ) $

Since $\cos 0^\circ = 1$ and $\cos 180^\circ = -1$:

$ W = 0.032 \times (1 - (-1)) $

$ W = 0.032 \times (1 + 1) $

$ W = 0.032 \times 2 = 0.064 \text{ J} $

Finding the Value of $\alpha$

The problem states that the work done is $\alpha \times 10^{-3} \text{ J}$.

$ W = \alpha \times 10^{-3} \text{ J} $

Equating the calculated work done to the given expression:

$ 0.064 \text{ J} = \alpha \times 10^{-3} \text{ J} $

Solving for $\alpha$:

$ \alpha = \frac{0.064}{10^{-3}} = 0.064 \times 10^3 = 64 $

Therefore, the value of $\alpha$ is 64.