A short bar magnet placed with its axis at 30° with an external field of 800 Gauss, experiences a torque of 0.016 N·m. The work done in moving it from most stable to most unstable position is α × 10⁻³ J. The value of α is ___

A short bar magnet placed with its axis at 30° with an external field of 800 Gauss experiences a torque of 0.016 N m

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

Correct Answer: 64

Explanation (Complete Step by Step Calculation)

Torque on a magnetic dipole placed in a uniform magnetic field is given by:

\[ \tau = m B \sin\theta \]

Given torque:

\[ \tau = 0.016 \, \text{N·m} \]

Magnetic field:

\[ B = 800 \, \text{Gauss} = 800 \times 10^{-4} = 0.08 \, \text{T} \]

Angle:

\[ \theta = 30^\circ, \quad \sin 30^\circ = \frac{1}{2} \]

Magnetic dipole moment:

\[ m = \frac{\tau}{B \sin\theta} \]

\[ m = \frac{0.016}{0.08 \times 0.5} \]

\[ m = \frac{0.016}{0.04} = 0.4 \, \text{A·m}^2 \]

Magnetic potential energy of a dipole in a magnetic field is:

\[ U = -mB\cos\theta \]

At most stable position, \( \theta = 0^\circ \):

\[ U_{\text{min}} = -mB \]

At most unstable position, \( \theta = 180^\circ \):

\[ U_{\text{max}} = +mB \]

Work done in moving from stable to unstable position:

\[ W = U_{\text{max}} - U_{\text{min}} = 2mB \]

\[ W = 2 \times 0.4 \times 0.08 \]

\[ W = 0.064 \, \text{J} = 64 \times 10^{-3} \, \text{J} \]

Comparing with \( \alpha \times 10^{-3} \) J:

\[ \alpha = 64 \]

Hence, the value of \( \alpha \) is 64.

Explanation (Complete Step by Step Calculation)

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