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: Use relation between torque and rate of change of potential energy

For a magnetic dipole in a uniform magnetic field, torque is related to potential energy by:

τ = − dU / dθ

Hence, the work done in rotating the dipole between two positions is equal to the area under the torque–angle curve.

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Step 2: Find the magnetic interaction energy MB

Given torque at θ = 30°:

τ = MB sin θ

0.016 = MB × sin 30°

0.016 = MB × 1/2

MB = 0.032 J

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Step 3: Work done as area under τ–θ curve

The torque varies with angle as:

τ = MB sin θ

Work done by an external agent in rotating the magnet from θ = 0° to θ = 180° is:

W = ∫₀^π MB sin θ dθ

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Step 4: Perform the integration

W = MB [ −cos θ ]₀^π

W = MB [ −cos π + cos 0 ]

W = MB [ −(−1) + 1 ]

W = 2MB

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Step 5: Substitute value of MB

W = 2 × 0.032

W = 0.064 J

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Final Answer:

The work done by the external agent is
0.064 J