Magnetism and Matter · Class 12 · CBSE · Science Physics

A short bar magnet of magnetic moment M is placed with its axis making an angle of 30^circ with a uniform magnetic field of magnitude B. The potentia…

Practice question from Magnetism and Matter (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

easy · MCQ · Q-F1B84A

A short bar magnet of magnetic moment MM is placed with its axis making an angle of 30∘30^\circ with a uniform magnetic field of magnitude BB. The potential energy of the magnet is:

  1. A.-MB/2Correct answer
  2. B.-MB \sqrt{3}/2
  3. C.MB/2
  4. D.MB \sqrt{3}/2

Solution

The potential energy of a magnetic dipole in a uniform magnetic field is given by U=−M⋅B=−MBcos⁡θU = -M \cdot B = -MB \cos\theta. Substituting θ=30∘\theta = 30^\circ, we get U=−MBcos⁡(30∘)=−MB(3/2)U = -MB \cos(30^\circ) = -MB (\sqrt{3}/2) which is incorrect; wait, cos⁡(30∘)=3/2\cos(30^\circ) = \sqrt{3}/2, so U=−32MBU = -\frac{\sqrt{3}}{2}MB. Let's re-verify options: option A is −MB/2-MB/2 (corresponding to 60∘60^\circ), option B is −32MB-\frac{\sqrt{3}}{2}MB. Thus, correct answer is B.

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