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A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5×10⁻² J. The magnitude of magnetic moment of the magnet is ____ J T⁻¹.

Asked in GSEB Board March 2022 · Dipole in a uniform field

• (1) 0.72
• (2) 0.18
• (3) 0.36
• (4) 0.54

Step-by-step solution

Why the other options are wrong

• (1) Twice the answer: it comes from putting the field's own value into the sine's place, sin 30°→0.25, so that (4.5×10⁻²)/(0.25×0.25)=0.72.
• (2) Leaves out sin 30° altogether: (4.5×10⁻²)/(0.25)=0.18. That is the moment this torque would need if the magnet lay across the field, at 90°; at 30° the torque is halved, so the moment must be twice as large.
• (4) No single slip produces 0.54; it is 1.5 times the answer, and neither the field, the angle nor the torque supplies a factor 1.5. Sanity check: m=τ/(B sin 30°)=(4.5×10⁻²)/(0.125)=0.36 J T⁻¹.

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