Bar Magnet As A Magnetic Dipole
A bar magnet of magnetic moment 0.8 A·m2 is held at 30∘ to a uniform magnetic field of 0.25 T; what is the magnitude of the torque experienced by the magnet?
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About This Question
- Subject
- physics
- Chapter
- magnetic effects of current and magnetism
- Topic
- bar magnet as a magnetic dipole
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
0.1 N·m
NCERT Class 12, Chapter 5 (Magnetism and Matter) treats a bar magnet as a magnetic dipole of moment m that, in a uniform field B, experiences a torque τ=mBsinθ, where θ is the angle between the dipole moment and the field. This torque tends to align the magnet with the field, analogous to an electric dipole in an electric field. Substituting m=0.8 A·m2, B=0.25 T, θ=30∘ so sin30∘=0.5: τ=(0.8)(0.25)(0.5)=0.1 N·m. The value 0.2 N·m wrongly uses sinθ=1. The value 0.17 N·m uses sin60∘ by confusing the angle. The value 0.05 N·m halves the moment. Plausibility check: torque is maximum at 90∘ and zero at 0∘, so a value at 30∘ should be well below the maximum of 0.2 N·m, and 0.1 N·m being exactly half fits sin30∘=0.5 perfectly.
This easy difficulty physics question is from the chapter magnetic effects of current and magnetism, covering the topic of bar magnet as a magnetic dipole. It appeared in the 2025 exam.
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