JEE Main 2026 — Magnetic Effects of Current Question with Solution
JEE Main 2026 (24 January Shift 1)
Question
A short bar magnet placed with its axis at with an external field of 800 Gauss, experiences a torque of . The work done in moving it from most stable to most unstable position is . The value of is .
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Show full solutionCorrect answer: 64
Correct answer
64
Step-by-step explanation
For a magnetic dipole in an external field, torque is
At 30°:
A·m
Work done from most stable (aligned, 0°) to most unstable (anti-aligned, 180°) position equals the change in potential energy:
J J
Therefore,
At 30°:
A·m
Work done from most stable (aligned, 0°) to most unstable (anti-aligned, 180°) position equals the change in potential energy:
J J
Therefore,
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