JEE Main 2022 Physics question
Not in JEE Advanced syllabus
A bar magnetic having a magnetic moment of , is placed along the direction of uniform magnetic field of magnitude T. The work done in rotating the magnet slowly through 60° from the direction of field is:
Answer: (1)
Solution
, so − = −MB cos60° + MB = MB(1 − cos60°) = MB/2 = $(2 ×
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