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CBSE Class 12-science Answered

A short bar magnet of magnetic moment 0.32 JT-1 is placed in a uniform external magnetic field of 0.15 T. Id the bar is free to rotate in the plane of the field, which orientation would correspond to stable and unstable equilibrium. Find potential energy in each case.
Asked by bjayanta | 18 Aug, 2018, 09:23: PM
answered-by-expert Expert Answer
Magnetic potential energy = - m×B×cosθ
 
when θ=0 (aligned parallel, direction of magnetic dipole moment is same as that of field), it is stable equilibrium.
Hence potential energy in stable equlibrium = -m×B = -0.32 × 0.15 = -0.048 J
 
when θ=180 (aligned parallel, direction of magnetic dipole moment is opposite to field), it is unstable equilibrium.
Hence potential energy in unstable equilibrium  = m×B = 0.32 × 0.15 = 0.048 J



Answered by Thiyagarajan K | 20 Aug, 2018, 03:52: PM

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