Q.

A short bar magnet of magnetic moment M=0.32JT1 is placed in a uniform magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case 4.8×10xJ, 4.8×10xJ  then x is

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answer is 2.

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Detailed Solution

 (a) θ=0, Since U=MBcosθU=4.8×102J 

(bθ=180, since U=MB cos180U=4.8×102J

x=2, & x=2

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