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A short bar magnet of magnetic moment m=0.32 JT^(-1) is placed in a uniform magnetic field of 0.15 T.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 magnetic in each case? |
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Answer» Solution :(i) For stable equilibrium of bar magnet placed in external UNIFORM magnetic field,weshould have `OVERSET(to)(m)|| overset(to) (B) rArr` angle between `overset(to) and overset(to) (B)` should be `theta=0^@`. Now, in this case, magnetostatic potential energy is given by formula, `U= - m B cos theta` `therefore U_("min") = - mB (because theta=0^@ rArr cos theta =1)` `therefore U_("min") = - (0.32) (0.15) ` `therefore U_("min") = -0.048` J (ii) For unstable equilibrium condition we should have `overset(to)(m) || (-overset(to)(B) ) rArr theta = 180^@`. In this case, `U= - m B cos theta ` `therefore U= mB cos (-1)` `therefore U_("max") = +mB` `=(0.32) (0.15)` `therefore U_("max") = 0.048` J |
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