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A bar magnet isplaced in a uniformmagnetic field whose strength is 0.8 T. Suppose the barmagnet orient an angle 30^(@) with the external field experiences a torque of 0.2 Nm .Calculate : (i)the magneticmoment of themagnet (ii)the work done by an appliedforce in moving it form moststable configuration to themost unstable configuration and alsocompute thework done by the applied magnetic field in this case . |
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Answer» Solution :Uniform MAGNETIC field strength B = 0.8 T Bar magnet orient an angle with magnetic field `theta = 30^(@)` Torque `tau ` = 0.2 NM (i) Magnetic MOMENT of the magnet torque `tau = P_(m) B sin theta` `therefore` Magnetic moment `, P_(m) = (tau)/("Bsin" theta )= (0.2 )/(0.8 xx Sin 30^(@) )= (0.2)/(0.4) ` `P = 0.5 Am^(2)` (ii) Work done by external torque is stored in the magnet as potential energy . W = U = `-P_(m) " B Cos " theta` here, applied force acting on magnet its moving from most stable `theta`. to most unstable `theta`. `theta. = 0^(@) and theta = 180^(@)` So, workdone W = U = -`P_(m) " B "( Cos theta - Cos theta^(@) )` = - `p_(m) B (Cos 180^(@)- Cos theta^(@)) = - 0.5 xx 0.8 ((-1)-1) = - 0.4 (-2) ` W = U = 0.8 W = 0.8 J |
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