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A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm. (a) What is the magnetic moment of the magnet ? (b) What is the work done in moving it from its most stable to most unstable position ? (c) The bar magnet is replaced by a solenoid of cross-sectional area 2 xx 10^(-4) m^(2) and 1000 turns, but of the same magneticmoment. Determine the current flowing through the solenoid. |
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Answer» Solution :Here, `B = 600 G = 6000 xx 10^(-4) T 6xx 10^(-2) T` `theta = 30^@` `tau = 0.012 Nm` `N= 1000` `A= 2 xx 10^(-4) m^(2)` (a) `tau = m B sin theta` `therefore m= (tau)/( B sin theta) = (0.012)/( 6 xx 10^(-2) xxsin 30^(@) ) = (0.2)/(½)` `m= 0.4 "Am"^(2)` (b) For most stable position `theta_(1) = 0^(@)` and for most unstable position `theta^(2) = 180^@`. `therefore W= - mB cos theta_(2) - (-mB cos theta_(1) ) ` `=-mB cos 180^(@) + m B cos 0^@` `=-mB (-1) + mB (1)` `=mB+ mB` `= 2mB` `2 xx 0.4 xx 6 xx 10^(-2)` `therefore W=0.048` J (C) Magnetic moment of solenoid, `m_(s) = NIA,"here" m_(s) = m = 0.4 "Am"^2` `I= (m)/( NIA) = (0.4)/( 1000 xx 2 xx 10^(-4) ) = 2A` |
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