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A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0xx10^(-2)T. The coil is free to turn about an axis in its plane perpendicular to the field direction . When the coil is turned slightly and released , it oscillates about its stable equilibrium with a frequency of 2.0 s^(-1) . What is the moment of inertia of the coil about its axis of rotation ? |
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Answer» Solution :`N = 16 , r = 10 CM = 10 xx10^(-2) m, I = 0.75A` `B = 5.0 xx10^(-2)T, upsilon=2.0s^(-1)` `T=2pisqrt(I/(mB))" or"upsilon=1/((2pi))sqrt((mB)/I)` `:.I=(mB)/(4pi^2upsilon^2)=(ANIB)/(4pi^2upsilon^2)=(pir^2xxNIB)/(4pi^2upsilon^2)=((10xx10^(-2))^2xx16xx0.75xx5.0xx10^(-2))/(4xx(22)/(7)xx4)` `=(10^(-2)xx16xx0.75xx5xx10^(-2)xx7)/(4xx4xx22)=(0.75xx35xx10^(-4))/22=1.2xx10^(-4)kg m^2` |
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