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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 5.0 xx 10^(-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 :It given that for circular coil `N = 16 , R= 10 cm = 0.1 m and I = 0.5 A` `therefore ` Magnetic moment of current carrying coil m = N IA = `NI pi R^2` Frequency of small anuglar oscillations of the coil is given by `v = 1/(2pi) sqrt((MB)/(I))` where I is the moment of inerita of the coil about the axis of rotation . THUS, we have `I = (mB)/(4pi^2 v^2) = ((NI pi R^2)B)/(4pi^2 v^2) = (NIR^2 B)/(4pi v^2)` `B = 5.0 xx 10^(-2) T and v = 2.0 s^(-1)` `thereforeI = (16xx0.75xx(0.1)^2 xx (5.0 xx 10^(-2))/(4xx3.14xx(2.0)^2) = 1.2 xx10^(-4) KG m^2` . |
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