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A magnetic needle is pivoted through its centre of mass and is free to rotate in a plane containing uniform magnetic field 200 xx 10^(-4)T. When it is displaced slightly from the equilibrium it makes 20 oscillations per second. If the moment of inertia of the needle about the axis of oscillation is 0.75 xx10^(-5) kg m^(2), find the magnetic moment of the needle. |
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Answer» Solution :Magnetic field induction , `B = 200 xx 10^(-4)T = 2 xx 10^(-2)T ` Frequency v=2Hz MOMENT of inertia , `I=0.75 xx 10^(-5) kgm^(2)` Frequency of oscillation , ` v=(1)/(2PI)sqrt((MB)/(I))` On squaring we can write . ` v^(2) = (1)/(4pi^(2)) xx (MB)/(I) IMPLIES M = (4pi^(2) xx v^(2) xx I)/(B)` `=(4(3.14)^(2) xx 2^(2) xx 0.75 xx 10^(-5))/(2 xx 10^(-2)) = 0.059` `:.` Magnetic moment of the needle , `M~= 0.06 Am^(2)` |
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