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Question
A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10−4 kg m2 and the earth's horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.
Solution
Given :
Time taken by the bar magnet to complete one oscillation, `T = pi/10 s`
Moment of inertia of the magnet about the axis of rotation, I = `1.2 xx 10^-4 "kgm"^2`
Horizontal component of Earth's magnetic field, BH = 30 μT
Time period of oscillating magnetometer (T) is given by
`T = 2pisqrt(I/(MB_H)`
Here ,
M = Magnetic moment of the magnet
On substituting the respective values, we get
`pi/10 = 2pisqrt((1.2 xx 10^-4)/(M xx 30 xx 10^-6)`
⇒ `(1/20)^2 = (1.2 xx 10^-4)/(M xx 30 xx 10^-6)`
⇒ `M = (1.2 xx 10^-4 xx 400)/(30 xx 10^-6)`
⇒ `M = 16 xx 10^2 "A-m"^2`
⇒ `M = 1600 "A-m"^2`
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