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Question
A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.
Solution
Given :
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and `B_H` be Earth's horizontal magnetic field.
According to the magnetometer theory,
`M/B_H = (4pi)/u_0 (d^2 - l^2)^2/(2d) tan θ`
For the short magnet ,
`M/B_H = (4pi)/u_0 xx d^4/(2d) tan θ`
⇒`M/B_H = (4pi)/(4pi xx 10^-7) xx (0.1)^3/2 xx tan 37^circ`
⇒`M/B_H = 0.5 xx 0.75 xx 1 xx 10^4`
⇒`M/B_H = 3.75 xx 10^3 "A-m"^2"/"T`
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