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Question
The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?
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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 BH 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) tanθ`
⇒`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"`
= 3.75 × 103 A-m2/T
Deflection in the magnetometer θ = 37°
From the magnetometer theory in Tan-B position, we have
`M/B_H = (4pi)/(u_0)(d^2 + l^2)^(3/2)` tan θ
Since for the short magnet / << d , we can neglect l w.r.t d.
Now ,
`M/B_H = (4pi)/(u_0) d^3 tanθ`
⇒ `3.75 xx 10^3 = 1/10^-7 xx d^3 xx 0.75`
⇒ `d^3 = (3.75 xx 10^3 xx 10^-7)/(0.75) = 5 xx 10^-4`
⇒ `d = root(3) (5 xx 10^-4)`
⇒ `d = 0.076 "m" = 7.9 "cm"`
Magnet will be at 7.9 cm from the centre.
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