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Question
A magnetic dipole of magnetic moment `1.44 "A m"^2`is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 μT.
Solution
Given :
Magnetic moment of the magnetic dipole, M = 1.44 A-m2
Horizontal component of Earth's magnetic field, BH = 18 μT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B) is given by
`vecB = (u_0M)/(4pid^3)`
This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, BH due to Earth.
Thus,
`vecB = (u_0M)/(4pid^3)`
⇒ `B = (10^-7 xx 1.44)/d^3`
⇒ `18 xx 10^-6 = (10^-7 xx 1.44)/d^3`
⇒ `d^3 = 8 xx 10^-3`
⇒ `d = 2 xx 10^-1 = 20 "cm"`
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