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Question
If the bar magnet is turned around by 180°, where will the new null points be located?
Solution
The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:
`"B"_1 = (μ_0 2"M")/(4pi ("d"_1)^3) = "H"` .......(1)
Where,
M = Magnetic moment
μ0 = Permeability of free space
H = Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be written as:
`"B"_2 = (μ_0 "M")/(4pi ("d"_2)^3) = "H"` .............(2)
Equating equations (1) and (2), we get:
`2/("d"_1)^3 = 1/("d"_2)^3`
`("d"_2/"d"_1)^3 = 1/2`
∴ `"d"_2 = "d"_1 xx (1/2)^(1/3)`
= 14 × 0.794
= 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
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