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Question
A short bar magnet of magnetic moment 5.25 × 10−2 J T−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Solution
Magnetic moment of the bar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10−4 T
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B = `(μ_0 "M")/(4pi"R"^3)`
Where,
μ0 = Permeability of free space = 4π × 10−7 T mA−1
When the resultant field is inclined at 45° with earth’s field, B = H
∴ `(μ_0"M")/(4pi"R"^3) = "H" = 0.42 xx 10^-4`
`"R"^3 = (μ_0"M")/(0.42 xx 10^-4 xx 4pi)`
= `(4pi xx 10^-7 xx 5.25 xx 10^-2)/(4pi xx 0.42 xx 10^-4)`
= 12.5 × 10−5
∴ R = 0.05 m = 5 cm
(b) The magnetic field at a distanced R' from the centre of the magnet on its axis is given as:
`"B'" = (μ_0 2"M")/(4pi"R"^3)`
The resultant field is inclined at 45° with earth’s field.
∴ B' = H
`(μ_0 2"M")/(4pi("R'")^3) = "H"`
`("R'")^3 = (μ_0 2"M")/(4pi xx "H")`
= `(4pi xx 10^-7 xx 2 xx 5.25 xx 10^-2)/(4 pi xx 0.42 xx 10^-4)`
= 25 × 10−5
∴ R' = 0.063 m = 6.3 cm
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