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प्रश्न
A magnetic dipole of magnetic moment `0.72sqrt(2) "Am"^2` is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 μT.
उत्तर
Given :
Magnetic moment of magnetic dipole,`M = 0.72sqrt(2) "A-m"^2`
Horizontal component of the earth's magnetic field,`B_H = 18 "uT"`
Let d be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet (B) on the equatorial line of the dipole is given by,
`B = (u_0)/(4pi) M/(d^3)`
⇒ `(4pi xx 10^-7)/(4pi) xx (0.72sqrt(2))/d^3 = 18 xx 10^-6`
⇒ `d^3 = (0.72 xx 1.414 xx 10^-7)/(18 xx 10^-6)`
⇒ `d^3 = 0.005656`
⇒ `d = root(3)(0.005656)`
⇒ `d ≈ 0.2 "m" = 20 "cm"`
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