Question

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.

Answer 1

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"`