A bar magnet of length `8 cm` and having a pole strengh of `1.0 A m` is placed vertically on a horizontal table with its south pole on the table. A neutral point is found on the table at a distance of 6.0 cm north of the magnet. Calculate the earth's horizontal magnetic field.

Explanation

Magnetic field due to single pole of a magnet is given by

`B = (mu_0)/(4pi) (M)/(d^(2))` where m is pole strength
Field due to north pole ,
`B_N = (mu_0)/(4pi)(M)/((d^(2))) ` along NP
Field due to south pole ,
`B_S = (mu_0)/(4pi) (M)/((d^(2) + 4l^(2))) ` along PS
Horizontal component of `B_S = B_S cos theta `
`= (mu_0)/(4pi)(M)/((d^(2) + 4l^(2)))*(d)/((d^(2) + 4l^(2))^(1//2))`
Resultant horizontal field
` = (mu_0 m)/(4pi) [ (1)/(d^(2)) -(d)/((d^(2) + 4l^(2))^(3//2))]`
or , `B_H = (mu_0 M)/( 4pi (2l)) [ (1)/(d^(2)) - (d)/((d^(2) + 4l^(2))^(3//2))]`
(where magnetic moment `M = mxx 2l` )
On substituting numerical values , we get
`B_H = 3.81 xx 10^(-5)` Tesla.