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

The situation is shown in figure. The magnetic field at `P` due to the south pole is
`B_s = (mu_0)/(4pi) m/d^2`
towards south and that due to the north pole is
`B_n = (mu_0)/(4pi) m/(d^2 + 4l^2)`
along NP. The horizontal component of this field will be towards north and will have the magnitude
`(mu_0)/(4pi) m/(d^2 + 4l^2) . d/((d^2 + 4l^2)^(1/2))`.
The resultant horizontal field due to the magnet is, therefore,
`(mu_0)/(4pi) m/d^2 - (mu_0)/(4pi) md/((d^2 + 4l^2)^(3/2))`
towards south. As `P` is a neutral point, this field should be equal in magnitude to the earth's megnetic field `B_H` which is towards north. Thus,
`B_H = (mu_0 m/(4pi) ([1/d^2 - d/(d^2 + 4l^2)^(3/2))])`
`= (mu_0 m)/(4pi) ([1/(36 cm^2) - (6 cm)/((36cm^2 + 64cm^2)^(3/2))]`
`=10^(-7) T m A^(-1) xx (1.0 A m) xx [1/(36cm^2) - 6/(1000 cm^2)]`
`= (10^(-6) T m^2) [ 0.028 - 0.006] xx 10^4 m^(-2)`
`= 22 xx 10^(-6) T = 22 muT`.