A short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East-West line drawn through the mid point of the magnet. What is the magnetic moment of the magnet in `Am^(2)`? (Given `m=10^(-7)` in SJ units and `B_(H)=` horizontal component of earth's magnetic field `=3.6xx10^(-5)` tesla)

Options

A. 9.7
B. 4.9
C. 19.4
D. 14.6

Answer

A

Explanation

Bar magnet is kept in such a way, that neutral points lie on the equatorial line of magnet. Let at point ‘P - neutral points is observed. Magnetic field for bar magnet on equatorial line at distance x
`B = (mu_0)/(4pi ) (M)/((x^2 + l^2)^(3//2))`
For short bar magnet `x^2 + l^2 = x^2`
`B = (mu_0)/(4pi ) (M)/(x^3)`
Thus, for neutral points it would be equal to horizontal component of earth.s magnetic field. (because it lies on equator of the earth)
` B_H = (mu_0)/(4pi) xx (M)/(x^3)`
` 3.6 xx10^(-5) = 10^(-7) xx (M)/((30)^2 xx 10^(-6))`
`M = 9.72 Am^2`