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

Answer

c

Explanation

(c) Here, `r=30cm=0.3cm`
we know `(mu_(0)M)/(4pir^(3))=B_(H)=3.6xx10^(-5)`
`rArr M=(3.6xx10^(-5))/(10^(-7))(0.3)^(3)`
Hence, `M-9.7Am^(2)`