A short magnet of moment `6.75 Am^(2)` produces a neutal point on its axis. If horizontal component of earth's magnetic field is `5xx10^(-5) "Wb"//m^(2)`, then the distance of the neutal point should be

Options

A. `10 cm`
B. `20 cm`
C. `30 cm`
D. `40 cm`

Answer

C

Explanation

At neutral point
`|"Magnetic field due to magnet'|=|"Magn etic field due to earth"|`
`mu_(0)/(4pi).(2M)/d^(3)=5xx10^(-5) implies 10^(-7)xx(2xx6.75)/d^(3)=5xx10^(-5)`
`implies d=0.3 m =30 cm`