A short bar magnet of magnetic moment 0,21 A -` m^(2)` is placed with its axis perpendicular to the direction of the horizontal component of the earth 's magnetic field .The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at `45^(@)` with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field `4.2xx10^(-5)`

Options

A. 12cm
B. 20cm
C. 5cm
D. 10cm

Answer

D

Explanation

Resultant field is Inclined at `45^(@)` to `B_(H)`, so
`B=B_(H)tan theta`
`implies (mu_(0))/(4pi)*(M)/(r_(3))=B_(H)impliesr_(3)=(Mxx10^(-7))/(B_(H))impliesr=10cm`