A short bar magnet has a magnetic moment of `0.65TJ ^(-1)`, then the magnitude and direction of the magnetic field produced by the magnet at a distance 8 cm from the centre of magnet on the axis is

Options

A. `2.5 xx 10^(-4)T`, along NS direction
B. `2.5 xx 10^(-4)T`, along SN direction
C. `4.5 xx 10^(-4)T`, along NS direction
D. `4.5 xx 10^(-4)T`, along NS direction

Answer

B

Explanation

Here, `"M = 0.65 J T"^(-1), d=8cm = 0.08m`
The field produced by magnet at axial point is given by `B=(mu_(0))/(4pi)(2M)/(d^(3))=10^(-7)xx(2xx0.65)/((0.08)^(3))=2.5xx10^(-4)" T along SN."`