A short bar magnet has a magnetic moment of `0.39 JT^(-1)`. The magnitude and direction of the magnetic field produced by the magnet at a distance of `20 cm` from the centre of the magnet on the equatorial line of the magnet is

Options

A. 0.049 G, N-S direction
B. 4.95 G, S-N direction
C. 0.0195 G, S-N direction
D. 19.5 G, N-S direction

Answer

A

Explanation

Here, `m=0.39"J T"^(-1), d = 20 cm = 0.2 m
On the equatorial line of magnet
`B=(mu_(0))/(4pi)*(m)/(d^(3))=10^(-7)xx(0.39)/((0.2)^(3))=(0.39xx10^(-7))/(2^(3)xx10^(-3))=(0.39)/(8)xx10^(-4)`
`" "=0.049xx10^(-4)` T, N-S direction
`" "=0.049 `G`, N-S direction