InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet. |
|
Answer» Here, `M=0.48JT^-1`, `B=?`, `d=10cm=0.1m` (i) On the axis of the magnet, `B=(mu_0)/(4pi)(2M)/(d^3)=10^-7xx(2xx0*48)/((0*1)^3)=0*96xx10^-4T` along S-N direction. (ii) On the equatorial line of the magnet `B=(mu_0)/(4pi)xx(M)/(d^3)=10^-7xx(0.48)/((0.1)^3)=0.48xx10^-4T`, along N-S direction. |
|