A short bar magnet of magnetic moment `0*5JT^-1` is placed with its magnetic axis in the magnetic meridian, with its north pole pointing geographic north. A neutral point is obtained at a distance of `0*1m` from the centre of the magnet. Find the horizontal component of earth's magnetic field.

Answer

`0*5xx10^-4T`

Explanation

Here, `M=0*5JT^-1`, `d=0*1m`, `H=?`
As north pole of magnet is pointing geograhic north, neutral points lie on equatorial line of magnet. At each neutral point,
`H=(mu_0)/(4pi)(M)/(d^3)`.
`H=10^-7xx(0*5)/((0*1)^3)=0*5xx10^-4T`