A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north south direction. Null points are found on the axis of the magnet at `14cm` from the centre of the magnet. The earth's magnetic field at the plane is `0*36G` and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null points (i.e. 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and apposite to the horizontal component of earth's magnetic field).

Explanation

Earth's magnetic at the given place, H=0.36G
The m agnetic field at a distance d, on the axis of the magnet is given as:
`B_(1)=(mu_(0))/(4pi)(2M)/(d^(3))=H` ....................(i)
where,
`mu_(0)=` permeatbility of free space
M=Magnetic moment
The maggnetic field at the same distance d, on the equatorial line of the magnet is given as:
`B_(2)=(mu_(0)M)/(4pid^(3))=(H)/(2)` [Using equation (i)]
Total magnetic field, `B=B_(1)+B_(2)`
`=H+(H)/(2)`
`=0.36+0.18=0.54G`
Hence, the magnetic field is 0.54G in the direction of earth's magnetic field.