A short magnet of moment `6.75 Am^(2)` produces a neutal point on its axis. If horizontal component of earth's magnetic field is `5xx10^(-5) "Wb"//m^(2)`, then the distance of the neutal point should be

The horizontal component of the earth's magnetic field is 3.6 xx 10^(-5) T where the dip is 60^@ . Find the magnitude of the earth's magnetic field.

Angle of dip at a place is 30^(@) . If the vertical component of earth's magnetic field at that point is 7.5 xx 10^(-5) T, then the total magnetic field of earth at the point will be .

At a point on earth surface horizontal component of earth's magnetic field is 40 mu T and dip angle is 30^(@) . Find the total magnetic field of earth at this point.

A short bar magnet is placed with its north pole pointing north. If magnetic moment of magnet is 0*4Am^2 and horizontal component of earth's magnetic field is 0*4 gauss, what is the distance of neutral point from the centre of the magnet?

The length of a magnet of moment 5Am^(2) is 14cm . The magnetic induction at a point, equidistant form both the poles is 3.2xx10^(-5)Wb//m^(3) . The distance of the point form either pole is

A compass needle whose magnetic moment is 60Am^2 pointing geographic north at a certain place, where the horizontal component of earth's magnetic field is 40mu Wb//m^2 , experience a torque 1*2xx10^-3Nm . What is the declination of the place?