A short bar magnet of magnetic moment `5.25 xx 10^(-2) JT^(-1)` is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet, the resultant field is inclined at `45^(@)` with earth's field on (a) Its normal bisector and (b) its axis. Magnitude of the earth's field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

A short bar magnet of magnetic moment 0,21 A - m^(2) is placed with its axis perpendicular to the direction of the horizontal component of the earth 's magnetic field .The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at 45^(@) with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field 4.2xx10^(-5)

A short bar magnet has a magnetic moment of 0.48 JT^(-1) . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equitorial lines (normal bisector) of the magnet.

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22^(@) with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place.

A bar magnet of magnetic moment vecM is placed in a magnetic field of induction vecB. The torque exerted on it is

The earth's magnetic field at the equator is approximately 0.4G. Estimate the earth's dipole moment.

A bar magnet of magnetic moment 1.5 JT^(-1) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of worl required by an external torque to turn the magnet so as to align its magnetic moment : (i) normal to the field direction, (ii) opposite to the field direction ?

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 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G 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-point (i.e., 14 cm) from the centre of the magnet ? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)