$A$ short bar magnet produces a magnetic field of $6.4 \times 10^{-5} \,T$ at a distance of $20 \,cm$ from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 \,cm$ from the centre of the magnet on the axis,is

Two identical short bar magnets,each having magnetic moment $M$,are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

Two identical short bar magnets are placed at $120^{\circ}$ as shown in the figure. The magnetic moment of each magnet is $M$. Then the magnetic field at the point $P$ on the angle bisector is given by

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ 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 $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?