Two equal bar magnets are kept as shown in the figure. The direction of the resultant magnetic field,indicated by the arrowhead at point $P$,is (approximately):

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?

$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^\circ$,the work done is $0.8\, J$. How much work is done in moving it $30^\circ$ further?

The ratio of magnetic fields due to a bar magnet at two axial points $P_1$ and $P_2$,which are separated from each other by $10 \ cm$,is $25: 2$. Point $P_1$ is situated at $10 \ cm$ from the center of the magnet. (Points $P_1$ and $P_2$ are on the same side of the magnet,and the distances of $P_1$ and $P_2$ from the center are much greater than the distance of the two ends of the magnet from the center). The magnetic length of the bar magnet is: (in $cm$)

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is