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

The axial field $(B_{A})$ and the equatorial field $(B_{E})$ due to a short bar magnet at equal distances are related as:

$A$ bar magnet with a magnetic moment of $3.0 \, A \cdot m^2$ is placed in a magnetic field of $2 \times 10^{-5} \, T$. If the force acting on each pole is $6 \times 10^{-4} \, N$,what is the length of the magnet in $m$?

If $L$ is the length of a bar magnet,then the separation between the two poles is nearly

Each atom of an iron bar $(5 \, cm \times 1 \, cm \times 1 \, cm)$ has a magnetic moment of $1.8 \times 10^{-23} \, A \cdot m^2$. Given that the density of iron is $7.78 \times 10^3 \, kg/m^3$,the atomic weight is $56 \, g/mol$,and Avogadro's number is $6.02 \times 10^{23} \, mol^{-1}$,calculate the magnetic moment of the bar in the state of magnetic saturation in $A \cdot m^2$.

The time period of a freely suspended magnet is $4 \ s$. If it is broken in length into two equal parts and one part is suspended in the same way,then its time period will be: (in $s$)