Calculate the force on a magnetic pole of strength $20\mathrm{Am}$ lying at a distance of $16\mathrm{cm}$ in front of a magnet of magnetic length $24\mathrm{cm}$ and pole strength $50\mathrm{Am}$.

The magnetic moment of a short magnet is $8{\mathrm{Am}}^2$. What is the magnetic induction at a point $20\mathrm{cm}$ away from its mid-point on its

(a) axial point

The magnetic moment of a short magnet is $8{\mathrm{Am}}^2$. What is the magnetic induction at a point $20\mathrm{cm}$ away from its mid-point on its

(b) equatorial point

Each atom of an iron bar of volume $5{\mathrm{cm}}^3$ has a magnetic moment $1.8\times {10}^{-23} {\mathrm{Am}}^2$. What will be the magnetic moment of the bar in the state of magnetic saturation.

[Density of iron$=7.8\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3$, Atomic weight of iron $56$, Avogadro's number, $N=6.02\times {10}^{23} /\mathrm{gm}-\mathrm{mole}$.]

Hint: magnetic moment at saturation= sum of atomic moment of all atoms.

The vertical component of the earth's magnetic field at a given place is $\sqrt{3}$ times its horizontal components. If the total intensity of the earth's magnetic field at the place is $0.4G$. Find the value of 

(i) angle of dip

(ii) the horizontal component of the earth's magnetic field.