A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence, the rod gains gravitational potential energy. The work required to do this comes from

In the product

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$=q\overrightarrow{v}\times \left(B\hat{\mathrm{i}}+B\hat{\mathrm{j}}+B_0\hat{\mathrm{k}}\right)$

For $q=1$ and $\overrightarrow{v}=2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+6\hat{\mathrm{k}}$ and $\overrightarrow{F}=4\hat{\mathrm{i}}-20\hat{\mathrm{j}}+12\hat{\mathrm{k}}$

What will be the complete expression for $\overrightarrow{B}$?

A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current-carrying coil in the shape of,

(i) an equilateral triangle of side $a$.
(ii) a square of side $a$.
The magnetic dipole moments of the coil in each case respectively are:

Three resistors having resistances $r_1, r_2$ and ${\mathrm{r}}_3$ are connected as shown in the given circuit. The ratio $\frac{i_3}{i_1}$ of currents in terms of resistances used in the circuit is :