An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of electric filed is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron, in comparison to the time of fall of the proton is

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 :