In an atom, electrons revolve around the nucleus along a path of radius $0.72 \text{ Å}$, making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [given, $e = 1.6 \times 10^{-19} \text{ C}$]. (in $\text{A}$)

$A$ long straight conductor is bent into the shape as shown. If it carries a current $i$ and the radius of the circular part is $R$,then find the magnetic field $B$ at the centre of the circular coil.

$A$ current of $i$ ampere is flowing in an equilateral triangle of side $a$. The magnetic induction at the centroid will be

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{8}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is ($\mu_0 =$ permeability of free space).

When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved,the magnetic field at its centre will become

Two concentric coils $X$ and $Y$ of radii $16 \, cm$ and $10 \, cm$ lie in the same vertical plane containing $N-S$ direction. Coil $X$ has $20$ turns and carries $16 \, A$. Coil $Y$ has $25$ turns and carries $18 \, A$. Coil $X$ has current in anticlockwise direction and coil $Y$ has current in clockwise direction for an observer looking at the coils facing the west. The magnitude of the net magnetic field at their common centre is: