An electron $(e)$ moves in a circular orbit of radius '$r$' with uniform speed '$V$'. It produces a magnetic field '$B$' at the center of the circle. The magnetic field $B$ is ($\mu_0 =$ permeability of free space).

In the figure shown,there are two semicircles of radii $r_1$ and $r_2$ in which a current $i$ is flowing. The magnetic induction at the centre $O$ will be

Write the formula for the magnetic field at a perpendicular distance $r$ from a finite current-carrying wire.

The magnetic induction at the centre $O$ of the current-carrying bent wire shown in the adjoining figure is

The magnetic field due to a current-carrying circular loop of radius $6 \ cm$ at a point on the axis at a distance of $8 \ cm$ from the centre is $216 \ \mu T$. Then the magnetic field at the centre of the ring is $\dots \ \mu T$.

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$