Calculate the magnetic field at point $M$ for the given current distribution.

The magnetic flux passing through the rectangular loop is proportional to:

The magnetic induction at any point due to a long straight wire carrying a current is

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,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

Let the current $I$ be associated with an electron of charge $e$ moving in a circular orbit of radius $r$ with speed $v$ around the positively charged nucleus. The ratio $\frac{r}{v}$ is

$A$ long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is $B$. If it is bent into a circular loop of radius $r_1$ having $n$ turns,the magnetic field at the centre of the coil for the same current is: