In the hydrogen atom,the electron is making $6.6 \times 10^{15} \, r.p.s.$ If the radius of the orbit is $0.53 \times 10^{-10} \, m,$ then the magnetic field produced at the centre of the orbit is (in $Tesla$):

The magnetic induction at the centre of a current-carrying circular coil of radius $r$ is

The ratio of the magnetic field at the center of a current-carrying circular coil of radius $R$ to the magnetic field at a distance of $3R$ from the center on its axis is:

An electric current is flowing through a circular coil of radius $R$. The ratio of the magnetic field at the centre of the coil and that at a distance $2\sqrt{2}R$ from the centre of the coil on its axis 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,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

The strength of the magnetic field at a point $r$ near a long straight current-carrying wire is $B$. The field at a distance $\frac{r}{2}$ will be