At what distance on the axis,from the centre of a circular current-carrying coil of radius $r$,does the magnetic field become $1/8$th of the magnetic field at the centre?

$A$ circular coil of radius '$r$' and number of turns '$n$' carries a current '$I$'. The magnetic fields at a small distance '$h$' along the axis of the coil $(B_a)$ and at the centre of the coil $(B_c)$ are measured. The relation between $B_c$ and $B_a$ is

The magnetic field at the centre of a circular coil of radius $r$,through which a current $I$ flows,is:

To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region,one can use the saddle coils illustrated in the figure. The loops are wrapped over a somewhat flattened tube. Assume the straight sections of wire are very long. The end view of the tube shows how the windings are applied. The overall current distribution is the superposition of two overlapping circular cylinders of uniformly distributed current,one toward you and one away from you. The current density $J$ is the same for each cylinder. The position of the axis of one cylinder is described by a position vector $\vec{a}$ relative to the other cylinder. The magnetic field inside the hollow tube is:

$A$ non-conducting disc of radius $R$ has a surface charge density which varies with distance from the centre as $\sigma(r) = \sigma_0 \left[1 + \sqrt{\frac{r}{R}}\right]$,where $\sigma_0$ is a constant. The disc rotates about its axis with angular velocity $\omega$. If $B$ is the magnitude of magnetic induction at the centre,then $\frac{B}{\mu_0 \sigma_0 \omega R}$ will be

What is the magnetic field at point $P$ in the given figure?