The radius of a circular current-carrying coil is $R$. At what distance from the centre of the coil on its axis,the intensity of the magnetic field will be $\frac{1}{2 \sqrt{2}}$ times that at the centre?

Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper,as shown. The variation of the magnetic field $B$ along the line $XX'$ is given by

The magnetic field at the centre of a circular coil of radius $R$ due to current $i$ flowing through it is $B$. The magnetic field at a point along the axis at a distance $R$ from the centre is

$B_{X}$ and $B_{Y}$ are the magnetic fields at the centre of two coils $X$ and $Y$ respectively,each carrying equal current. If coil $X$ has $200$ turns and $20 \ cm$ radius and coil $Y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{X}$ and $B_{Y}$ is:

Two concentric coils each of radius equal to $4 \pi \text{ cm}$ are placed at right angles to each other. If $10 \text{ A}$ and $24 \text{ A}$ are the currents flowing through the coils,respectively,the magnetic induction at the center of the coils will be

$A$ particle carrying a charge equal to $1000$ times the charge on an electron is rotating at $1$ rotation per second in a circular path of radius $r \ m$. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum $\mu_0$,the radius $r$ in $m$ is: $[e = 1.6 \times 10^{-19} \ C], [x = 2 \times 10^{-16}]$