$A$ long straight wire of radius $r$ carries a steady current $I$. The current is uniformly distributed over its cross-section. The ratio $\left(\frac{B}{B^1}\right)$ of the magnetic field $B$ and $B^1$ at radial distances $\frac{r}{2}$ and $3r$ respectively,from the axis of the wire 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:

$A$ current $I$ flows in a circular arc of radius $r$ subtending an angle $\theta$ at the centre $O$ as shown in the figure. Find the magnetic field at the centre $O$ of the circle.

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^\circ$. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is:

$A$ current of $0.1\, A$ circulates around a coil of $100$ turns and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $({\mu _0} = 4\pi \times {10^{ - 7}}\,T\cdot m/A)$.

The dimensions of the ratio of magnetic flux $(\phi)$ and permeability $(\mu)$ are