The magnetic induction at a point $P$ which is distant $4 \, cm$ from a long current-carrying wire is $10^{-8} \, T$. The magnetic field induction at a distance $12 \, cm$ from the same current would be:

If an electron revolves around a nucleus in a circular orbit of radius $R$ with frequency $n$,then the magnetic field produced at the centre of the nucleus will be

$A$ direct current is passing through a wire. It is bent to form a coil of one turn. Now,it is further bent to form a coil of two turns but at a smaller radius. Find the ratio of the magnetic induction at the centre of this two-turn coil to that at the centre of the one-turn coil.

$A$ current $I=5 \text{ A}$ flows along a thin wire shaped as shown in the figure. The radius of the curved part of the wire is $R=100 \text{ mm}$, and the angle $2\phi=90^{\circ}$. The magnitude of the magnetic field at point $O$ is approximately:
$\left[\text{Use, } \frac{\mu_0}{4\pi}=10^{-7} \text{ T m A}^{-1}\right]$ (in $\mu\text{T}$)

$A$ circular coil $A$ has a radius $R$ and the current flowing through it is $I$. Another circular coil $B$ has a radius $2R$ and the current flowing through it is $2I$. The ratio of the magnetic fields at the centre of the circular coils $(B_A : B_B)$ is: (in $:1$)

$A$ current of $1.5 \, A$ is flowing through an equilateral triangle of side $9 \, cm$. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.)