The current is flowing in the south direction along a power line. The direction of the magnetic field above the power line (neglecting the Earth's field) is

The magnetic field at the centre of an equilateral triangular loop of side $2L$ and carrying a current $i$ is

$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?

$A$ uniform wire is bent in the form of a circle of radius $R$. $A$ current $I$ enters at $A$ and leaves at $C$ as shown in the figure. If the length $ABC$ is half of the length $ADC$,the magnetic field at the centre $O$ will be

$A$ linear small part of wire $PQ$ is situated on the $y$-axis from $y = -a/2$ to $y = +a/2$, and a current $I$ is flowing through it. The magnetic field produced due to wire $PQ$ at the point $y = +a$ will be:

$A$ thin ring of radius $R$ metre has charge $q$ coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolution/s. The value of magnetic induction at the centre of the ring in $\text{Wb/m}^2$ is ($\mu_0$ = permeability of free space)