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 Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$. This indicates that the magnetic field $\vec{B}$ produced by an electron moving with velocity $\vec{v}$ is such that:

$A$ thin ring of radius $R$ carries a uniformly distributed charge. The ring rotates at a constant speed $N$ r.p.s. about its axis perpendicular to the plane. If $B$ is the magnetic field at the centre,the charge on the ring is ($\mu_0 =$ permeability of free space).

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).

Due to $10 \ A$ of current flowing in a circular coil of $10 \ cm$ radius,the magnetic field produced at its centre is $3.14 \times 10^{-3} \ Wb/m^2$. The number of turns in the coil will be:

Define the law for finding the direction of a magnetic field due to a circular current loop.