The magnetic field at the centre of a circular coil of radius $r$,through which a current $I$ flows,is:

At what distance from a long straight wire carrying a current of $12 \, A$ will the magnetic field be equal to $3 \times 10^{-5} \, Wb/m^2$?

State the Biot-Savart law.

The magnetic field at the centre of a circular current-carrying conductor of radius $r$ is $B_{c}$. The magnetic field on its axis at a distance $r$ from the centre is $B_{a}$. The value of $B_{c} : B_{a}$ will be

The magnetic induction at the centre of a current-carrying circular coil of radius $8 \ cm$ is $6 \sqrt{6}$ times the magnetic induction at a point on its axis. Then the distance of the point from the centre of the coil in $cm$ is $(\sqrt{5} = 2.236)$.

$A$ wire of length $L = 1 \ m$ carries a constant current $I$. The wire is bent to form a single circular loop of radius $R$. The magnetic field at the centre of this loop is $B$. If the same wire is now bent to form a circular loop with $n = 4$ turns,what will be the magnetic field $B'$ at the centre of this new loop?