$A$ coil of $12$ turns is made from a wire of constant length carrying a current $I$. If the number of turns is changed to $3$,what is the percentage change in the magnetic field produced at its centre?

How can we determine the direction of the magnetic field using the Biot-Savart law?

If a positive ion is moving away from an observer with some acceleration,then the lines of force of magnetic induction will be

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

$Weber/m^2$ is equal to

Two parallel wires of equal lengths are separated by a distance of $3 \,m$ from each other. The currents flowing through the first and second wires are $3 \,A$ and $4.5 \,A$ respectively in opposite directions. The resultant magnetic field at the mid-point of both the wires is ($\mu_{0} =$ permeability of free space).