$A$ vertical straight conductor carries a current vertically upwards. $A$ point $P$ lies to the east of it at a small distance and another point $Q$ lies to the west at the same distance. The magnetic field at $P$ is

On connecting a battery to the two corners of a diagonal of a square conductor frame of side $a$,the magnitude of the magnetic field at the centre will be:

Two identical circular loops $P$ and $Q$ each of radius $r$ are lying in parallel planes such that they have a common axis. The currents through $P$ and $Q$ are $I$ and $4I$ respectively in the clockwise direction as seen from $O$. The net magnetic field at $O$ is:

$A$ long wire lies along the $X$-axis and carries a current of $40 \, A$ in the positive $x$-direction. $A$ second long wire is perpendicular to the $xy$-plane, passes through the point $(3.0 \, m) \hat{j}$, and carries a current along the positive $z$-direction. If the magnitude of the resultant magnetic field at the point $(2.0 \, m) \hat{j}$ is $R=5 \times 10^{-6} \, T$, then the current in the second wire is (Permeability of free space, $\mu_0=4 \pi \times 10^{-7} \, T \cdot m/A$) (in $A$)

Two long parallel wires $P$ and $Q$ are held perpendicular to the plane of the paper with a distance of $5 \; m$ between them. If $P$ and $Q$ carry currents of $2.5 \; A$ and $5 \; A$ respectively in the same direction,then the magnetic field at a point half-way between the wires is:

State the Biot-Savart law.