$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the mid-point between them. The net magnetic field at $M$ is $B$. Now,the current $2\,A$ is switched off. The field at $M$ now becomes

Five very long,straight wires are bound together to form a small cable. Currents carried by the wires are $I_1 = 20\,A, I_2 = -6\,A, I_3 = 12\,A, I_4 = -7\,A, I_5 = 18\,A.$ The magnetic induction at a distance of $10\,cm$ from the cable is

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^\circ$. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is:

Describe the arrangement of iron filings sprinkled around a straight current-carrying wire.

$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

Two protons $A$ and $B$ move parallel to the $x$-axis in opposite directions with equal speeds $v$. At the instant shown,the ratio of magnetic force and electric force acting on the proton $A$ is ($c=$ speed of light in vacuum).