The resultant force on the current loop $PQRS$ due to a long current-carrying conductor is:

Three long,straight parallel wires carrying currents are arranged as shown. The wire $C$ which carries a current of $5.0 \text{ A}$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is (in $cm$)

Two free parallel wires carrying currents in opposite directions:

$A$ square loop of length $L$ is placed with its edges parallel to the $XY$-axes. The loop carries a current $I$. If the magnetic field in the region varies as $B = B_0 \left(1 + \frac{xy}{L^2}\right) \hat{k}$, then the magnitude of the net force on the loop will be:

Two long parallel conductors,separated by a distance $d$,carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2I$ and its direction is reversed. The distance between them is also increased to $3d$. The new value of the force between them is:

$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)