In an experiment,setup $A$ consists of two parallel wires which carry currents in opposite directions as shown in the figure. $A$ second setup $B$ is identical to setup $A$,except that there is a metal plate between the wires. Let $F_A$ and $F_B$ be the magnitude of the force between the two wires in setup $A$ and setup $B$,respectively.

$A$ conducting circular loop of radius $r$ carries a constant current $i$. It is placed in a uniform magnetic field $B$,such that $B$ is perpendicular to the plane of the loop. The net magnetic force acting on the loop is

If the distance between two current-carrying parallel wires is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the magnitude of the force between them (the same current is passing through both wires) is:

$A$ conductor $ABCDE$,shaped as shown,carries a current $i$. It is placed in the $xy$ plane with the ends $A$ and $E$ on the $x$-axis. $A$ uniform magnetic field of magnitude $B$ exists in the region. The force acting on it will be

The circuit in the figure consists of wires at the top and bottom and identical springs as the left and right sides. The wire at the bottom has a mass of $10 \, g$ and is $5 \, cm$ long. The wire is hanging as shown in the figure. The springs stretch $0.5 \, cm$ under the weight of the wire and the circuit has a total resistance of $12 \, \Omega$. When the lower wire is subjected to a static magnetic field,the springs stretch an additional $0.3 \, cm$. The magnetic field is

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