For the circuit shown in the figure,the direction and magnitude of the force on the segment $PQR$ is:

$A$ horizontal wire carries $160 \ A$ current. Below it, another wire with a linear mass density of $10 \ g \ m^{-1}$ is kept at a distance of $4 \ cm$. If the lower wire hangs in the air, what is the current in this wire when the direction of current in both wires is the same (in $A$)? $(g=10 \ m \ s^{-2} \text{ and } \mu_0=4 \pi \times 10^{-7} \ T \ m \ A^{-1})$

$A$ current of $10 \ A$ is flowing in two straight parallel wires in the same direction. The force of attraction between them is $1 \times 10^{-3} \ N$. If the current is doubled in both the wires,the force will be:

$A$ square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow{F}$,the net force on the remaining three arms of the loop is

$A$ square loop of side '$a$' carrying a current '$I$' is suspended from an insulating hanger of a spring balance as shown in the figure. The transverse magnetic field '$B$' directed into the paper occurs only at the bottom side of the loop. When the direction of current in the loop is reversed, the change in the reading of the spring balance is:

$A$ square coil $ABCD$ of side $L$ is carrying a current $I_1$ in the clockwise direction. $A$ straight conductor carrying current $I_2$ (upward direction) is kept parallel to side $AB$ at a distance $\frac{L}{3}$ in the plane of $ABCD$. The net force on the coil $ABCD$ is ($\mu_0 =$ magnetic permeability).