Two wires $2 \text{ mm}$ apart supply current to a $100 \text{ V}$,$1 \text{ kW}$ heater. The force per metre between the wires is (Given $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$)

The resultant force on the current loop $PQRS$ due to a long current-carrying conductor will be $..... \times 10^{-4} \text{ N}$.

$A$ straight horizontal conducting rod of length $0.45\; m$ and mass $60\; g$ is suspended by two vertical wires at its ends. $A$ current of $5.0\; A$ is set up in the rod through the wires.
$(a)$ What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
$(b)$ What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) $g = 9.8\; m s^{-2}.$

$A$ straight wire of length $0.5\,m$ carrying a current of $1.2\,A$ is placed in a uniform magnetic field of induction $2\,T$. The magnetic field is perpendicular to the length of the wire. The force on the wire is.......$N$.

$A$ 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:

$A$ rigid wire consists of a semicircular portion of radius $R$ and two straight sections. The wire is partially immersed in a perpendicular magnetic field $B = B_0 \hat{k}$ as shown in the figure. The magnetic force on the wire if it carries a current $i$ is: