$A$ $4.0 \ cm$ long straight wire carrying a current of $8 \ A$ is placed perpendicular to a uniform magnetic field of strength $0.15 \ T$. The magnetic force on the wire is . . . . . . $mN$.

$A$ 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.) $g = 9.8\; m s^{-2}.$

Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is

$A$ current of $10 \ A$ flows through two long parallel wires. The magnetic force per unit length on each wire is $2 \times 10^{-3} \ N/m$. If their currents are doubled and the separation between them is halved,then the magnetic force per unit length of each wire becomes $...... \times 10^{-3} \ N/m$.

In the given figure,the magnetic force on the wire $ABC$ will be $(B = 2 \, T, I = 2 \, A)$.

$A$ wire $AB$ is carrying a steady current $I_1$ and is kept on a table. Another wire $CD$ carrying current $I_2$ is held directly above it at a distance $r$. When the wire $CD$ is left free and it remains suspended at its position, find its mass per unit length. (Given: $g$ = acceleration due to gravity, $\mu_0$ = permeability of free space)