$A$ 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 $...... \times 10^{-3} \ N/m$.

$A$ 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)

Two very long,straight,parallel conductors $A$ and $B$ carry currents of $5\,A$ and $10\,A$ respectively and are at a distance of $10\,cm$ from each other. The direction of current in the two conductors is the same. The force acting per unit length between the two conductors is: $(\mu_0 = 4\pi \times 10^{-7} \text{ SI unit})$

The force acting on a current-carrying wire joining two fixed points $A$ and $B$ in a uniform magnetic field is:

$A$ current of $10\, A$ is flowing in a wire of length $1.5\, m$. $A$ force of $15\, N$ acts on it when it is placed in a uniform magnetic field of $2\, T$. The angle between the magnetic field and the direction of the current is.....$^o$

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