$A$ wire carrying current '$I$' along the $x$-axis has length '$L$' and is kept in a magnetic field $\vec{B} = B(\hat{i} + 2\hat{j} - 2\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is:

$A$ wire of length $L$ carries current $I$ along the $x$-axis. $A$ magnetic field $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) \text{ T}$ acts on the wire. The magnitude of the magnetic force acting on the wire is:

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

Write the magnetic force equation on a current-carrying element $I\vec{dl}$ inside a magnetic field $\vec{B}$. Write the law used to determine the direction of the magnetic force.

Two free parallel wires carrying currents in opposite directions:

Two parallel wires of length $9 \, m$ each are separated by a distance $0.15 \, m$. If they carry equal currents in the same direction and exert a total force of $30 \times 10^{-7} \, N$ on each other,then the value of current must be ........ $A$. (in $.5$)