As shown in the figure, a uniform straight wire of length $30 \sqrt{3} \text{ cm}$ is bent in the form of an equilateral triangle $ABC$. $A$ uniform magnetic field $2 \text{ T}$ is applied parallel to the side $BC$. If the current through the wire is $2 \text{ A}$, the magnitude of the force on the side $AC$ is ($\overline{B}$ represents the direction of the magnetic field).

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

Through two parallel wires $A$ and $B$,$10 \, A$ and $2 \, A$ of currents are passed respectively in opposite directions. If the wire $A$ is infinitely long and the length of the wire $B$ is $2 \, m$,the force on the conductor $B$,which is situated at a $10 \, cm$ distance from $A$,will be:

If the distance between two current-carrying parallel wires is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the magnitude of the force between them (the same current is passing through both wires) is:

$A$ semi-circular current-carrying wire having radius $R$ is placed in the $x-y$ plane with its centre at the origin $O$. $A$ non-uniform magnetic field $\vec B = \frac{B_o x}{2R} \hat k$ (where $B_o$ is a positive constant) exists in the region. The magnetic force acting on the semi-circular wire will be along:

$P, Q$ and $R$ are long parallel straight wires in air,carrying currents as shown in the figure. The direction of the resultant force on wire $R$ is