Two long parallel conductors,separated by a distance $d$,carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2I$ and its direction is reversed. The distance between them is also increased to $3d$. The new value of the force between them is:

What is the net force on the square coil?

$A$ current of $3\, A$ is flowing in a linear conductor of length $40\, cm$. The conductor is placed in a magnetic field of strength $500\, G$ and makes an angle of $30^\circ$ with the direction of the field. The magnitude of the force experienced by the conductor is:

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

Two thin long parallel wires separated by a distance $b$ are carrying a current $i$ $A$ each. The magnitude of the force per unit length exerted by one wire on the other is

$A$ straight wire is placed in a magnetic field that varies with distance $x$ from the origin as $\vec{B} = B_0 \left( 2 - \frac{x}{a} \right) \hat{k}$. The ends of the wire are at $(a, 0)$ and $(2a, 0)$ and it carries a current $I$ in the positive $x$-direction. If the force on the wire is $\vec{F} = IB_0 \left( \frac{ka}{2} \right) \hat{j}$, then the value of $k$ is: