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:

The magnetic field existing in a region is given by $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. $A$ square loop of edge $l$ and carrying a current $i$ is placed with its edges parallel to $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

The force between two long parallel wires is $F$ when each one of them carries a certain current $I$. If the current in each is halved,the force between them would be . . . . . . .

The figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field perpendicular to its plane. The part $ABC$ is the $(3/4)^{th}$ portion of the square of side length $l$. The part $ADC$ is a circular arc of radius $R$. The points $A$ and $C$ are connected to a battery which supplies a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ is

Two long wires carrying currents $I_1$ and $I_2$ are arranged as shown in the figure. The wire carrying current $I_1$ is along the $x$-axis. The other wire carrying current $I_2$ is parallel to the $y$-axis,passing through the point $(0, 0, d)$. Find the magnetic force exerted on a segment of length $dl$ of the second wire at the point $O_2(0, 0, d)$ due to the wire carrying current $I_1$.

$A$ magnetic field is in the $+Y$ direction. $A$ current $i$ flows through the wire $PQRSTU$. Each side has a length $L$. What is the net force on the wire?