Three long straight wires are carrying current as shown in the diagram. The magnetic force per unit length on wire $Q$ is $..... \times 10^{-7} \ N/m$.

The force exerted by a magnetic field on a wire having length $L$ and current $I$ is perpendicular to the wire and given as $|F| = IL|B|$. An experimental plot shows $|F|$ as a function of $L$. The plot is a straight line with a slope $S = (10 \pm 1) \times 10^{-5} \ T$. The current in the wire is $I = (15 \pm 1) \ mA$. The percentage error in $B$ is:

$A$ metallic rod of mass per unit length $0.5 \,kg \,m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. $A$ magnetic field of strength $0.25 \,T$ is acting on it in the vertical direction. When a current $I$ is flowing through it, the rod is not allowed to slide down. The quantity of current required to keep the rod stationary is (in $\,A$)

Two long straight wires $A$ and $B$ carrying equal current $I$ were kept parallel to each other at distance $d$ apart. The magnitude of the magnetic force experienced by length $L$ of wire $A$ is $F$. If the distance between the wires is halved and the currents are doubled,the force $F_2$ on length $L$ of wire $A$ will be:

$ABCD$ is a rectangular loop made of uniform wire. If $AD = BC = 2 \text{ cm}$,what is the magnetic force per unit length acting on wire $DC$ due to wire $AB$ if the ammeter reads $20 \text{ A}$? (The lengths of $AB$ and $DC$ are large in comparison with the other two sides).

In an experiment,setup $A$ consists of two parallel wires which carry currents in opposite directions as shown in the figure. $A$ second setup $B$ is identical to setup $A$,except that there is a metal plate between the wires. Let $F_A$ and $F_B$ be the magnitude of the force between the two wires in setup $A$ and setup $B$,respectively.