$A$ square loop $ABCD$ of side $L$ carrying a current $I_1$ is placed at a distance $(L/3)$ from a straight conductor $XY$ carrying current $I_2$. The loop and the conductor are coplanar. The net force on the loop will be ($\mu_0$ = magnetic permeability).

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then:

Two thin,long,parallel wires,separated by a distance $d$,carry a current of $i$ in the same direction. They will

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:

$A$ metal ring of radius $r = 0.5 \, m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2 \, T$ carries a current $I = 100 \, A$. The tension in newtons developed in the ring is:

$A$ long straight wire is carrying current $I_1$ in the $+z$ direction. The $x-y$ plane contains a closed circular loop carrying current $I_2$ and does not encircle the straight wire. The net force on the loop will be: