$A$ straight horizontal conducting rod of length $L$ and mass $M$ is suspended by two vertical wires at its ends. If $I$ is the current passing through the rod,then in order that the tension in the wires is zero,the magnetic field set up normal to the conductor is (Neglect the mass of the wire,$g$ = acceleration due to gravity).

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 . . . . . . .

$P, Q$ and $R$ are long parallel straight wires in air,carrying currents $20 \ A, 40 \ A$ and $60 \ A$ as shown in the figure. What is the direction of the resultant force on $Q$?

$A, B$ 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:

$A$ conducting wire bent in the form of a parabola $y^2=2x$ carries a current of $4 \ A$ as shown in the figure. This wire is placed in a uniform magnetic field of induction $\vec{B}=+6 \hat{k} \ T$. The magnetic force on the wire is

$A$ rectangular loop of wire,supporting a mass $m$,hangs with one end in a uniform magnetic field $\overrightarrow{B}$ pointing into the plane of the paper. $A$ clockwise current $i$ is set up such that $i > mg/Ba$,where $a$ is the width of the loop. Then: