Three long,straight and parallel wires carrying currents are arranged as shown in the figure. The wire $C$ which carries a current of $5.0\, A$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is (in $, cm$)

$A$ straight wire of mass $0.2 \,kg$ and length $1.5 \,m$ carries a current of $2 \,A$, as shown in the figure. It is suspended in mid-air by a uniform magnetic field $B$ pointing into the plane of the paper. Calculate the magnitude of the magnetic field. (Ignore Earth's magnetic field and assume $g = 10 \,m/s^2$) (in $\,T$)

As shown in the figure,a uniform straight wire of length $30 \sqrt{3} \text{ cm}$ is bent in the form of an equilateral triangle $ABC$. $A$ uniform magnetic field $2 \text{ T}$ is applied parallel to the side $BC$. If the current through the wire is $2 \text{ A}$,the magnitude of the force on the side $AC$ is ($\overline{B}$ represents the direction of the magnetic field).

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 . . . . . . .

$A$ 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 $\,A$)

$A$ horizontal semi-circular wire of radius $r$ is connected to a battery through two similar springs $X$ and $Y$. The battery sends a current $I$ through the wire. $A$ uniform magnetic field $B$ is applied perpendicular to the plane of the wire,as shown in the figure. What is the force acting on each spring?