Three infinite straight wires $A, B$ and $C$ carry currents as shown in the figure. The resultant force on wire $B$ is directed:

$A$ massless square loop of wire of resistance $R$ supporting a mass $M$ hangs vertically with one of its sides in a uniform magnetic field $B$ directed outwards in the shaded region. $A$ d.c. voltage $V$ is applied to the loop. For what value of $V$ will the magnetic force exactly balance the weight of the supporting mass $M$? (side of loop $= L, g =$ acceleration due to gravity)

The magnetic force between the wires as shown in the figure is:

$A$ thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction,as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is

Two parallel very long straight wires carrying current of $5 \text{ A}$ each are kept at a separation of $1 \text{ m}$. If the currents are in the same direction,the force per unit length between them is . . . . . . $\text{N/m}$. $(\mu_0 = 4\pi \times 10^{-7} \text{ SI units})$

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)