The force per unit length on a straight wire carrying a current of $8 \ A$ making an angle of $30^{\circ}$ with a uniform magnetic field of $0.15 \ T$ is: (in $N \ m^{-1}$)

$A$ $100$ turn rectangular coil $ABCD$ (in $xy$-plane) is hung from one arm of a balance (Figure). $A$ mass of $500 \text{ g}$ is added to the other arm to balance the weight of the coil. $A$ current of $4.9 \text{ A}$ passes through the coil and a constant magnetic field of $0.2 \text{ T}$ acting inward (in $xz$-plane) is switched on such that only arm $CD$ of length $1 \text{ cm}$ lies in the field. How much additional mass '$m$' must be added to regain the balance?

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \ cm$ length of wire $Q$ is . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A)$

$A$ current of $10 \ A$ is flowing in two straight parallel wires in the same direction. The force of attraction between them is $1 \times 10^{-3} \ N$. If the current is doubled in both the wires,the force will be:

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

$A$ 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 $\text{ g}$)