Two parallel,long wires are kept $0.20 \, m$ apart in a vacuum,each carrying a current of $x \, A$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} \, N$,then the value of $x$ is approximately:

Two long straight wires $A$ and $B$ carrying equal current $I$ were kept parallel to each other at distance $d$ apart. The magnitude of the magnetic force experienced by length $L$ of wire $A$ is $F$. If the distance between the wires is halved and the currents are doubled,the force $F_2$ on length $L$ of wire $A$ will be:

$A$ $2 \, A$ current carrying straight metal wire of resistance $1 \, \Omega$, resistivity $2 \times 10^{-6} \, \Omega m$, area of cross-section $10 \, mm^2$ and mass $500 \, g$ is suspended horizontally in mid-air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is . . . . . . . $\times 10^{-1} \, T$ (given, $g=10 \, m/s^2$).

Two thin long parallel wires, $W_1$ and $W_2$ separated by distance '$a$',carry currents $i$ and $3i$ respectively in the same direction. The magnitude of the force per unit length exerted by wire $W_1$ on wire $W_2$ is

$A$ current-carrying rectangular loop $PQRS$ is made of uniform wire. The lengths are $PR = QS = 5\,cm$ and $PQ = RS = 100\,cm$. If the ammeter current reading changes from $I$ to $2I$,the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively,$f_{PQ}^{I} : f_{PQ}^{2I}$,is:

Wires $1$ and $2$ carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$?