$A$ metal ring of radius $r = 0.5 \, m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2 \, T$ carries a current $I = 100 \, A$. The tension in newtons developed in the ring is:

$A$ $3.0 \; cm$ wire carrying a current of $10 \; A$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27 \; T$. What is the magnetic force on the wire?

Two long straight wires are arranged parallel to each other and are kept $20$ cm apart in vacuum. They carry currents of $2$ $A$ and $4$ $A$ respectively in the same direction. What will be the magnetic force on a length of $10$ cm of either wire? $(\mu_0 = 4\pi \times 10^{-7} \text{ SI units})$

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:

$A$ conducting wire $PQ$ carries a current $10 \ A$ as shown in the figure. It is placed in a uniform magnetic field $5 \ T$ which is acting normally outwards from the paper. The net force experienced by the wire is:

$A$ $2 \, A$ 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, $g=10 \, m/s^2$).