The net charge in a current-carrying wire is zero,yet a magnetic field exerts a force on it. This is because a magnetic field exerts a force on:

$A$ conductor $ABCDE$,shaped as shown,carries a current $i$. It is placed in the $xy$ plane with the ends $A$ and $E$ on the $x$-axis. $A$ uniform magnetic field of magnitude $B$ exists in the region. The force acting on it will be

What is the force per unit length between the two wires shown in the figure? [${\mu _0} = 4\pi \times {10^{ - 7}} \text{ T}\,m/A$]

The force exerted by a magnetic field on a wire having length $L$ and current $I$ is perpendicular to the wire and given as $|F| = IL|B|$. An experimental plot shows $|F|$ as a function of $L$. The plot is a straight line with a slope $S = (10 \pm 1) \times 10^{-5} \ T$. The current in the wire is $I = (15 \pm 1) \ mA$. The percentage error in $B$ is:

$A$ 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:

$A$ ring of radius $a$ carries a current $i$. $A$ magnetic field $B$ is present at the center of the ring,directed radially outward in the plane of the ring. What is the net magnetic force on the ring?