Two long and parallel wires are at a distance of $0.1\, m$ and a current of $5\, A$ is flowing in each of these wires. The force per unit length due to these wires will be

If two streams of protons move parallel to each other in the same direction,then they

$A$ horizontal rod of mass $10 \, g$ and length $10 \, cm$ is placed on a smooth plane inclined at an angle of $60^\circ$ with the horizontal,with the length of the rod parallel to the edge of the inclined plane. $A$ uniform magnetic field of induction $B$ is applied vertically downwards. If the current through the rod is $1.73 \, A$,then the value of $B$ for which the rod remains stationary on the inclined plane is......$T$

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^o$. Another straight thin wire with a steady current $I_1$ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current $I_1$ at the origin:

Three long,straight and parallel wires carrying currents are arranged as shown in the figure. The force experienced by $10\, cm$ length of wire $Q$ is

$A$ 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: