$A$ rectangular loop of sides $25 \text{ cm}$ and $10 \text{ cm}$ carrying a current of $10 \text{ A}$ is placed with its longer side parallel to a long straight conductor $10 \text{ cm}$ apart carrying a current of $25 \text{ A}$. The net force on the loop is

$A$ 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:

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 $1$?

If a current is passed through a spring,then the spring will:

Define $1 \ A$ (Ampere) using the force between two current-carrying parallel wires.

$A$ metallic rod of mass per unit length $0.5 \,kg \,m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. $A$ magnetic field of strength $0.25 \,T$ is acting on it in the vertical direction. When a current $I$ is flowing through it, the rod is not allowed to slide down. The quantity of current required to keep the rod stationary is (in $\,A$)