$AB$ and $CD$ are smooth parallel rails,separated by a distance $l$,and inclined to the horizontal at an angle $\theta$. $A$ uniform magnetic field of magnitude $B$,directed vertically upwards,exists in the region. $EF$ is a conductor of mass $m$,carrying a current $i$. For $EF$ to be in equilibrium,

Two parallel wires in free space are $10 \,cm$ apart and each carries a current of $10 \,A$ in the same direction. The force exerted by one wire on the other [per unit length] is

$A$ square loop of side '$a$' carrying a current '$I$' is suspended from an insulating hanger of a spring balance as shown in the figure. The transverse magnetic field '$B$' directed into the paper occurs only at the bottom side of the loop. When the direction of current in the loop is reversed, the change in the reading of the spring balance is:

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)

$A$ current of $1 \ A$ is flowing along the positive $x$-axis through a straight wire of length $0.5 \ m$ placed in a region of a magnetic field given by $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$. The magnitude and the direction of the force experienced by the wire are, respectively:

$A$ straight rod of mass $m$ and length $L$ is suspended from two identical springs as shown in the figure. The springs are stretched by a distance of $x_0$ due to the weight of the rod. The circuit has a total resistance $R$. When a magnetic field perpendicular to the plane of the paper is switched on,the springs are observed to extend further by the same distance $x_0$. The magnetic field strength is: