The force per unit length on a straight wire carrying a current of $8 \ A$ making an angle of $30^{\circ}$ with a uniform magnetic field of $0.15 \ T$ is: (in $N \ m^{-1}$)

$A$ straight horizontal conducting rod of length $0.45\; m$ and mass $60\; g$ is suspended by two vertical wires at its ends. $A$ current of $5.0\; A$ is set up in the rod through the wires.
$(a)$ What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
$(b)$ What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) $g = 9.8\; m s^{-2}.$

$A$ straight wire of mass $200 \;g$ and length $1.5 \;m$ carries a current of $2 \;A$. It is suspended in mid-air by a uniform horizontal magnetic field $B$ (Figure). What is the magnitude of the magnetic field (in $T$)?

$AB$ 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 $i$. If the conductor is in equilibrium,find the relation between the given parameters.

Two parallel,long wires are kept $0.20 \, m$ apart in a vacuum,each carrying a current of $x \, A$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} \, N$,then the value of $x$ is approximately:

$A$ square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow{F}$,the net force on the remaining three arms of the loop is