$A$ 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. The rod is prevented from sliding down by passing a current through it,while a magnetic field of induction $0.25\; T$ acts on it in the vertical direction. The current flowing in the rod to keep it stationary is: (in $; A$)

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \ cm$ length of wire $Q$ is . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A)$

Two long conductors separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is:

What is the magnitude of magnetic force per unit length (in $N \;m^{-1}$) on a wire carrying a current of $8 \;A$ and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of $0.15 \;T$?

$A$ current of $3\, A$ is flowing in a linear conductor of length $40\, cm$. The conductor is placed in a magnetic field of strength $500\, G$ and makes an angle of $30^\circ$ with the direction of the field. The magnitude of the force experienced by the conductor is:

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then: