$A$ $4.0 \ cm$ long straight wire carrying a current of $8 \ A$ is placed perpendicular to a uniform magnetic field of strength $0.15 \ T$. The magnetic force on the wire is . . . . . . $mN$.

$A$ conductor of length $l$ and mass $m$ is placed along the east-west line on a table. Suddenly,a certain amount of charge is passed through it and it is found to jump to a height $h$. The earth's magnetic induction is $B$. The charge passed through the conductor is:

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 . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A)$

The resultant force on the current loop $PQRS$ due to a long current-carrying conductor is:

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 $1$?

An infinitely long,straight conductor $AB$ is fixed and a current $i_1$ is passed through it. Another movable straight wire $CD$ of finite length carrying current $i_2$ is held perpendicular to it and released. Neglect the weight of the wire.