The magnetic force on a current-carrying wire in an external uniform magnetic field depends on:

$A$ wire carrying a current $I$ is placed inside a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. The shape of the wire is parabolic and has the equation $y = 2x - x^2$. The force on the wire will be:

Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is

The force between two long parallel wires is $F$ when each one of them carries a certain current $I$. If the current in each is halved,the force between them would be . . . . . . .

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 $1$?

$A$ wire of length $L$ carries a current $I$ along the $X$-axis. The magnetic force acting on the wire is given by $\vec{F} = I B_0 L(\hat{k} - \hat{j})$. The existing magnetic field $\vec{B}$ is