$A$ vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed towards the north. The wire will experience a force directed towards:

Two very long,straight,parallel conductors $A$ and $B$ carry currents of $5\,A$ and $10\,A$ respectively and are at a distance of $10\,cm$ from each other. The direction of current in the two conductors is the same. The force acting per unit length between the two conductors is: $(\mu_0 = 4\pi \times 10^{-7} \text{ SI unit})$

Three long current-carrying wires $P, Q,$ and $R$ are placed perpendicular to the plane of the paper as shown. The magnetic force per unit length on wire $R$ is:

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:

Write the magnetic force equation on a current-carrying element $I\vec{dl}$ inside a magnetic field $\vec{B}$. Write the law used to determine the direction of the magnetic force.

Two long wires with no contact are placed perpendicular to each other. $i_1$ and $i_2$ are currents flowing through these wires respectively. The magnetic force on a small length '$d$' of the second wire situated at a distance '$l$' from the first wire is proportional to,