Question

A long straight wire AB carries a current I. A proton P levels with a speed V, parallel to the wire at a distance d from it in a direction opposite to the current. What is the force experienced by the proton and what is its direction?

Answer 1

Magnetic field due to current carrying wire is perpendicular to plane of paper-
downward.
i.e., $\overrightarrow{B}=-\frac{{\mu }_{0}I}{2\pi d}\overrightarrow{k}$
Force $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}=e(-v\hat{j})\times (-\frac{{\mu }_{0}I}{2\pi d}\overrightarrow{k})=\frac{{\mu }_{0}evI}{2\pi d}\overrightarrow{i}$
That is the magnetic force has magnitude $\frac{{\mu }_{0}evI}{2\pi d}\overrightarrow{i}$ and is directed along positive $X-$axis i.e.,
in the plane of paper perpendicular to direction of $\overrightarrow{v}$ and to the right.