Q. Two very long straight parallel wires carry current $i\ and\ 2\ i$ in opposite directions. The distance between the wires is r. At a certain instant of time a point charge q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $\overrightarrow{V}\ $is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is
A) zero
B) $\frac{3\mu_{0}}{2\pi}\frac{i\ q\ v}{r}$
C) $\frac{\mu_{0}}{\pi}\frac{i\ q\ v}{r}$
D) $\frac{\mu_{0}}{2\pi}\frac{i\ q\ v}{r}$
Explanation
(a) The magnetic field induction at $P$ due to currents through both wise is
$B = \frac{\mu_{0}}{4\pi}\ \frac{2i}{r/2}$ + $\frac{\mu_{0}}{4\pi}\ \frac{2(2i)}{r/2}$
$= \ \frac{\mu_{0}}{4\pi} \times \ \frac{12i}{r}$ acting perpendicular to plane of wire inwards.
Now, $\overrightarrow{B}$ and $\overrightarrow{V}$ are acting in the same direction $ie,\ \theta = 0^{0}.$
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Force on charged particle is $F = qvB\sin{\theta = q\ v\ B \times 0 = 0.}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.