Question

Two very long straight parallel wires carry steady currents $i$ and $2i$ in opposite directions. The distance between the wires is $d$. 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 whose magnitude is $v$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

• A. $\frac{{\mu }_{0}iqv}{2\pi d}$
• B. $\frac{{\mu }_{0}iqv}{\pi d}$
• C. $\frac{3{\mu }_{0}lqv}{2\pi d}$
• D. Zero

Answer 1

The correct option is D Zero

Using the Right Hand Rule, the direction of magnetic field along the middle axis due to both of these wires is into the plane (-z direction).


Force acting on the charge is given by
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
$=qvB\sin \theta \hat{n}$

Given that velocity of charge is also perpendicular to the plane, we can say $\theta =0^{\circ }\text{or}{180}^{\circ }\text{i.e}\sin \theta =0$
$\therefore \overrightarrow{F}=0$. Net force acting on the charge due to the magnetic field is zero.