Question

Two very long straight, parallel wires carry steady currents $I$ and $-I$ respectively. The distance between the wires is $d$. AT a certain instant of time, at a point charge $q$ is at a point equidistant from the two wires, in the plane of the wires. its instantaneous velocity $v$ is perpendicular to the plane of wires. the magnitude of force due to the magnetic field acting on the charge at the instant is

• A. $\frac{3{\mu }_{0}Iqv}{2\pi d}$
• B. $\frac{{2\mu }_{0}Iqv}{\pi d}$
• C. $\frac{{\mu }_{0}Iqv}{2\pi d}$
• D. zero

Answer 1

The correct option is D zero

Given: Current are I & -I ,distance between wires =d ,charge=q, velocity =v

Solution: If we let that I is inwards and -I is outwards and distance between wires is d then force on the charge which is placed equidistant from two wires is:

$F=q(\overrightarrow{v}\times \overrightarrow{B})=qvBsin\theta$

Remember here that the angle between B& v is ${180}^0$ because according to the question velocity is perpendicular to the plane of wires and magnetic field of both the wires directs in same direction.

We know that sin${180}^0$=0

So, F=0

Hence the correct answer is D