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Chapter 15: Problem 64

A long wire having linear charge density \(\lambda\) moving with constant velocity \(v\) along its length. A point charge moving with same speed in opposite direction and at that instant, it is \(r\) distance from the wire. The net force acting on the charge is given by (A) \(\frac{\lambda q}{2 \pi r}\left[\frac{1}{\varepsilon_{0}}+v^{2} \mu_{0}\right]\) (B) \(\frac{\lambda q}{2 \pi r}\left[\frac{1}{\varepsilon_{0}}-\mu_{0} v^{2}\right]\) (C) \(\frac{\lambda q}{2 \pi r} \sqrt{\left(\frac{1}{\varepsilon_{0}}\right)^{2}+v^{4} \mu_{0}^{2}}\) (D) Zero

Short Answer

Expert verified
The correct answer is (A) \( \frac{\lambda q}{2 \pi r}\left[\frac{1}{\varepsilon_{0}}+v^{2} \mu_{0}\right] \). The net force acting on the charge is the electric force, as the magnetic force does not do any work on the charge.

Step by step solution

01

- Understanding the Electric Field

The electric field (\( E \)) around an infinite wire with linear charge density \( \lambda \) is given by \( E = \frac{\lambda }{2 \pi \epsilon_{0} r} \), where \( \epsilon_{0} \) is the permittivity of free space and \( r \) is the distance from the wire. The electric field points radially out from the wire. Therefore, the electric force on the point charge is \( F_{e} = qE = \frac{\lambda q}{2 \pi \epsilon_{0} r} \).
02

- Understanding the Magnetic Field

As the wire is moving, it also creates a magnetic field. According to right hand rule, the direction of the magnetic field (\( B \)) is circular, perpendicular to the direction of current. The magnetic field around a current carrying wire is given by \( B = \mu_{0} \lambda v / 2 \pi r \), where \( \mu_{0} \) is the permeability of free space. But the magnetic force acts perpendicular to the velocity of the point charge, so the magnetic force can't do any work on the point charge and doesn't contribute in the net force.
03

- Calculating the net force

Since the magnetic force doesn't contribute to the net force, the net force acting on the point charge is caused by the electric field only. Therefore the net force is \( F = F_{e} =\frac{\lambda q }{2 \pi \epsilon_{0} r}\)

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