Question

A point charge $+q$rests halfway between two steady streams of positive charge of equal charge per unit length $\lambda$, moving opposite directions and each at relative to point charge.With equal electric forces on the point charge, it would remain at rest. Consider the situation from a frame moving right at .(a) Find the charge per unit length of each stream in this frame.(b) Calculate the electric force and the magnetic force on the point charge in this frame, and explain why they must be related the way they are. (Recall that the electric field of a line of charge is $\lambda /2\pi {\epsilon }_{0}r$, that the magnetic field of a long wire is ${\mu }_{0}I/2\pi r$, and that the magnetic force is $qv\times B$. You will also need to relate $\lambda$and the current l.)

Short answer

(a) The charge per unit length of each stream are $\frac{\sqrt{8}}{3}{\lambda }_{\text{moving}}$and $\frac{5\sqrt{8}}{12}{\lambda }_{\text{moving}}$respectively.

(b) The electric and the magnetic force is$\frac{\sqrt{8}q{\lambda }_{\text{moving}}}{24\pi {\epsilon }_{0}r}$ and both are related because of the current and the length of the unit charge.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The value of the point charge is $+q$.
• The value of the length of the unit charge is$\lambda$ .
• The distance between the charges are $c/3$.

Significance of the electric field

The electric field is described as a region that helps an electrically charged particle to exert force on another particle. The electric field is mainly influenced by the electric charge.

(a) Determination of the charge per unit length

Here, the length per unit length is inversely proportional to the length which shows that $\lambda \propto \frac{1}{L}\propto \frac{{\gamma }_v}{L_0}$. Here, with the decrease in the length, the density of charge also increases by gamma.

The equation of the stationary length is expressed as:

${\lambda }_{\text{stationary}}=\frac{{\lambda }_{\text{moving}}}{{\gamma }_{\frac{c}{3}}}$ …(i)

The equation of the gamma factor is also expressed as:

${\gamma }_{\frac{c}{3}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Substitute $\frac{c}{3}$for V in the above equation.

$\begin{array}{rcl}{\gamma }_{\frac{c}{3}}& =& \frac{1}{\sqrt{1-\frac{c^2}{9c^2}}}\\ & =& \frac{1}{\sqrt{1-\frac{1}{9}}}\\ & =& \frac{1}{\sqrt{\frac{8}{9}}}\\ & =& \frac{3}{\sqrt{8}}\\ & & \end{array}$

Substitute the value of the above equation in the equation (i).

${\lambda }_{\text{stationary}}=\frac{\sqrt{8}}{3}{\lambda }_{\text{moving}}$

The equation of the initial velocity relative to the observer is expressed as:

$u\text{'}=\frac{u-v}{1-\frac{uv}{c^2}}$

Substitute $-\frac{c}{3}$for u and $\frac{c}{3}$for v in the above equation

$\begin{array}{rcl}u\text{'}& =& \frac{-\frac{c}{3}-\frac{c}{3}}{1-\frac{-\frac{c}{3}\times \frac{c}{3}}{c^2}}\\ & =& \frac{-\frac{2c}{3}}{1+\frac{1}{9}}\\ & =& \frac{-\frac{2c}{3}}{\frac{10}{9}}\\ & =& -0.6c\\ & & \end{array}$

The equation of the gamma is expressed as:

${\gamma }_{0.6c}=\frac{1}{\sqrt{1-\frac{{u\text{'}}^2}{c^2}}}$

Substitute$-0.6c$for $u\text{'}$in the above equation.

$\begin{array}{rcl}{\gamma }_{0.6c}& =& \frac{1}{\sqrt{1-\frac{{\left(-0.6c\right)}^2}{c^2}}}\\ & =& \frac{1}{\sqrt{1-\frac{\left(0.36c^2\right)}{c^2}}}\\ & =& \frac{1}{\sqrt{1-0.36}}\\ & =& \frac{5}{4}\\ & & \end{array}$

The equation of the length of the unit charge is expressed as:

${\lambda }_{0,6c}={\gamma }_{0,6c}{\lambda }_{\text{stationary}}$

Substitute the values in the above equation.

$\begin{array}{rcl}{\lambda }_{0,6c}& =& \frac{5}{4}\times \frac{\sqrt{8}}{3}{\lambda }_{\text{moving}}\\ & =& \frac{5\sqrt{8}}{12}{\lambda }_{\text{moving}}\\ & & \end{array}$

Thus, the charge per unit length of each stream are$\frac{\sqrt{8}}{3}{\lambda }_{\text{moving}}$ and $\frac{5\sqrt{8}}{12}{\lambda }_{\text{moving}}$respectively.

(b) Determination of the electric and magnetic force 

The equation of the electric field is expressed as:

$\begin{array}{rcl}{F}_E& =& qE\\ & =& \frac{q}{2\pi \epsilon }\frac{\lambda }{r}\\ & & \end{array}$

Substitute the value of the part (a) in the above equation

$\begin{array}{rcl}{F}_E& =& qE\\ & =& \frac{q}{2\pi {\epsilon }_0}\frac{1}{r}\left(\frac{5\sqrt{8}}{12}{\lambda }_{\text{moving}}-\frac{\sqrt{8}}{3}{\lambda }_{\text{moving}}\right)\\ & =& \frac{\sqrt{8}q}{24\pi {\epsilon }_{0}r}{\lambda }_{\text{moving}}\\ & & \end{array}$

The equation of the magnetic field is expressed as:

$F_B=q{v}_{\text{charge}}\times B$

Substitute $\frac{\mu I}{2\pi r}$for B in the above equation.

$F_B=q{v}_{\text{charge}}\times \frac{\mu I}{2\pi r}$

Substitute $\frac{\sqrt{8}}{4}c{\lambda }_{\text{moving}}$for l in the above equation.

$\begin{array}{rcl}{F}_B& =& q{v}_{\text{charge}}\times \frac{\mu }{2\pi r}\times \frac{\sqrt{8}}{4}c{\lambda }_{\text{moving}}\\ & =& q\left(\frac{c}{3}\right)\frac{\mu c{\lambda }_{\text{moving}}}{\sqrt{8}\pi r}\\ & =& \frac{\sqrt{8}q\mu c^2{\lambda }_{\text{moving}}}{24\pi r}\\ & & \end{array}$

Substitute $\frac{1}{{\epsilon }_0\mu }$for $c^2$in the above equation.

$F_B=\frac{\sqrt{8}q{\lambda }_{\text{moving}}}{24\pi {\epsilon }_{0}r}$

Thus, the electric and the magnetic force is$\frac{\sqrt{8}q{\lambda }_{\text{moving}}}{24\pi {\epsilon }_{0}r}$ and both are related because of the current and the length of the unit charge.