Question

(a) Use Eq. 10.75 to calculate the electric field a distanced from an infinite straight wire carrying a uniform line charge .$\lambda$, moving at a constant speed down the wire.

(b) Use Eq. 10.76 to find the magnetic field of this wire.

Short answer

Answer

(a) The expression of the electric field is$\frac{2\lambda }{4\pi {\epsilon }_{0}d}$ .

(b) The expression for magnetic field is$\frac{{\mu }_0}{4\pi }\frac{2I}{d}$.

Step-by-step solution

Write the given data from the question.

The uniform line charge is$\lambda$.

The constant speed is V.

Determine the equation to calculate the electric field and magnetic field.

The equation to calculate the electric field (Eq. 10.75) is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\frac{1-\frac{v^2}{c^2}}{{\left(1-\frac{v^2}{c^2}si{n}^2\theta \right)}^{\frac{3}{2}}}\right)\frac{\hat{\mathbf{R}}}{{\mathbf{R}}^{\mathbf{2}}}$ …… (1)

Here, q is the charge, v is the speed, c is the velocity of light, R is the vector from the present location of the particle to and r is the angle between the R and V.

The expression for the magnetic field (Eq. 10.76) is given as follows.

$\mathit{B}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\left(v\times E\right)$ …… (2)

Calculate the expression for the electric field.

Consider the figure given below

Here d and are R the distances.

The expression of sin of angle is given by.

$\sin \theta =\frac{d}{R}$

Take a square of above equation.

$$\begin{gathered} {\sin }^2\theta =\frac{d^2}{R^2} \\ \frac{1}{R^2}=\frac{{\sin }^2\theta }{d^2} \end{gathered}$$

The value of cosine of angle is given by,

$$\begin{gathered} \cos \theta =-\frac{x}{d} \\ x=-d\cos \theta \end{gathered}$$

Here x is the distance on infinite long wire.

$$\begin{gathered} dx=-d\left(-\cos e{c}^2\theta \right)d\theta \\ dx=\frac{d}{{\sin }^2\theta }d\theta \end{gathered}$$

Multiply$\frac{1}{R^2}$ both the sides of the above equation.

$\frac{1}{R^2}dx=\frac{1}{R^2}\frac{d}{{\sin }^2\theta }d\theta$

Substitute$\frac{{\sin }^2\theta }{d^2}$ for$\frac{1}{R^2}$ into above equation.

$$\begin{gathered} \frac{1}{R^2}dx=\frac{{\sin }^2\theta }{d^2}\frac{d}{{\sin }^2\theta }d\theta \\ \frac{1}{R^2}dx=\frac{1}{d}d\theta \\ dx=\frac{R^2}{d}d\theta \end{gathered}$$

The vertical component of is R .

The electric field due to the line charge is given by,

$E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\int \frac{\hat{R}}{R^2}\frac{dx}{{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}}$

Substitute$\frac{R^2}{d}d\theta$ for dx into above equation.

role="math" localid="1658292396875" $$\begin{gathered} E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right){\int }_0^{\pi }\frac{\hat{y}}{R^2}\frac{\frac{R^2}{d}d\theta \sin \theta }{{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}} \\ E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\frac{\hat{y}}{d}{\int }_0^{\pi }\frac{\sin \theta d\theta }{{\left(1-\frac{v^2}{c^2}{\sin }^2\theta \right)}^{\frac{3}{2}}} \end{gathered}$$

…….. (3)

Let assume,

$$\begin{gathered} z=\cos \theta \\ dz=-\sin \theta d\theta \end{gathered}$$

Substitute $1-z^2$for ${\sin }^2\theta$ into equation (3).

$$\begin{gathered} E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\frac{\hat{y}}{d}{\int }_{-1}^1\frac{dz}{{\left(1-\frac{v^2}{c^2}\left(1-z^2\right)\right)}^{\frac{3}{2}}} \\ E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\frac{\hat{y}}{d}{\int }_{-1}^1\frac{dz}{{\left(1-\frac{v^2}{c^2}+\frac{v^2}{c^2}{z}^2\right)}^{\frac{3}{2}}} \\ E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\frac{\hat{y}}{d}{\left[\frac{1}{{\left(\frac{v}{c}\right)}^3}\frac{z}{\left({\left(\frac{c}{v}\right)}^2-1\right)\sqrt{{\left(\frac{c}{v}\right)}^2-1+z^2}}\right]}_{-1}^1 \\ E=\frac{\lambda }{4\pi {\epsilon }_0}\left(1-\frac{v^2}{c^2}\right)\frac{\hat{y}}{d}\left[\frac{c}{v}\right]\frac{1}{\left(1-\frac{v^2}{c^2}\right)}\frac{2}{\sqrt{{\left(\frac{c}{v}\right)}^2}} \end{gathered}$$

Solve further as,

$$\begin{gathered} E=\frac{\lambda }{4\pi {\epsilon }_{0}d}\left(1-\frac{v^2}{c^2}\right)\frac{c}{v}\frac{1}{\left(1-\frac{v^2}{c^2}\right)}\frac{2}{\sqrt{{\left(\frac{c}{v}\right)}^2}}\hat{y} \\ E=\frac{2\lambda }{4\pi {\epsilon }_{0}d}\hat{y} \end{gathered}$$

Hence the expression of the electric field is $\frac{2\lambda }{4\pi {\epsilon }_{0}d}$.

Determine the expression for the magnetic field.

Calculate the expression for magnetic field.

Substitute $\frac{2\lambda }{4\pi {\epsilon }_{0}d}\hat{y}$for E into equation (2).

$$\begin{gathered} B=\frac{1}{c^2}\left(v\hat{x}\times \frac{2\lambda }{4\pi {\epsilon }_{0}d}\hat{y}\right) \\ B=\frac{1}{c^2}\frac{2\lambda v}{4\pi {\epsilon }_{0}d}\left(\hat{x}\times \hat{y}\right) \\ B=\frac{1}{c^2}\frac{1}{4\pi {\epsilon }_0}\frac{2\lambda v}{d}\hat{z} \end{gathered}$$

Substitute I for$\lambda v$ and ${\mu }_0{\epsilon }_0$for $\frac{1}{c^2}$ into above equation.

$$\begin{gathered} B={\mu }_0{\epsilon }_0\frac{1}{4\pi {\epsilon }_0}\frac{2I}{d}\hat{z} \\ B=\frac{{\mu }_0}{4\pi }\frac{2I}{d}\hat{z}\backslash \end{gathered}$$

Hence the expression for magnetic field is$\frac{{\mu }_0}{4\pi }\frac{2I}{d}$.