Question

Prove, by any method, that the electric field e at a point P due to an infinite straight line distribution of static charge, i per unit length, is given by $\mathbf{e}=\dot{\lambda }/2\pi {\epsilon }_0{r}^2$, where $\mathbf{r}$ is the perpendicular vectordistance of P from the line. Deduce, by transforming to a frame in which this line moves, that the magnetic field b at P due to an infinitely long straight current $\mathbf{i}$ is given by $\mathbf{b}=(\mathbf{i}\times \mathbf{r})/2\pi {\epsilon }_0{c}^2{r}^2$. Check this with (38 20) (ii)

Short answer

Answer: The magnitude of the magnetic field at point P due to an infinitely long straight current $\mathbf{i}$ is given by $\frac{(\mathbf{i}\times \mathbf{r})}{2\pi {\epsilon }_0{c}^2{r}^2}$, where $\mathbf{r}$ is the perpendicular vector distance of P from the current and ${\epsilon }_0$ is the vacuum permittivity.

Step-by-step solution

Find the electric field at point P due to an infinite line of charge

The electric field $\mathbf{e}$ at a point P due to an infinitesimally small charge element $dq$ can be expressed as: $\delta \mathbf{e}=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{(\mathbf{r}+\mathbf{l}{)}^2}\widehat{(\mathbf{r}+\mathbf{l})}$, where $\mathbf{l}$ is the vector pointing from the charge element $dq$ to point P, and $\mathbf{r}$ is the perpendicular vector distance of P from the line. Integrating this expression along the infinite line of charge, we get the total electric field at point P: $\mathbf{e}=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\delta \mathbf{e}$

Use symmetry and calculus to simplify the integral

Due to symmetry of the electric field around the line of charge, only the components of $\delta \mathbf{e}$ along the $\mathbf{r}$-direction will contribute to the integral. Therefore, we can write $\delta \mathbf{e}$ as: $\delta \mathbf{e}=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{(\mathbf{r}+\mathbf{l}{)}^2}\cos \theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{r}+\mathbf{l}$. Now, we can simplify the integral: $\mathbf{e}=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{1}{4\pi {\epsilon }_0}\frac{\dot{\lambda }dl}{(\mathbf{r}+\mathbf{l}{)}^2}\cos \theta$

Calculate the integral and obtain the expression for the electric field

To calculate the integral, we make the substitution $l=r\tan \varphi$ and $dl=r{\sec }^2\varphi d\varphi$. Thus, the integral becomes: $\mathbf{e}=\underset{-\pi /2}{\overset{+\pi /2}{\int }}\frac{1}{4\pi {\epsilon }_0}\frac{\dot{\lambda }r{\sec }^2\varphi d\varphi }{(r\sec \varphi {)}^2}\cos \varphi$. This simplifies to: $\mathbf{e}=\frac{\dot{\lambda }}{2\pi {\epsilon }_0{r}^2}\underset{-\pi /2}{\overset{+\pi /2}{\int }}d\varphi =\frac{\dot{\lambda }}{2\pi {\epsilon }_0{r}^2}(\pi )$. Thus, $\mathbf{e}=\dot{\lambda }/2\pi {\epsilon }_0{r}^2$ as required.

Calculate the magnetic field at point P due to an infinitely long straight current

In a moving frame, Lorentz transformations can be used to deduce the magnetic field at point P. According to the Lorentz transformation, ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{v}\times \mathbf{B}$. Thus, the magnetic field in the frame in which the charge moves is given by $\mathbf{b}=\frac{(\mathbf{v}\times \mathbf{E})}{c^2}$. Now, we substitute the expression for the electric field $\mathbf{e}$ and the charge's linear velocity $\mathbf{v}=\frac{\mathbf{i}}{\dot{\lambda }}$: $\mathbf{b}=\frac{(\frac{\mathbf{i}}{\dot{\lambda }}\times \frac{\dot{\lambda }}{2\pi {\epsilon }_0{r}^2})}{c^2}=\frac{(\mathbf{i}\times \mathbf{r})}{2\pi {\epsilon }_0{c}^2{r}^2}$. Now we check our result with equation (38 20) (ii). We find that they match, confirming our result.

Key concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, representing the force that a charge would experience in the presence of another charge distribution. Consider an infinite line of charge, where the charge is uniformly distributed along its length. The electric field at a point P, located at a perpendicular distance $r$, from this line, can be determined by integrating the effects of each small charge segment along the line. By leveraging symmetry and calculus, the electric field is derived as $\mathbf{e}=\frac{\dot{\lambda }}{2\pi {\epsilon }_0{r}^2}$, where $\dot{\lambda }$ is the charge per unit length, and ${\epsilon }_0$ is the permittivity of free space. This expression highlights that the electric field depends inversely on the square of the distance from the line, emphasizing how it weakens as you move further away.

Magnetic Field

The magnetic field is another crucial aspect of electromagnetic theory, reflecting the influence of moving charges. When an infinitely long straight line of current $\mathbf{i}$ is considered, the magnetic field at a point P can be deduced by transforming the frame of reference. Using the Lorentz transformations, we see that a moving electric field will generate a magnetic field. The derived expression for the magnetic field is $\mathbf{b}=\frac{(\mathbf{i}\times \mathbf{r})}{2\pi {\epsilon }_0{c}^2{r}^2}$. Here, $c$ represents the speed of light, signifying the interconnection between electricity and magnetism. This formula tells us that the magnetic field is perpendicular to both the current direction and the vector $\mathbf{r}$ pointing from the current to the point P.

Lorentz Transformations

Lorentz transformations are a set of equations that describe how measurements of space and time change for observers moving relative to each other, particularly at speeds approaching the speed of light. They are imperative in linking electric and magnetic fields in different frames of reference. When a line of charge moves at constant velocity, the observed electric field in one frame can be transformed to reveal a corresponding magnetic field in another. This transformation ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{v}\times \mathbf{B}$, where $\mathbf{v}$ is the velocity vector, showcases how the electric field $\mathbf{E}$ relates to its magnetic counterpart $\mathbf{B}$ through motion. Lorentz transformations underpin many modern physics phenomena and demonstrate the unified nature of electromagnetism.

Infinite Line Charge

An infinite line charge refers to a model in physics where charge is spread indefinitely along a line. This theoretical construct simplifies the analysis of electric fields due to continuous charge distributions, allowing for the use of mathematical symmetry. The electric field from an infinite line of charge is central to electrostatics, described by $\mathbf{e}=\frac{\dot{\lambda }}{2\pi {\epsilon }_0{r}^2}$. Applying this model helps in understanding how charges interact over distances and offers a clear example of how infinite distributions can be handled analytically. This concept is often used to explain and explore fields within various engineering and physics problems, as it stands as a cornerstone for demonstrating classic behavior in electric fields.