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 \(\mathbf{e}=\dot{\lambda} / 2 \pi \varepsilon_{0} r^{2}\), 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 \(\mathbf{b}=(\mathbf{i} \times \mathbf{r}) / 2 \pi \varepsilon_{0} c^{2} r^{2}\). Check this with (38 20) (ii)

The electric field \(\mathbf{e}\) at a point P due to an infinitesimally small charge element \(dq\) can be expressed as: \(\delta \mathbf{e} = \frac{1}{4\pi \varepsilon_{0}}\frac{dq}{(\mathbf{r}+\mathbf{l})^2} \widehat{(\mathbf{r}+\mathbf{l})}\), 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: \(\mathbf{e}=\int\limits_{-\infty}^{+\infty} \delta \mathbf{e}\)

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: \(\delta \mathbf{e} = \frac{1}{4\pi \varepsilon_{0}}\frac{dq}{(\mathbf{r}+\mathbf{l})^2} \cos{\theta}\), where \(\theta\) is the angle between \(\mathbf{r}\) and \(\mathbf{r}+\mathbf{l}\). Now, we can simplify the integral: \(\mathbf{e} = \int\limits_{-\infty}^{+\infty} \frac{1}{4\pi \varepsilon_{0}}\frac{\dot{\lambda} dl}{(\mathbf{r}+\mathbf{l})^2} \cos{\theta}\)

To calculate the integral, we make the substitution \(l = r\tan{\phi}\) and \(dl = r\sec^2{\phi} d\phi\). Thus, the integral becomes: \(\mathbf{e} = \int\limits_{-\pi/2}^{+\pi/2} \frac{1}{4\pi \varepsilon_{0}}\frac{\dot{\lambda} r\sec^2{\phi} d\phi}{(r\sec{\phi})^2} \cos{\phi}\). This simplifies to: \(\mathbf{e} = \frac{\dot{\lambda}}{2\pi \varepsilon_{0} r^2}\int\limits_{-\pi/2}^{+\pi/2} d\phi = \frac{\dot{\lambda}}{2\pi \varepsilon_{0} r^2}(\pi)\). Thus, \(\mathbf{e}=\dot{\lambda} / 2 \pi \varepsilon_{0} r^{2}\) as required.

In a moving frame, Lorentz transformations can be used to deduce the magnetic field at point P. According to the Lorentz transformation, \(\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B}\). 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 \(\mathbf{v} = \frac{\mathbf{i}}{\dot{\lambda}}\): \( \mathbf{b} = \frac{(\frac{\mathbf{i}}{\dot{\lambda}} \times \frac{\dot{\lambda}}{2\pi \varepsilon_{0} r^2})}{c^2} = \frac{(\mathbf{i} \times \mathbf{r})}{2\pi \varepsilon_{0} c^2 r^2}\). Now we check our result with equation (38 20) (ii). We find that they match, confirming our result.