91Ó°ÊÓ

a)

b) You may try resolving via defining the charge in a tiny volume $dV$:

$dq=\mathrm{ÒÏ}dV$

$dV=2r\mathrm{Ï€}Ldr$(differentiate the volume of cylinder)

$dq=\frac{{\mathrm{ÒÏ}}_{0}r}{R}â‹\dots 2r\mathrm{Ï€}Ldr$ (try substituting$dV$and $\mathrm{ÒÏ}$)

$Q=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}L}{R}{∫}_0^R r^{2}dr$( To obtain net charge integrating it from $0$to $r$)

$Q=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}L}{R}{\left[\frac{r^3}{3}\right]}_0^R$

$Q=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}L}{R}\left[\frac{R^3}{3}\right]$

$Q=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}L{R}^2}{3}$

In additional, you knew about localid="1648884695068" $Q=\mathrm{λ}L$.

$Q=\mathrm{λ}L$

$\mathrm{λ}L=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}L{R}^2}{3}$(try substituting$Q$)

localid="1648884669435" ${\mathrm{ÒÏ}}_0=\frac{3\mathrm{λ}}{2R^2\mathrm{Ï€}}$(show${\mathrm{ÒÏ}}_0$)

c) For obtain an electromagnetic field at $r≤R$, you need to adopt Gauss equation. Its helical Probabilistic distribution was selected.

$\text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Ï\mu }}_0}$

$\text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}=EA$(A cylinder is a Gaussian surface)

$EA=\frac{Q_{in}}{{\mathrm{Ï\mu }}_0}$

$E=\frac{Q_{in}}{A{\mathrm{Ï\mu }}_0}$

$A=2r\mathrm{Ï€}l$(cylinder's surface area)

$Q_{in}=∫\mathrm{ÒÏ}dV$(charge in the surface)

Then you will obtain $Q_{\text{in}}$using the same calculation as seen at portion a).

$Q_{\text{in}}=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}l}{R}{∫}_0^r r^{2}dr$ (integrate it from localid="1648907852327" $0$to localid="1648907900366" $r$)

$Q_{in}=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}l}{R}{\left[\frac{r^3}{3}\right]}_0^r$ (solve the integral)

$Q_{in}=\frac{2{\mathrm{ÒÏ}}_0\mathrm{Ï€}l}{R}\left[\frac{r^3}{3}\right]$

$Q_{in}=\frac{2\mathrm{Ï€}l{r}^3}{3R}â‹\dots \frac{3\mathrm{λ}}{2R^2\mathrm{Ï€}}$ (try substituting ${\mathrm{ÒÏ}}_0$)

$Q_{in}=\frac{\mathrm{λ}l{r}^3}{R^3}$

$E=\frac{\mathrm{λ}l{r}^3}{R^3}â‹\dots \frac{1}{2r\mathrm{Ï€}l{\mathrm{Ï\mu }}_0}$(put $A$and $Q_{in}$to formula for $E$)

$E=\frac{\mathrm{λ}{r}^2}{2R^3\mathrm{Ï€}{\mathrm{Ï\mu }}_0}$

d) To $r=R$you obtain,

$E=\frac{\mathrm{λ}{R}^2}{2R^3\mathrm{Ï€}{\mathrm{Ï\mu }}_0}$

$E=\frac{\mathrm{λ}}{2R\mathrm{Ï€}{\mathrm{Ï\mu }}_0}$

Considering every lengthy electrified cable, you see an equation to such electromagnetic current.