Statement
8.1.1. [Insert the problem statement]
Solution
The electric current is defined by:
$i = \frac{dq}{dt}$
but $dq = e dn$, where e is the fundamental electrical charge and $dt = ds/c$, where c is the speed of light.
$i = ce \frac{dn}{ds}$
we can approximate $\frac{dn}{ds}$ to $\frac{n}{\ell}$, son
$i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$
b) Again, electric current can be expressed as
$i = \frac{dq}{dt} \simeq \frac{e}{t}$ (1)
Applying Newton Second Law:
$\frac{e^2}{4\pi\varepsilon r^2} = \frac{m_e v^2}{r}$
where $v = \frac{2\pi r}{t}$,
$t = \sqrt{\frac{16 (\pi r)^3 m_e}{e^2}}$ (2)
Putting (2) into (1),
$i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$
Answer
a) $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$
b) $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$
[Insert a concise answer or boxed result]