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