Новое решение

Luisito правка от 4 июн. 2026 г., 17:57

правка #18265 позже →

en/8.1.1.md +37 −0

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		+	### 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
		+
		+	[Insert a concise answer or boxed result]

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			### 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

			[Insert a concise answer or boxed result]