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Luisito edited Jun 4, 2026, 6:37 PM

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revision #18267 newer →

en/10.1.2.md +30 −0

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		+	### Statement
		+
		+	$10.1.2.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Inside the magnetic field, the electron moves following a circular trajectory with constant speed.
		+	From Newton Second Law:
		+
		+	$\frac{m_e v^2}{R} = e \|\vec{v} \times \vec{B}\|$
		+
		+	but $\vec{v}$ is perpendicular to $\vec{B}$, so
		+
		+	$\frac{m_e v^2}{R} = e v B$
		+
		+	$R = \frac{m_e v}{eB}$ (1)
		+
		+	Applying Energy Conservation Law:
		+
		+	$\frac{m_e v^2}{2} = e U$
		+
		+	$v = \sqrt{\frac{2eU}{m_e}}$ (2)
		+
		+	Substituting (2) into (1)
		+
		+	$R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,30 @@
			### Statement

			$10.1.2.$ [Insert the problem statement]

			### Solution

			Inside the magnetic field, the electron moves following a circular trajectory with constant speed.
			From Newton Second Law:

			$\frac{m_e v^2}{R} = e \|\vec{v} \times \vec{B}\|$

			but $\vec{v}$ is perpendicular to $\vec{B}$, so

			$\frac{m_e v^2}{R} = e v B$

			$R = \frac{m_e v}{eB}$ (1)

			Applying Energy Conservation Law:

			$\frac{m_e v^2}{2} = e U$

			$v = \sqrt{\frac{2eU}{m_e}}$ (2)

			Substituting (2) into (1)

			$R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$

			#### Answer

			[Insert a concise answer or boxed result]