Правка разделов «Statement», «Solution», «Answer»

Alexphysics правка от 17 июн. 2026 г., 15:33

правка #18993 предыдущая #18992 ← раньше позже →

en/14.3.23.md +31 −21

		@@ -1,20 +1,29 @@
			### Statement

		−	$14.3.23.$ ~~[Insert the problem statement]~~
		+	$14.3.23.$
		+	Solve problem 14.3.22 in the case of a capacitor moving at a velocity βc
		+	pendicular to the plates

			### Solution

			We start analysing Rest frame S'

			In the system S' the solenoid is at rest:

		−	The solenoid is "long and flat": its turns define a plane; the magnetic moment ~~\$mathbf{M} $is perpendicular to that plane (normal direction). Let us take $\mathbf{M}' = M\,\hat{\mathbf{z}~~}$
		−	~~$There is no electric moment$~~ \mathbf{p}' = ~~0$.~~
		+	The solenoid is "long and flat": its turns define a plane; the magnetic moment $\mathbf{M}$
		+	is perpendicular to that plane (normal direction). Let us take
		+	take $\mathbf{M}' =M\,\hat{\mathbf{z}}$

		−	In the laboratory system S, the solenoid moves with velocity$ \mathbf{v} = v\,\hat{\mathbf{x}}$ parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$
		+	There is no electric moment
		+	$\mathbf{p}' = 0$.

		−	~~Under a boost with velocity$ \mathbf{v}$ the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:~~
		+	In the laboratory system S, the solenoid moves with velocity $\mathbf{v} = v\,\hat{\mathbf{x}}$
		+	parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$

		+	Under a boost with velocity
		+	$ \mathbf{v}$
		+	the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:
		+
			$\begin{aligned}
			\mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\
			\mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right).
			\end{aligned}$

			In our case:$ \mathbf{p}' = 0$ and $\mathbf{v} \cdot \mathbf{m}' = 0 $(because they are perpendicular)
			Therefore:

		@@ -26,35 +35,36 @@Solution
			$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad
			\mathbf{m} = \gamma\,\mathbf{M}$

		−	~~The magnetic moment in S increases by a factor of \gamma, an~~d an electric moment appears perpendicular to $\mathbf{v} and to \mathbf{M}$
		+	The magnetic moment in S increases by a factor of $\gamma$
		+	and an electric moment appears perpendicular to $\mathbf{v}$ and to $\mathbf{M}$

			For $ v \ll c (\gamma \approx 1)$

			In CGS: $\displaystyle \mathbf{p} \approx \frac{1}{c}\,\mathbf{v} \times \mathbf{M}$.

		−	In SI: the factor is ~~$1/c^2$~~
		+	In SI: the factor is
		+	$1/c^2$
			$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M} $
		−	(with $\mathbf{M} in A·m²$)

		+	(with $\mathbf{M}$ in $A·m²$)
		+
			Magnitude (if $\mathbf{v} \perp \mathbf{M})$:
		−	~~$ p = \dfrac{v}{c}\,M (CGS)$ or~~ p = ~~\dfrac~~{v}{c^2}\,M$ (SI).
		−	Direction:
		+	$ p = \frac{v}{c}\,M (CGS)$
		+	or $ p = \frac{v}{c^2}\,M$(SI)
		+
		+	Direction:
			perpendicular to$ \mathbf{v} $and to $\mathbf{M}$,
			that is, parallel to the flat surfaces but perpendicular to the motion.

		−	5. Final result

		−	\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad
		−	\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI}).}
		+	#### Answer

		−	In the low-velocity limit (\gamma \to 1):
		+	$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad
		+	\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$

		−	p \approx \frac{v}{c}M \;\;(\text{CGS}), \qquad
		−	p \approx \frac{v}{c^2}M \;\;(\text{SI}).
		+	In the low-velocity limit $(\gamma \to 1)$:

		−	The electric moment appears because, when in motion, the currents that produce the magnetic moment are no longer simultaneous in the new system, generating an effective dipolar charge distribution.
		+	$p \approx \frac{v}{c}M \;\;(\text{CGS}), \qquad
		+	p \approx \frac{v}{c^2}M \;\;(\text{SI})$

		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]
		+	The electric moment appears because, when in motion, the currents that produce the magnetic moment are no longer simultaneous in the new system, generating an effective dipolar charge distribution.

	@@ -1,20 +1,29 @@
	### Statement		### Statement

	$14.3.23.$ ~~[Insert the problem statement]~~		$14.3.23.$
			Solve problem 14.3.22 in the case of a capacitor moving at a velocity βc
			pendicular to the plates

	### Solution		### Solution

	We start analysing Rest frame S'		We start analysing Rest frame S'

	In the system S' the solenoid is at rest:		In the system S' the solenoid is at rest:

	The solenoid is "long and flat": its turns define a plane; the magnetic moment ~~\$mathbf{M} $is perpendicular to that plane (normal direction). Let us take $\mathbf{M}' = M\,\hat{\mathbf{z}~~}$		The solenoid is "long and flat": its turns define a plane; the magnetic moment $\mathbf{M}$
	~~$There is no electric moment$~~ \mathbf{p}' = ~~0$.~~		is perpendicular to that plane (normal direction). Let us take
			take $\mathbf{M}' =M\,\hat{\mathbf{z}}$

	In the laboratory system S, the solenoid moves with velocity$ \mathbf{v} = v\,\hat{\mathbf{x}}$ parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$		There is no electric moment
			$\mathbf{p}' = 0$.

	~~Under a boost with velocity$ \mathbf{v}$ the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:~~		In the laboratory system S, the solenoid moves with velocity $\mathbf{v} = v\,\hat{\mathbf{x}}$
			parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$

			Under a boost with velocity
			$ \mathbf{v}$
			the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:

	$\begin{aligned}		$\begin{aligned}
	\mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\		\mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\
	\mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right).		\mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right).
	\end{aligned}$		\end{aligned}$

	In our case:$ \mathbf{p}' = 0$ and $\mathbf{v} \cdot \mathbf{m}' = 0 $(because they are perpendicular)		In our case:$ \mathbf{p}' = 0$ and $\mathbf{v} \cdot \mathbf{m}' = 0 $(because they are perpendicular)
	Therefore:		Therefore:

	@@ -26,35 +35,36 @@Solution
	$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad		$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad
	\mathbf{m} = \gamma\,\mathbf{M}$		\mathbf{m} = \gamma\,\mathbf{M}$

	~~The magnetic moment in S increases by a factor of \gamma, an~~d an electric moment appears perpendicular to $\mathbf{v} and to \mathbf{M}$		The magnetic moment in S increases by a factor of $\gamma$
			and an electric moment appears perpendicular to $\mathbf{v}$ and to $\mathbf{M}$

	For $ v \ll c (\gamma \approx 1)$		For $ v \ll c (\gamma \approx 1)$

	In CGS: $\displaystyle \mathbf{p} \approx \frac{1}{c}\,\mathbf{v} \times \mathbf{M}$.		In CGS: $\displaystyle \mathbf{p} \approx \frac{1}{c}\,\mathbf{v} \times \mathbf{M}$.

	In SI: the factor is ~~$1/c^2$~~		In SI: the factor is
			$1/c^2$
	$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M} $		$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M} $
	(with $\mathbf{M} in A·m²$)

			(with $\mathbf{M}$ in $A·m²$)

	Magnitude (if $\mathbf{v} \perp \mathbf{M})$:		Magnitude (if $\mathbf{v} \perp \mathbf{M})$:
	~~$ p = \dfrac{v}{c}\,M (CGS)$ or~~ p = ~~\dfrac~~{v}{c^2}\,M$ (SI).		$ p = \frac{v}{c}\,M (CGS)$
	Direction:		or $ p = \frac{v}{c^2}\,M$(SI)

			Direction:
	perpendicular to$ \mathbf{v} $and to $\mathbf{M}$,		perpendicular to$ \mathbf{v} $and to $\mathbf{M}$,
	that is, parallel to the flat surfaces but perpendicular to the motion.		that is, parallel to the flat surfaces but perpendicular to the motion.

	5. Final result

	\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad		#### Answer
	\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI}).}

	In the low-velocity limit (\gamma \to 1):		$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad
			\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$

	p \approx \frac{v}{c}M \;\;(\text{CGS}), \qquad		In the low-velocity limit $(\gamma \to 1)$:
	p \approx \frac{v}{c^2}M \;\;(\text{SI}).

	The electric moment appears because, when in motion, the currents that produce the magnetic moment are no longer simultaneous in the new system, generating an effective dipolar charge distribution.		$p \approx \frac{v}{c}M \;\;(\text{CGS}), \qquad
			p \approx \frac{v}{c^2}M \;\;(\text{SI})$

			The electric moment appears because, when in motion, the currents that produce the magnetic moment are no longer simultaneous in the new system, generating an effective dipolar charge distribution.
	#### Answer

	[Insert a concise answer or boxed result]