Новое решение

Alexphysics правка от 17 июн. 2026 г., 15:16

правка #18992 позже →

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		+	### Statement
		+
		+	$14.3.23.$ [Insert the problem statement]
		+
		+	### 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$.
		+
		+	In the laboratory system S, the solenoid moves with velocity$ \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:
		+
		+	$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad
		+	\mathbf{m} = \gamma\,\mathbf{M}$
		+
		+	The magnetic moment in S increases by a factor of \gamma, 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 $1/c^2$
		+	$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M} $
		+	(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:
		+	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}).}
		+
		+	In the low-velocity limit (\gamma \to 1):
		+
		+	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]

	@@ -0,0 +1,60 @@
			### Statement

			$14.3.23.$ [Insert the problem statement]

			### 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$.

			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:

			$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad
			\mathbf{m} = \gamma\,\mathbf{M}$

			The magnetic moment in S increases by a factor of \gamma, 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 $1/c^2$
			$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M} $
			(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:
			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}).}

			In the low-velocity limit (\gamma \to 1):

			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]