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en/14.3.23.md
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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] | ||
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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] | |||