| ### Statement | | ### Statement |
| | | |
| $14.3.25.$ | | $14.3.25.$ |
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| Solve problem 14.3.24 for a round long solenoid | | Solve problem 14.3.24 for a round long solenoid |
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| ### Solution | | ### Solution |
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| first Analyse the rest frame S' | | first Analyse the rest frame S' |
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| We have a long cylindrical solenoid, with axis along z'. | | We have a long cylindrical solenoid, with axis along z'. |
| rest magnetic moment | | rest magnetic moment |
| $\mathbf{M}' = M\,\hat{\mathbf{z}}'$ | | $\mathbf{M}' = M\,\hat{\mathbf{z}}'$ |
| There is no electric dipole moment: $\mathbf{p}' = 0$ | | There is no electric dipole moment: $\mathbf{p}' = 0$ |
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| In the laboratory system S, the solenoid moves with velocity | | In the laboratory system S, the solenoid moves with velocity |
| $\mathbf{v} = v\,\hat{\mathbf{x}}$ | | $\mathbf{v} = v\,\hat{\mathbf{x}}$ |
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| This is the same situation as in the problem 14.3.24 but with the round solenoid | | This is the same situation as in the problem 14.3.24 but with the round solenoid |
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| Under a boost with velocity $\mathbf{v}$ | | Under a boost with velocity $\mathbf{v}$ |
| the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as | | the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as |
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| $\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),\\ |
| so this simplifies to | | so this simplifies to |
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| $\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}.$ |
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| Expression in SI | | Expression in SI |
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| And finaly In the International System, the equivalent transformation is | | And finaly In the International System, the equivalent transformation is |
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| $\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad | | $\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad |
| \mathbf{m} = \gamma\,\mathbf{M}$ | | \mathbf{m} = \gamma\,\mathbf{M}$ |
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| #### Answer | | #### Answer |
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| The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. | | The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. |
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| $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad | | $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad |
| \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$ | | \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$ |
