New solution

Alexphysics edited
View solution History
revision #19007 newer →
en/14.3.25.md +51 −0
@@ -0,0 +1,51 @@
+### Statement
+
+$14.3.25.$ [Insert the problem statement]
+
+### Solution
+
+Analyse the rest frame S'
+
+We have a long cylindrical solenoid, with axis along z'.
+rest magnetic moment
+$\mathbf{M}' = M\,\hat{\mathbf{z}}'$
+There is no electric dipole moment: $\mathbf{p}' = 0$
+
+In the laboratory system S, the solenoid moves with velocity
+$\mathbf{v} = v\,\hat{\mathbf{x}} $
+
+This is the same situation as in 14.3.24 but with the round solenoid
+
+Under a boost with velocity $\mathbf{v}$
+the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as
+
+$\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}$
+
+With of course
+$\mathbf{p}' = 0$ $\mathbf{v}\cdot\mathbf{m}' = 0 $
+
+so this simplifies to
+
+$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad
+\mathbf{m} = \gamma\,\mathbf{M}.$
+
+Expression in SI
+
+And finaly In the International System, the equivalent transformation is
+
+$\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad
+\mathbf{m} = \gamma\,\mathbf{M}$
+
+Conclusion
+
+The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. In both cases, a moving magnetic dipole acquires an electric dipole moment given by:
+
+\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
+
+[Insert a concise answer or boxed result]