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+### Statement
+
+$14.3.24.$ [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
+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 = \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.
+
+
+#### Answer
+
+$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad
+\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$
+
+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]