@@ -6,65 +6,43 @@Statement ### Solution − We start analysing Rest frame S'+ 1. − In the system S' the solenoid is at rest:+ Plates parallel to the x'y' plane, separated in z' by a distance h. + Uniform electric field: + $ \displaystyle \mathbf { E } ' = (0,\,0,\,E_0), with E_0 = \dfrac { \sigma }{ \varepsilon _0 } (SI) $ .+ Uniform leakage current: $ \mathbf { j } ' = (0,\,0,\,j) $ . + There is no magnetic field $ ( \mathbf { B } ' = 0) $ by symmetry. − 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 } } $ + The capacitor moves with velocity $ \mathbf { v } = v\, \hat { \mathbf { z }} (v = \beta c) $ , parallel to the electric field and the leakage current. + $ \gamma = 1/ \sqrt { 1- \beta ^2 } $ − There is no electric moment− $ \mathbf { p } ' = 0 $ .+ Electromagnetic fields (boost in z): − 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 ).+ E_z &= E'_z = E_0, \quad \mathbf {E}_\perp = \gamma (\mathbf {E}'_\perp + \mathbf {v}\times\mathbf {B}') = 0,\\ + B_z &= B'_z = 0, \quad \mathbf {B}_\perp = \gamma \!\left (\mathbf {B}'_\perp - \frac {\mathbf {v}}{c^2}\times \mathbf {E }'\right ) = 0. \end {aligned}$ − In our case:$ \mathbf { p } ' = 0 $ and $ \mathbf { v } \cdot \mathbf { m } ' = 0 $ (because they are perpendicular) − Therefore: + $ \boxed { \mathbf { E } =(0,\,0,\,E_0), \qquad \mathbf { B } = 0 } $ − $ \boxed { \mathbf { p } = \frac { \gamma }{ c } \, \mathbf { v } \times \mathbf { M }} , \qquad − \mathbf {m} = \gamma \,\mathbf {M}$ + The electric field does not change relative to the rest system. − The magnetic moment in S increases by a factor of $ \gamma $ − and an electric moment appears perpendicular to $ \mathbf { v } $ and to $ \mathbf { M } $ + The four-current $ (c \rho , \mathbf { j } ) transforms as a four-vector. In S': \rho ' = 0, \mathbf { j } ' = (0,0,j) $ − For $ v \ll c ( \gamma \approx 1) $ + $ \begin { aligned } + \rho &= \gamma \!\left (\rho ' + \frac {v}{c^2}j'_z\right ) = \gamma\frac {v}{c^2}j,\\+ j_z &= \gamma \,(j'_z + v\rho ') = \gamma \,j,\\ + j_x &= j'_x = 0,\quad j_y = 0. + \end {aligned}$ − In CGS: $ \displaystyle \mathbf { p } \approx \frac { 1 }{ c } \, \mathbf { v } \times \mathbf { M } $ . + The volumetric current density increases: + $ j_z = \gamma \,j $ .− In SI: the factor is − $ 1/ c^2 $ − $ \displaystyle \mathbf { p } = \frac { 1 }{ c^2 } \, \mathbf { v } \times \mathbf { M } $ + A net charge density appears in the dielectric: + $ \rho = \gamma v j / c^2 $ − (with $ \mathbf { M } $ in $ A·m² $ ) + However, the total current I through a cross-section remains invariant because the area of the plates (perpendicular to the motion) contracts: + $ A = A'/ \gamma $ .− 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. + $ I = \int j_z\,dA = ( \gamma j) \left ( \frac { A' }{ \gamma } \right ) = j A' = I' $
