(17.8) Transformation of the field of a solenoid moving in a direction perpendicular to its axis
Inside a long solenoid, $B=\mu_{0} N^{\prime} I$ if $N^{\prime}$ is the number of turns per meter and the current is $I$. The solenoid moves at a velocity $\mathscr{V}$ in a direction, perpendicular to its length.
(a) Calculate $\boldsymbol{E}$ and $\boldsymbol{B}$, both inside and outside the solenoid, as measured by a stationary observer. The axis of the solenoid is the $z^{\prime}$-axis, and $\mathscr{V}=\mathscr{V} \hat{x}$
(b) You can also calculate this field by transforming the potentials. First show that, in the frame of the solenoid, the vector potential
$$
A^{\prime}=-\frac{B^{\prime} y^{\prime}}{2} \hat{x}^{\prime}+\frac{B^{\prime} x^{\prime}}{2} \hat{y}^{\prime}
$$
gives the correct $\boldsymbol{B}^{\prime}$. Note that there exists an infinite number of possible expressions for $\boldsymbol{A}^{\prime} .$ For example, we could have set $\boldsymbol{A}^{\prime}=B^{\prime} x^{\prime} \boldsymbol{y}^{\prime}$.
(c) Set $V^{\prime}=0$. Now calculate $\boldsymbol{A}, V, \boldsymbol{E}$, and $\boldsymbol{B}$ inside the solenoid. Both $V$ and $A$ depend on the expression that we chose arbitrarily for $A^{\prime}$. Nonetheless, the relations $\boldsymbol{E}=-\boldsymbol{\nabla} V-\partial \boldsymbol{A} / \partial t$ and $\boldsymbol{B}=\boldsymbol{\nabla} \times \boldsymbol{A}$ always apply.
At points outside the solenoid, in its own frame, $\boldsymbol{A} \neq 0$, as we shall see in the cxample in Sec. 19.1.