Assuming that the magnet frame and the conductor frame are related by a
Galilean transformation, it is straightforward to compute the fields and forces in both frames. This will demonstrate that the induced current is indeed the same in both frames. As a byproduct, this argument will
also yield a general formula for the electric and magnetic fields in one frame in terms of the fields in another frame. In reality, the frames are
not related by a Galilean transformation, but by a
Lorentz transformation. Nevertheless, it will be a Galilean transformation
to a very good approximation, at velocities much less than the speed of light. Unprimed quantities correspond to the
rest frame of the magnet, while primed quantities correspond to the rest frame of the conductor. Let
v be the velocity of the conductor, as seen from the magnet frame.
Magnet frame In the rest frame of the magnet, the magnetic field is some fixed field
B(
r), determined by the structure and shape of the magnet. The electric field is zero. In general, the force exerted upon a particle of charge
q in the conductor by the
electric field and
magnetic field is given by (SI units): \mathbf{F} = q \left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right), where q is the charge on the particle, \mathbf{v} is the particle velocity and
F is the
Lorentz force. Here, however, the electric field is zero, so the force on the particle is \mathbf{F} = q \mathbf{v} \times \mathbf{B}.
Conductor frame In the conductor frame, there is a time-varying magnetic field
B′ related to the magnetic field
B in the magnet frame according to: \mathbf{B}'(\mathbf{r'},t') = \mathbf{B}(\mathbf{r_{t'}}) where \mathbf{r_{t'}} = \mathbf{r'} + \mathbf{v}t' In this frame, there
is an electric field, and its curl is given by the
Maxwell-Faraday equation: \boldsymbol{\nabla} \times \mathbf{E}' = -\frac{\partial \mathbf{B}'}{\partial t'}. This yields: \mathbf{E}' = \mathbf{v}\times \mathbf{B}. To make this explicable: if a conductor moves through a B-field with a gradient \partial B_z/\partial z, along the z-axis with constant velocity v_z = \partial z/\partial t, it follows that in the frame of the conductor \frac{\partial B'_z}{\partial t} = v_z \frac{\partial B_z}{\partial z} = -(\nabla \times \mathbf{E'})_z = \frac{\partial E'_x}{\partial y} - \frac{\partial E'_y}{\partial x}. It can be seen that this equation is consistent with \mathbf{E'} = \mathbf{v}\times \mathbf{B} = v_z B_x \hat{y} - v_z B_y \hat{x}, by determining \partial E'_x/\partial y and \partial E_y'/\partial x from this expression and substituting it in the first expression while using that \nabla \cdot \mathbf{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0. Even in the limit of infinitesimal small gradients \partial B_z/\partial z these relations hold, and therefore the
Lorentz force equation is also valid if the magnetic field in the conductor frame is not varying in time. At relativistic velocities a correction factor is needed, see below and
Classical electromagnetism and special relativity and Lorentz transformation. A charge
q in the conductor will be at rest in the conductor frame. Therefore, the magnetic force term of the Lorentz force has no effect, and the force on the charge is given by \mathbf{F}' = q\mathbf{E}' = q\mathbf{v} \times \mathbf{B}. This demonstrates that
the force is the same in both frames (as would be expected), and therefore any observable consequences of this force, such as the induced current, would also be the same in both frames. This is despite the fact that the force is seen to be an electric force in the conductor frame, but a magnetic force in the magnet's frame.
Galilean transformation formula for fields A similar sort of argument can be made if the magnet's frame also contains electric fields. (The
Ampere-Maxwell equation also comes into play, explaining how, in the conductor's frame, this moving electric field will contribute to the magnetic field.) The result is that, in general, \begin{align} \mathbf{E}' &= \mathbf{E} + \mathbf{v}\times \mathbf{B} \\[1ex] \mathbf{B}' &= \mathbf{B} - \frac{1}{c^2} \mathbf{v} \times \mathbf{E}, \end{align} with
c the
speed of light in
free space. By plugging these transformation rules into the full Maxwell's equations, it can be seen that if Maxwell's equations are true in one frame, then they are
almost true in the other, but contain incorrect terms proportional to the quantity
v/c raised to the second or higher power. Accordingly, these are not the exact transformation rules, but are a close approximation at low velocities. At large velocities approaching the speed of light, the
Galilean transformation must be replaced by the Lorentz transformation, and the field transformation equations also must be changed, according to the expressions given below. ==Transformation of fields as predicted by Maxwell's equations==