Electromagnetism question, what formula

[fluidistic]

I'd like to know what formula to use in order to find the following :
Imagine a uniform magnetic field. I am moving with a constant velocity perpendicularly through it. Do I see only an electric field? A magnetic field? Both? Or both of them?
What if I move in the sense of the magnetic field with a constant velocity?

What if my velocity changes with time?
What if, instead of the magnetic field, it's an electric field?

I do not expect the answer to all these questions, rather I'd like a formula to check it out myself. But if you have to say a word or even give the answer, I'm all ears.
Thanks in advance.

 

[Pythagorean]

What velocity are you traveling with? If it's much lower than the speed of light:

 

[fluidistic]

What velocity are you traveling with? If it's much lower than the speed of light:

Thanks a lot. I'm a bit confused, if B is uniform, if I move, \frac{\partial B}{\partial t}=0 or I'm wrong?

 

[jtbell]

See the page below for the transformation equations for E and B between two inertial reference frames:

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html

 

[fluidistic]

See the page below for the transformation equations for E and B between two inertial reference frames:

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html

Ok thank you very much. I think I'm done for now.

 

[kcdodd]

To look at EM in another velocity you need to use a lorentz transformation of the field. If you are familiar with vectors, J.D. Jackson p. 558 gives a useful form:

\vec{E}' = \gamma (\vec{E} + \vec{\beta} \times \vec{B}) - \frac{\gamma^2}{\gamma + 1} \vec{\beta} (\vec{\beta} \cdot \vec{E})

\vec{B}' = \gamma (\vec{B} - \vec{\beta} \times \vec{E}) - \frac{\gamma^2}{\gamma + 1} \vec{\beta} (\vec{\beta} \cdot \vec{B})

\vec{\beta} = \vec{v}/c
\gamma = \frac{1}{\sqrt{1 - \beta^2}}

So, you see if you have one frame where E = 0, and only B != 0. Then in every other frame E' != 0, and B' != 0, since B appears in both terms.

 

[fluidistic]

To look at EM in another velocity you need to use a lorentz transformation of the field. If you are familiar with vectors, J.D. Jackson p. 558 gives a useful form:

\vec{E}' = \gamma (\vec{E} + \vec{\beta} \times \vec{B}) - \frac{\gamma^2}{\gamma + 1} \vec{\beta} (\vec{\beta} \cdot \vec{E})

\vec{B}' = \gamma (\vec{B} - \vec{\beta} \times \vec{E}) - \frac{\gamma^2}{\gamma + 1} \vec{\beta} (\vec{\beta} \cdot \vec{B})

\vec{\beta} = \vec{v}/c
\gamma = \frac{1}{\sqrt{1 - \beta^2}}

So, you see if you have one frame where E = 0, and only B != 0. Then in every other frame E' != 0, and B' != 0, since B appears in both terms.

Ok thank you very much. I'm going to try to grasp this. I took vector calculus. I'm taking the 1 year EM course on next term (in March). I thought I would need it for the intro to EM course, but I'm not sure now. Anyway, I'll do an effort to learn this part.