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Unified field theory says the Magnetic field and Electric field are proven to be two manifestations of the same thing just viewed from different perspectives. That is the magnetic field is essentially electrical in nature.

That being the case, why does a Faraday cage (which blocks everything electrical) not block the magnetic field (which is ultimately electrical)?

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    $\begingroup$ Well, why should it? (Saying "the magnetic field is ultimately electrical" is not an argument for that, since if it behaved exactly the same as the electric field, they wouldn't be two different manifestations, they'd be the same) $\endgroup$ – ACuriousMind Oct 8 '15 at 17:53
  • $\begingroup$ I never claimed they were the same thing. It is quite possible to have two things that are both electrical, yet different, e.g. voltage and current. A Faraday cage would block either, as it blocks all electrical phenomena. I appreciate your time in replying, but there is no logical flaw in my question. $\endgroup$ – chris-mail Oct 9 '15 at 10:32
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An at rest Faraday cage does not block static magnetic field, but a moving Faraday cage can block a static B that arises from frame where there is only an static E field outside the at rest cage.

Suppose there is a copper box and only an external E field. Electrons move to positions in the copper so that their field cancels the E field within the box. The B field is zero everywhere both outside and inside the box.

Now give yourself a velocity with respect to the box. Now you see the outside fields are E' and B' both of which are non-zero and different than before. The E' and B' inside the box are both still zero. How can this be? How can the charge distribution on the copper box cancel out the B' field in the box? It is because the previously at rest charges are now moving with the box and have become currents, which make a magnetic field that exactly cancels the external B' penetration into the box.

It must work this way because Maxell's equations are invariant under boosts. A physical situation ($E,B,j_\mu$) that satisfies Maxwell's equations in one frame, will satisfy Maxwell's equation in all other frames related by a velocity boost (or rotation) even though there will be new ($E',B',j'_\mu$).

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