So suppose I have a regular circuit with a battery connected to a resistor and a lightbulb.
Suppose now somehow the battery is inside a metal box (faraday cage) but the rest of the circuit is outside of it so the wire is maybe poked through a tiny hole in the box.
Since energy flow through a circuit is due to the electromagnetic field as described by the Poynting vector, since the field cannot penetrate through the faraday cage, will current flow through the circuit?
Good question, related to a controversial Veritasium video. My answer is yes, current will still flow and the bulb will still light. While the region outside the box is shielded from the field inside, there is no reason the portion of the wire outside the box can't generate its own E & B fields.
BTW, while I believe that Veritasium (Derek) is correct in spirit, I disagree with his answer to the multiple choice question. I believe that the answer to his question is none of the above. The current won't ramp up appreciably until about one RL-time constant elapses.
Yes, it will. The Faraday cage won't stop the current flowing through the wire around the circuit.
Yes. From the Maxwell equation: $$\nabla \times \mathbf B = \mu_0 \mathbf I$$ we can know the directions of the $\mathbf B$ vector field using the right hand rule. Inside the battery, the E-field is from + to - ($\mathbf E = - \nabla V $). In the external resistance it has the same direction of the current.
Using the right hand rule for the Poynting vector ($\mathbf E \times \mathbf B$), it is easy to see that it is to outside in the battery and to inside in the resistor.
When we say that it flows from the battery to outside, it doesn't mean that some stuff is really flowing and reaching the resistor, after travelling through the air (as it would be the case for EM waves). It is only the expression of a vector field that only exist inside the components. So the Faraday cage doesn't affect the energy flow.
