Why is magnetic field zero for open circuit and electric field zero for short circuit?

Question

I read this statement in a book for a simple circuit (1 resistor connected to a voltage source). If we short out the resistor, the E field is said to be zero and if the circuit is broken (open circuit), the magnetic field is said to be zero. Why is this ? How can I prove this to myself ?

Answer 1

If the resistor is shorted, then for the $E$ field since $V=IR$, if $R=0$ then $V=0$ so $E=0$.

If the circuit is open $I=0$, then for the magnetic field $B=(uI)/(2\pi r)$ and if $I=0$ then $B=0$.

Answer 2

The comment below is restricted to stationary or very low frequency circuits where any electric field is localized within a resistive wire via Ohm's law or between capacitor plates, while the magnetic field can be found only in a coil.

The short circuit case is easier to visualize. Assuming an ideal short such that its resistance is zero then the electric field parallel within the wire must also be zero for any finite current density. When you have a coil (or just a piece conducting wire) and a current passing through it you will have a magnetic field surrounding the coil, so for that to be zero the current must be zero and that means an open circuit for any finite voltage across the terminals.

For an arbitrary distributed parameter circuit it is conventional to say that an ideal conducting wall provides an "electric wall", ie., one so that the E field has no tangential component with the wall. This is the case of a capacitor where the static field is always orthogonal to the metal plates but the idea carries over to arbitrary frequencies. By analogy it is conventional to call magnetic wall a surface to which the H field is orthogonal, ie., the tangential H component is zero. This can never happen over a metallic surface ("metallic short") at any frequency but can be made to happen by some symmetric excitation, for example. At any rate, it happens in empty space, ie., an "open circuit".