A changing magnetic field passes through a wire loop but the loop itself is not in the field. Is an EMF induced in the loop? [duplicate]

Question

$$\nabla \times \vec{E} =-\frac{\partial{\vec{B}}}{\partial{t}}$$

Applying Stokes' theorem:

$$\oint_{loop} \vec{E} \cdot d\vec{l}=\int_S -\frac{\partial{\vec{B}}}{\partial{t}} \cdot d\vec{S}$$

Where $S$ is any open surface with the loop as its perimeter. Considering the simplicity of the derivation I struggle to see why this shouldn't hold in all cases. However, consider the setup described below:

A uniform magnetic field parallel to the $z$ axis crosses the $x$-$y$ plane in a circular region radius $r$ centred at the origin. A wire loop of radius $R>r$ is also centred at the origin in the $x$-$y$ plane. The field changes with time (but remains parallel to the $z$ axis). Is an EMF induced in the loop?

According to Faraday's law as expressed above, the answer surely should be yes. The integral on the right hand side will not return 0, so therefore the left hand side cannot be 0 and an EMF should be induced in the coil.

Physically, however, the answer surely should be no. An EMF is induced in the wire ultimately because of the magnetic force that acts on the charges in the loop itself, moving them along the loop and producing a potential difference. If the loop itself is not in the magnetic field, then how can the field exert a force on the charges in it?

There is an error in reasoning somewhere here. Where am I going wrong?

Answer 1

(a) It's not the magnetic field that exerts forces on the charges in the loop, but the electric field given by your second equation. [For one thing the charges in the wire are (on average) stationary, so wouldn't experience forces from a magnetic field even if it were local to the charges.]

(b) The line integral of the electric field around the loop gives the emf in the loop, but the concept of potential difference doesn't apply because the field is not conservative: if we take two points A and B on the loop, the work done by a charge going from A to B depends on the route by which it goes from A to B.