# Cause of electromagnetic induction?

The rate of change of magnetic flux through a surface (open) is related with the line integral over the closed loop binding the selected surface by one of the Maxwell's equation. But that means even if a changing magnetic flux is present anywhere on the surface such that the electric and magnetic field at the chosen surface's loop is zero. For example, for a perfect toroidal solenoid, all the magnetic field and hence flux is confined within the windings. Therefore, for a wire making a loop surrounding the toroid and passing through the centre, the fields (electric and magnetic) at the wire is zero yet the changing flux produces an emf and a current in the loop-wire!

Is this not spookier than the conventional "action at a distance". So what induces the EMF in this wire?

The most general, integral form of Faraday's Law is (see this physics.SE question: Faraday's law for a current loop being deformed) \begin{align} \int_{C_t} (\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol \ell = - \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a \end{align} Where $C_t$ is some closed curve that can depend on time, $\Sigma_t$ is a surface with $C_t$ as its boundary, $\mathbf E$ and $\mathbf B$ are the electromagnetic fields as measured in some inertial frame, and $\mathbf v$ is the velocity of a point on the curve resulting from its time-dependence.

Now if we consider the situation you describe, then the $\mathbf v\times\mathbf B$ terms goes away if we choose a stationary loop $C=C_t$, and we get \begin{align} \int_{C}\mathbf E\cdot d\boldsymbol \ell = - \frac{d}{dt}\int_{\Sigma}\mathbf B\cdot d\mathbf a \end{align} Now you say that

all the magnetic field and hence flux is confined within the windings.

This is true. However you also say that

Therefore, for a wire making a loop surrounding the toroid and passing through the centre, the fields (electric and magnetic) at the wire is zero

This is not quite right. If the right hand side (the rate of change of the flux) is nonzero, then the line integral of the electric field around the loop must be nonzero.
\begin{align} \int_{C}\mathbf E\cdot d\boldsymbol \ell \neq 0 \end{align} In particular, this means that the electric field itself cannot vanish along the loop, otherwise we would have a contradiction. In other words, it may be the case that there is no magnetic field along the loop (at least at the initial instant before any current is generated), but there is an electric field along the loop, and this pushes charges around (if the loop is a conductor with charges in it). As a side note, once the charges start moving, they create their own magnetic field even in the absence of a magnetic field produced by the solenoid.

• This means that there is an electric field in the absence of any charge densities? Or is there a charge density created in a subtle way? – Satwik Pasani Aug 22 '13 at 6:19
• Ultimately, the charges that lead to the magnetic field inside of the solenoid also are responsible for the existence of the electric field outside of the solenoid. – joshphysics Aug 22 '13 at 6:37

What you say here actually contradicts Maxwell's equations. The relevant equation is $$\nabla\times B-\frac{\partial E}{\partial t} = J$$ where, B is the magnetic field, E is the electric field and J is the current density and we are using units in which the $\mu_0, \epsilon_0 = 1$.

If the electric field is zero, the curl of the magnetic field has to be non-zero or there is no current, which means that since any real wire is of finite thickness the magnetic field must be non-zero somewhere inside the wire.