When Faraday's law drives a current around a loop, something must push the charges along the wire. A magnetic force needs moving charges, but here the charges may start at rest. The push is an electric field, made not by charges but by the changing magnetic field itself. It exists whether or not a wire is there to feel it.
Key idea
$$\oint \vec{E}\cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}$$
around any closed path, in matter or in empty space. Inside a long solenoid of radius $R$ whose field changes, the induced field circles the axis with $E = \dfrac{r}{2}\dfrac{dB}{dt}$ ($r < R$) and $E = \dfrac{R^2}{2r}\dfrac{dB}{dt}$ ($r > R$).
Why it works
The emf of a loop is the work per unit charge done around it, $\oint \vec{E}\cdot d\vec{\ell}$. Faraday's law says this equals $-d\Phi_B/dt$. Unlike the field of still charges, this field curls around: its integral around a closed loop is not zero, so it has no potential (you cannot speak of the voltage between two points). Around a solenoid, symmetry says the field lines are circles, so $\oint\vec{E}\cdot d\vec{\ell} = E(2\pi r)$, which equals the rate of change of the flux inside the circle. This is the method of Ampère's law, used for $\vec{E}$.
Worked examples
Worked example: Inside a solenoid
The field inside a solenoid of radius $5.0\ \text{cm}$ grows at $0.20\ \text{T/s}$. Find the induced field $2.0\ \text{cm}$ from the axis.
- Circle of radius $r$: $E(2\pi r) = \pi r^2\dfrac{dB}{dt}$, so $E = \dfrac{r}{2}\dfrac{dB}{dt}$.
- $E = \dfrac{0.020}{2}(0.20) = 2.00\ \text{mV/m}$.
Answer: $E = 2.00\ \text{mV/m}$
Worked example: Outside, where B is zero
Find the induced field $10\ \text{cm}$ from the axis of the same solenoid, outside it.
- The circle encloses all of the changing flux, $\pi R^2 B$: $E(2\pi r) = \pi R^2\dfrac{dB}{dt}$.
- $E = \dfrac{R^2}{2r}\dfrac{dB}{dt} = \dfrac{(0.050)^2(0.20)}{2(0.10)} = 2.50\ \text{mV/m}$, in a place where the magnetic field itself is nearly zero.
Answer: $E = 2.50\ \text{mV/m}$
Worked example: Any loop around it
A wire ring of any size encircles the solenoid. What emf is induced?
- $|\varepsilon| = \dfrac{d\Phi}{dt} = \pi R^2\dfrac{dB}{dt} = \pi(0.050)^2(0.20) = 1.57\ \text{mV}$.
- Bigger rings feel a weaker field over a longer path, and the product stays the same.
Answer: $|\varepsilon| \approx 1.57\ \text{mV}$
Common mistake
"The induced field needs a wire to exist." The wire only provides free charges for the field to push, so that a current flows. The field itself is made by the changing magnetic field, in empty space as well as in metal. Electrons circling in a vacuum tube can be sped up this way; a machine called the betatron did exactly that. Check: Faraday's law mentions a closed path, not a conductor.