Can a local magnetic field when changed introduce an electric field till infinity? Here is a question from my school textbook:

Here is the solution which they expect : 
Change in magnetic flux is $\pi a^2 B $, and since $\oint\vec{E}.\vec{dl} = -\frac{d\phi}{dt}$, the total tangential electric force acting on the rim is $(\pi a^2 B \lambda)/(\delta t)$, and multiplied by torque and divided by moment of inertia gives the angular acceleration which multiplied by $\delta t$ gives the angular velocity as $\frac{\lambda B \pi a^2}{MR}$.
Doubt : If the magnetic field is limited to radius $a$ from the centre of the wheel, is it right that the electric field induced by it when it is diminishing reaches out till the rim of the wheel?
 A: Yes, consider the equation relating the path integral of $\vec{E}$ to $d\phi/dt$;  this equation holds for any closed  path in space, not just where the line of charge is.  From this you could evaluate the induced EMF, i.e. the electric field at any point in space:


*

*In this discussion, I'll ignore the charged ring; in this problem it just serves as
test charge.

*Note that the problem setup is rotationally symmetric, thus we know that the
the electric field is also rotationally symmetric.

*Write the electric field in terms of the radial and azimuthal unit vectors as $\vec{E}=E_r(r) \hat{r} + E_{\theta}(r) \hat{\theta}$; note that both of the field configurations are both independent of $\theta$.

*I leave it to the reader to prove that $\nabla \times (f(r) \hat{r})=0$; so we can ignore the radial component.

*Now use the path-integral relationship: take a circular path at some radius $R>a$.

*Since the path is circular, we have $\hat{\theta} \parallel d \vec{l}$, and $\hat{\theta} \cdot d \vec{l} = R d\theta$so we can
do the intgral and obtain
\begin{equation}
\begin{aligned}
\oint \vec{E} \cdot d\vec{l}&=\int E_\theta(R) R d\theta= 
(2 \pi R) E_\theta(R) = \frac{d\phi}{dt}
\end{aligned}
\end{equation}

*So the electric field points along circles, and has a magnitude that decays like $R^{-1}$:
\begin{equation}
E_\theta(R) = \frac{1}{ 2 \pi R }\frac{ d\phi}{dt }
\end{equation}

