# Does the induced Electric field develop instantly or lags by $\frac rc$?

Consider a loop of radius $r_0=3 \times 10^8$ cm. A thin bar magnet is passed through its center. This implies that the magnetic flux through the loop will change. Now according to Faraday's law a non conservative electric field must develop inside the wire of that loop.

Question:

Let the magnet be cross though the centre at time $t_0$. Then would the electric field develop in the wire at $t_0$ or at $t_0+\dfrac {r_0}c$?

Commentary

According to the Faraday's Law of Electromagnetic induction the electric field $E(t)$ developed inside the wire is given by the following mathematical relationship,

$$\oint \mathbf{E}(t_0) \cdot d\boldsymbol{\ell} = - \dfrac{d \Phi(t)}{dt}|_{t=t_0}$$

According to this, the electric field developed at time $t_0$, $E(t_0)$ has a magnitude which directly depends upon the rate of change of magnetic flux at time $t_0$. This implies that the information of the motion of the magnet can be transmitted at any distance instantly. I might be wrong. If I am not wrong then this fact can be used to create paradox by applying special theory of relativity.

The second case can be that the electric field doesn't develop instantly and lages its cause by $\dfrac {r_0}{c}$. If this is the case then the current in the loop will develop after a time interval of $\dfrac {r_0}{c}$, then this current will create its own magnetic field which will cause a change in magnetric flux, and a net change in magnetic flux will be observed after $\dfrac {r_0}{c}$, this net change in the flux will cause its effect on $E$ at time $\dfrac {2r_0}{c}$. But we know that in the analysis of a pure conductor, V-I characteristic is solved by differential equations which gives a solution free of $c$. So either this second case is incorrect or the V-I characteristic of the pure inductor.

Thank You.

• Your first option certainly isn't true, the E field cannot develop instaneously. The solutions of Maxwell's equations are retarded as in your second option. I cannot see why you think Lenz's law causes any paradox. Any B field set up in the wire opposes the original changing magnetic flux. – ProfRob Jan 31 '15 at 13:52
• @RobJeffries There will be a paradox if my first option is correct. If my second option is correct then the Faraday's law must be modified and the analysis of self induction and mutual induction should also be modified. I will also have to make the steady state analysis. Can you provide an analysis of V-I characteristic of a pure inductor, considering the r/c delay. All the differential equations will have to be changed too. – user32348 Jan 31 '15 at 14:00
• That sounds like a difficult problem, but that is what must happen. Faraday's law, as you have written it, cannot apply. But of course Maxwell's equations do. – ProfRob Jan 31 '15 at 14:07
• @RobJeffries Your first comment is not satisfactory. Faraday's Law does not appear to account for retardation. Also Note that we don't specify the spatial configuration of the surface whose boundary is the loop. The surface can be distorted to any shape, including shapes in which parts of the surface are very far from the loop. (I typed this while you were submitting your second comment.) – garyp Jan 31 '15 at 14:09
• @seuser32111, the differential form is local; it relates the curl of $\vec E$ at a point to the time rate of change of $\vec B$ at the same point and time. – Alfred Centauri Jan 31 '15 at 15:20

This should be considered a provisional answer.

First, let's make the setup concrete.

• There is a thin conducting circular loop in the $xy$ plane with radius $r_0$.
• There is an ideal magnetic dipole aligned with and located on the $z$ axis and initially at rest at $z = z_0$.
• At time $t = t_0$, the dipole begins accelerating along the $z$ axis.

Now, let the elapsed time since $t = t_0$ be

$$\Delta t = t - t_0$$

and the (static) magnetic field at time $t = t_0$

$$\vec B_0 = \vec B(t_0)$$

Then, according to special relativity, the changing magnetic field and associated electric field cannot reach the loop until an elapsed time of

$$\Delta t_l = \frac{\sqrt{r^2_0 + z^2_0}}{c}$$

For $\Delta t < \Delta t_l$, the magnetic field outside of a sphere of radius $c\Delta t$ and centered on $(0,0,z_0)$ is just $\vec B_\text{outside} = \vec B_0$ while inside, the magnetic field is $\vec B_\text{inside} = \Delta \vec B(t) + \vec B_0$.

This means that, while $\Delta t < \Delta t_l$, no magnetic field lines of $\Delta \vec B$ thread the conducting loop. $^1$

So, here's the provisional answer: the flux of $\Delta \vec B$ through the surface bounded by the conducting loop is proportional to the number of lines of $\Delta \vec B$ threading the loop.

Thus, for the surface bounded by the conducting loop,

$$-\frac{d\Phi(t)}{dt} = 0\; ,t < (t_0 + \Delta t_l)$$

$^1$ To be clear, I mean that no field line of $\Delta B$ can cut the loop before $t = t_0 + \Delta t_l$

• @Timaeus, I disagree with you. For there to be a net flux of $\Delta \vec B$ through the surface bounded by the ring, at least one field line must have 'cut' the ring and that cannot be the case at $t = t_0 + \frac{z_0}{c}$. – Alfred Centauri Mar 1 '15 at 21:36
• @Timaeus, I edited the answer to clarify what I mean by "thread the loop". Thanks! – Alfred Centauri Mar 1 '15 at 22:04