I Calculating Induced Electric field

1. May 2, 2017

Arman777

In class we were trying to calculate the induced electric field created by changing the magnetic field stregth.
Imagine theres a circular surface which magnetic field out of the screen.Since we are changing the magnetic field from Faraday's Law there should be a induced current or charge flow simply.To create this motion we need electric field.So he drew another circle inside the outer surface with radius r Here is the pic

Then he said lets suppose theres a charge on the point P.And he explained the Electric Field and direction etc.And He said lets suppose It rotates once the circle

Now then He did something like this;

$W=qε=\int {\vec{F}⋅d\vec{r}}$
$W=qε=Eq2πr$
$ε=E2πr$ Then he used Faraday's Law and we found the E field.
I am stucked cause
$W=qε=\int {\vec{F}⋅d\vec{r}}$ should be zero.Cause it comes to same point.
$W=\int_p^p {\vec{F}⋅d\vec{r}}=0$
He never used $\oint$
What am I missing ?
If were used closed integral like $\oint_p^p q\vec {E}⋅d\vec{r}=Eq2πr$ ?

I think He should use closed integral.

Last edited: May 2, 2017
2. May 2, 2017

The induced electric field from a changing magnetic field is not a conservative field. Electrostatic fields (conservative fields) satisfy $\nabla \times E=0$ so that $\oint E \cdot ds=0$ by Stokes law. This is not the case for the induced $E$ field because $\nabla \times E=-\frac{dB}{dt}$ so that by Stokes law $\oint E \cdot ds=-\frac{d \Phi_m}{dt}$. And yes, your instructor should use $\oint$ for this integral.

3. May 2, 2017

Arman777

I understand , thanks a lot :)

4. May 5, 2017

vanhees71

Well, of course in Stokes's theorem the line integral is around a closed path. Otherwise it's wrong. Why one should need an extra symbol, I however don't know ;-).

5. May 5, 2017

Arman777

Its more nice,I like it :p