Calculating Induced Electric field

[Arman777]

In class we were trying to calculate the induced electric field created by changing the magnetic field stregth.
Imagine there's 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 let's suppose there's a charge on the point P.And he explained the Electric Field and direction etc.And He said let's 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.

 

[Charles Link]

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.

 

[Arman777]

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.

I understand , thanks a lot :)

 

[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 ;-).

 

[Arman777]

Why one should need an extra symbol, I however don't know ;-).

Its more nice,I like it :p