Confusion about applying Faraday's law to a circuit

[eyeweyew]

Homework Statement

Analyzing circuit with Faraday's law

Relevant Equations

$$\oint_C {E \cdot d\ell} = -\frac{d}{{dt}} \int_S {B_n dA} = -\frac{d\Phi_{B}}{{dt}}$$

If all electric fields generated by electrostatic charges, then we know $$\oint_C {E \cdot d\ell} = 0$$ so in the following circuit, $$\oint_C {E \cdot d\ell} = -V+IR = 0$$

In cases where not all electric fields generated by electrostatic charges, then according Faraday's law, we know $$\oint_C {E \cdot d\ell} = -\frac{d}{{dt}} \int_S {B_n dA} = -\frac{d\Phi_{B}}{{dt}}$$

So in the following circuit, we have for loop2:
$$\oint_C {E \cdot d\ell} = -V+I_tR_t=-\frac{d\Phi_{B}}{{dt}}=-L\frac{dI_L}{{dt}}$$

and we have for loop3:
$$\oint_C {E \cdot d\ell} = -I_1R_1=-\frac{d\Phi_{B}}{{dt}}=-L\frac{dI_L}{{dt}}$$

However, for loop1 we have:
$$\oint_C {E \cdot d\ell} = -V+I_1R_1=-\frac{d\Phi_{B}}{{dt}}=0$$

which is just like all electric fields generated by electrostatic charges but it is not the case in this circuit and in loop1. So does that mean $$\oint_C {E \cdot d\ell}$$ can be equal 0 even not all electric fields are generated by electrostatic charges around the closed loop?

 

[BvU]

There is no magnetic flux in your circuits.
$\Phi_B = 0$ so $-\frac{d\Phi_{B}}{{dt}}=0$ and thereby $\oint_C {E \cdot d\ell}=0$

Your equation for loop 1 misses a term $+I_tR_t$

$\$

 

[TSny]

So in the following circuit, we have for loop2: $$\oint_C {E \cdot d\ell} = -V+I_tR_t=-\frac{d\Phi_{B}}{{dt}}=-L\frac{dI_L}{{dt}}$$ and we have for loop3:
$$\oint_C {E \cdot d\ell} = -I_1R_1=-\frac{d\Phi_{B}}{{dt}}=-L\frac{dI_L}{{dt}}$$

These look correct for setting up the loop equations according to Faraday's law.

I think it's important to emphasize that when setting up the equations this way, whenever the closed path of integration of the E-field includes the inductor, the path is taken to stay within the conducting wire of the inductor. So, the path stays within the helical windings of the inductor. It is this part of the path where the B-field of the inductor produces the magnetic flux through the path and this is the reason for the appearance of the term $-L\frac{dI_L}{dt}$ on the right side of the loop equation for loops 2 and 3.

However, for loop1 we have: $$\oint_C {E \cdot d\ell} = -V+I_1R_1=-\frac{d\Phi_{B}}{{dt}}=0$$ which is just like all electric fields generated by electrostatic charges but it is not the case in this circuit and in loop1. So does that mean $$\oint_C {E \cdot d\ell}$$ can be equal 0 even not all electric fields are generated by electrostatic charges around the closed loop?

Yes. If the electric field is nonconservative, there can nevertheless be closed paths for which $$\oint_C {E \cdot d\ell} = 0.$$ This will be true for any path having zero magnetic flux through it. For example, suppose there is an increasing magnetic field into the page that is confined to the gray area shown below.

The induced electric field is nonconservative. But $\oint_C {E \cdot d\ell} = 0$ for the brown path since there is no magnetic flux through the brown path.

For loop 1 in your circuit, there is no magnetic flux through the loop.

 

[eyeweyew]

There is no magnetic flux in your circuits.
$\Phi_B = 0$ so $-\frac{d\Phi_{B}}{{dt}}=0$ and thereby $\oint_C {E \cdot d\ell}=0$

Your equation for loop 1 misses a term $+I_tR_t$

$\$

Yes, you are right. The correct equation for loop1 should be: $$\oint_C {E \cdot d\ell} = -V+I_tR_t+I_1R_1=-\frac{d\Phi_{B}}{{dt}}=0$$