Faraday's Law and the distance from another wire

[Biocalculus]

Hello, I have a conceptual question regarding Faraday's Law: emf = d/dt(Φ), where |Φ|=|B*A|. My question is does Faraday's Law take into account the distance between the solenoid producing the non-coulombic electrical field (Enc) and a wire circling the solenoid, which now have an emf due to the Enc. According to the textbook, the non-coulombic electrical field decreases at a rate of 1/r outside of the solenoid, so that when calculating the emf of the encircling wire we should take into account the distance from the solenoid.

 

[vanhees71]

What is the specific setup you are discussing? It's not clear to me from your description. Maybe it's sufficient to tell us which textbook you are referring to.

 

[Biocalculus]

Hello, the setup is that from a bird's eye view there is a wire connected to an ammeter encircling a solenoid that is vertically positioned such that we only see a small circle (solenoid) within a larger "circle" (wire + ammeter) when looking down. There is a gap between the perimeter of the solenoid and the boundary of the outer wire. The distance between the inner and outer circle is not specified which confused me because non coulombic E is proportional to 1/r.
This question is an example problem from "Matter and Interaction II: electricic and magnetic interactions" by Chabay and Sherwood, 3rd edition on page 954, titled "A Circuit Surrounding a Solenoid".

 

[vanhees71]

The great thing with Faraday's Law is that all these details don't matter at all!

All you need to know is that an ammeter measures the current through the wire, and this current is due to the induced electric field. Integrating along the wire (running through the ammeter, which doesn't matter either), you get
$$\mathcal{E}=R i.$$
Then you know that this line integral of $\vec{E}$ equals the time derivative of the magnetic flux encircled by the path along you've taken the line integral. Since for a very long solenoid you can neglect the magnetic field outside of it and in the interior you have the homogeneous magnetic field $B$. Thus the magnetic flux simply is
$$\Phi=B A_{\text{solenoid}},$$
and thus
$$R i=\pm \dot{B} A_{\text{solenoid}}.$$
I don't know the book you quote, but if it's calculus based you can get the sign right too:

You give an arbitrary direction, running along the wire with the ammeter. With the right-hand rule that defines the direction of the surface-element normal vectors along an arbitrary surface bounded by the closed loop through the wire. It doesn't matter which particular surface you choose. Just use the simplest, the one, whose part crossing the solenoid is just the plane cross section of it. Then the right-hand rule specifies the constant normal unit vector $\vec{n}$, and the proper sign for the magnetic flux is $\Phi=\vec{B} \cdot \vec{n} A_{\text{solenoid}}$. Then the correct sign of the current is given by
$$i=-\frac{1}{R} A_{\text{solenoid}} \vec{n} \cdot \dot{\vec{B}},$$
with the minus sign from Faraday's Law.

 

[Biocalculus]

The great thing with Faraday's Law is that all these details don't matter at all!

All you need to know is that an ammeter measures the current through the wire, and this current is due to the induced electric field. Integrating along the wire (running through the ammeter, which doesn't matter either), you get
$$\mathcal{E}=R i.$$
Then you know that this line integral of $\vec{E}$ equals the time derivative of the magnetic flux encircled by the path along you've taken the line integral. Since for a very long solenoid you can neglect the magnetic field outside of it and in the interior you have the homogeneous magnetic field $B$. Thus the magnetic flux simply is
$$\Phi=B A_{\text{solenoid}},$$
and thus
$$R i=\pm \dot{B} A_{\text{solenoid}}.$$
I don't know the book you quote, but if it's calculus based you can get the sign right too:

You give an arbitrary direction, running along the wire with the ammeter. With the right-hand rule that defines the direction of the surface-element normal vectors along an arbitrary surface bounded by the closed loop through the wire. It doesn't matter which particular surface you choose. Just use the simplest, the one, whose part crossing the solenoid is just the plane cross section of it. Then the right-hand rule specifies the constant normal unit vector $\vec{n}$, and the proper sign for the magnetic flux is $\Phi=\vec{B} \cdot \vec{n} A_{\text{solenoid}}$. Then the correct sign of the current is given by
$$i=-\frac{1}{R} A_{\text{solenoid}} \vec{n} \cdot \dot{\vec{B}},$$
with the minus sign from Faraday's Law.

I understand the concept much better now, thank you!