# Homework Help: Induced electric field

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1. Dec 24, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

The emf gets induced due to the changing flux.

The flux through the rod remains 0. So, there is no induced emf.

I don’t know how to calculate induced electric field.

But as there is no induced emf, so there is no induced electric field. Hence, I chose option (a).

2. Dec 24, 2017

I believe this is incorrect. One way to get to the correct answer is to treat the case of a bar vertical on the paper and moving to the right along two horizontal rails with a space between them, and a second vertical bar at the origin that is stationary. Have a constant magnetic field out of the paper. Because the area of the rectangle is changing, there is an EMF $\mathcal{E}=-\frac{d \Phi}{dt}$ in the rectangular region. The EMF occurs in the moving bar.

3. Dec 24, 2017

### Pushoam

I think you are talking about the above problem.

In this case, the area covered by the loop ABCD is changing due to the motion of BD. So, there is a changing flux and induced emf.

But in the question I am solving, there is no loop and so no covered area by the rod as the rod is thin. Even if the rod has some area, the area covered by it in the presence of magnetic field remains same, so is the flux through this area.

Even if I consider a circular loop concentric with the rod and whose radius is half of the length of the rod, then the area covered by the semi – circular loop and the rod remains same. So, the flux through the area is same. But I do not see why I should consider this.

4. Dec 24, 2017

5. Dec 24, 2017

### cnh1995

Using the formula Charles gave in #4, what can you say about the linear velocity at each point along the length of the rod? What can you say about the induced electric field at each point?

6. Dec 24, 2017

### Pushoam

Considering cylinderical coordinate system with origin at the center of the rod,

The linear velocity of a pnt is along $\hat \phi$ direction. $\vec v = \vec \omega \times \vec r$ , where r is the r coordinate of the point of the rod.

So, $|\vec v \times \vec B| = \omega B r$ .

So, the strength of the induced electric field is maximum at the end and minimum at the center.

This I understood.

I am trying to understand how one gets the relation $\vec E = \vec v \times \vec B$ .

7. Dec 24, 2017

The EMF $\mathcal{E} =\int E \cdot dl=E L$, if $E$ is constant. Meanwhile for the moving bar of length $L$ in the diagram, $\frac{dA}{dt}=v L$, so that $\mathcal{E}=- \frac{d \Phi}{dt}=B v L$. (Note $\Phi=BA$).$\\$ Since $\mathcal{E}=EL$,(the only possible source of the EMF in the rectangle is in the moving bar), $E=vB$, and more precisely with a vector cross product, the induced electric field is $E=v \times B$.

Last edited: Dec 24, 2017
8. Dec 24, 2017

### Pushoam

This is valid for moving bar problem. How can one say that the relation is true for any system?
Because in case of the rod problem, there is no loop. So, the concept of flux cannot be used here.

9. Dec 24, 2017

The loop you make is a mathematical construction. You don't need to have actual material there to make such a loop. $\mathcal{E}=\oint E \cdot dl=-\frac{d \Phi}{dt}$. Any EMF that occurs must be happening in the part of the loop that is moving.

10. Dec 24, 2017

### cnh1995

E=dΦ/dt is the universal flux rule that applies everywhere. E=vBL is the motional emf developed in a moving rod and it is the same flux rule in disguise. Here, Φ=BA, where A is the area "swept by" the moving rod. For a rod following rectilinear motion, the area swept by the rod dA=l.dx and the induced emf E=dΦ/dt=BdA/dt=Bldx/dt=Blv.

11. Dec 24, 2017

### cnh1995

@Pushoam, as a fellow Indian, I believe you should read Concepts of Physics by Dr H.C.Verma (part 2). The electromagnetics section is wonderfully written.

12. Dec 24, 2017

### TSny

Consider the forces that act on a free charge carrier in the rod from the point of view of the lab frame. Then you don't need to use the concept of magnetic flux Φ.

13. Dec 26, 2017

### Pushoam

I am referring to Griffiths.
And Griffiths talks of a loop while using flux rule.
Since there is no loop, I cannot use of the flux rule.
$\vec v \times \vec B$ is the force per unit charge acting on the rod.
How do we know that this is the induced electric field?
In this way, is it that induced electric field is another name for magnetic force per unit charge?

As I undesrstand, motional emf is established in a moving loop due to magnetic force.
Since, there is no loop, we can't talk of emf as for existence of motional emf a loop is needed.

In equation 7.11, the motional emf is defined as the integral of the magnetic force per unit charge around the loop. This is the other thing that the integral is non - zero only along the moving part.

And induced electric field exists due to changing magnetic field. Here, in the question, magnetic field is not changing, so there is no induced electric field.

Last edited: Dec 26, 2017
14. Dec 26, 2017

Just to give a little additional information on EMF's that came out of a previous discussion on Physics Forums about a year ago, the EMF that occurs in an inductor supplies a voltage to the circuit. One thing that is slightly puzzling about the voltage that arises from an EMF as opposed to one that arises from an electrostatic potential is that the formula for EMF is $\mathcal{E}= \int E \cdot dl$, while in determining an electrostatic voltage at a given point $V=- \int E \cdot dl$. The EMF behaves differently in some ways from the electrostatic $E$ because it is not a conservative field, and although the EMF is a voltage, it can not be considered to be a potential. Let me see if I can provide a "link" to that discussion... Yes, here it is: https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/ Professor Lewin's video is also of interest,(see post 32 in the thread,... Editing: post 32 is the notes to the lecture=post 52 contains the video), and might give some insight into how changing magnetic fields can effect electrical circuits via Faraday's law. His video created a somewhat lively debate in this thread, but IMO, it was more a discussion of semantics, because they all agreed on the overall effect that a changing magnetic field can have.

Last edited: Dec 26, 2017
15. Dec 26, 2017

### Pushoam

Could you please give me a link where it is said explicitly that $\vec v \times \vec B$ is an induced electric field?
From Griffiths, I understand that I can talk of induced electric field only when there is changing magnetic field
and the concept of emf is valid only when there is a loop as we say emf developed in a loop. Emf is not potential differencce between two points. So, without loop, I cannot talk of emf.
Yes, we can use the concept of induced electic field even when there is no loop.

Is the above understanding right?

16. Dec 26, 2017

See the "link" in post 4.

17. Dec 26, 2017

### Pushoam

From Griffiths, I believe that the inductor acts as a battery of EMF $- L \frac {dI}{dt}$. I don't know why it is so.

18. Dec 26, 2017

Please study the"link" in post 14. I think you have some very similar questions to the OP of that thread. It was a rather lengthy discussion, but I think you will likely find it of much educational value. And yes, you are correct. The inductor acts as a battery of voltage $V=-L \, dI/dt$. $\\$ Editing: I just gave it a little thought on why the electrostatic voltage is $E=-\int E \cdot dl$ while the voltage from an EMF $\mathcal{E}=+\int E \cdot dl$. It could be it is because the electrostatic voltage is being probed from outside the source to figure out the effect it has, (i.e. what voltage it creates), while the EMF is determined from inside the source. This is at least worth considering. I'll leave this part to you to determine whether that explanation makes sense. Alternatively an EMF inside an open circuited conductive bar must necessarily generate an electrostatic $E_{electrostatic}$ field opposite the $E_{induced}$ , so that $E_{total}=0$. Perhaps this the voltage from this electrostatic field is really what the voltmeter is measuring. I'll let you decide which explanation makes the most sense.

Last edited: Dec 26, 2017
19. Jan 18, 2018

### Pushoam

Please, first read the red line, if you understand that then go ahead to the next red line. If you find the red line not clear, then you can go for the black and then for the textbook pages. This way your time will get saved from reading unnecessarily the whole post.

As I understand, for me, it is confusing to use the phrase " voltage from an EMF " as it is not defined in any standard textbook which I have read till now. So, let's use directly the term, EMF.

According to Griffith's, Introduction to electrodynamics, 3rd edition, the emf is defined according to equation (7.9).

EMF is defined for a whole circuit, not between two points of the circuit.

When there is a current through a circuit, and the total force acting on each charge particle is $\vec f$ , then emf is defined as the line integral of this force through the whole circuit.

So, let's first get clear about the difference between electrostatic potential difference and emf.

[ I am not using the term voltage as the book uses potential difference. I guess voltage is another name for potential difference, is it?]

Electrostatic potential difference is defined between two points, $V = - \int_{ a}^{ b} \vec E \cdot d\vec l$, while emf is defined for a whole circuit.

Now, for the circuit being considered in the book, it turns out that the emf is equal to the potential difference of the terminals of the battery. This doesn’t mean that emf is the potential difference (and this meaning creates conceptual confusion later).

Sometimes emf is interpreted as work done per unit charge, but this interpenetration does not express explicitly that the work done has to be calculated for the whole circuit, not between two points.

Just before equation (7.10), Griffith says that the total electric field inside an ideal conductor remains 0. This means that if there is current through the ideal conductor, it will remain so forever and if there is not, there will be no current. So, when the wire is not attached to the battery, there is no current. When the wire is attached to the battery, there is current. What happens when we attach the battery to the wire,( as during this small time interval the current happens in the circuit) is not understood by me. But this is not the part of this discussion. I just put it so that if you have any thought upon it, you can share it.

Let's consider a different circuit, a conducting loop and let's change the magnetic field through this loop which induces current into the loop.

Now, according to what I have understood from Griffith book, Faraday wanted to know what kind of force induces current in the circuit. As the loop is at rest, magnetic force cannot work on it. So, the force must be electric and hence he concluded that an electric field is present in the region and changing magnetic field induces electric field.

So, there is induced electric field inside the circuit and it is this field which brings out current in the circuit. Is this correct? Or to oppose this field (as the total field in the conductor should remain 0) electrostatic field of the loop increases and this brings out the current ( and as long as the induced field remains greater than electrostatic field the current goes on increasing)?

So, the force per unit charge in the circuit is equal to the induced electric field and by definition ( according to equation. 7.9) emf is given as $emf = \oint \vec E \cdot d \vec l$. .....(1)

Here, I have doubt on (1). I am writing another statement below as (1.1) which looks to me more correct. Please check it.
So, the force per unit charge in the circuit is equal to the induced electric field + the electrostatic field and by definition ( according to equation. 7.9) emf is given as $emf = \oint \vec E \cdot d \vec l = \oint \{\vec E_{in} +\ vec E_{es} \}\cdot d \vec l =\oint \vec E_{in} \cdot d \vec l$. .....(1.1)

Now, it is taken as an empirical fact that emf = $- \frac { d\phi } {dt}$. .....(2)

Then, it turns out by calculation that $\nabla \times \vec E = - \frac{\partial { \vec B } }{\partial { t} }$ .....(3)

So, now the questions are 1) How is the emf measured? What does practically emf mean? It could not be measured using its definition in (1) as $\vec E_{in}$ is not well – defined till now.

Griffith says that Faraday found (2) empirically (without saying how could emf be measured). So, in the book he has taken (2) for granted for reaching (3) from (1).

When magnetic field changes, it induces electric field, this induced electric field creates current. To oppose this, electric field due to the loop changes itself to make the total electric field inside the conductor 0. So, ultimately the current becomes finite.

This process takes time. Meanwhile, induced electric field remains stronger than electric field due to loop and so the current happens. ...(4)

So, when the current is finite, the electric field due to loop is electrostatic and equal and opposite to the induced electric field.

Hence, emf = $\oint \vec E_{in} \cdot d\vec l = \oint \vec E_{es} \cdot d\vec l = 0$ , true as the induced electric field itself is o.

The electrostatic potential difference between any two points of the loop is not always 0 as there remains non - zero electrostatic field till there is induced electric field.

Now, the question is if I put the voltmeter at the two points of the loop, what will it measure?
It will measure the potential difference due to the electrostatic field.

The next question is : Is this potential difference equal to the emf?
Using (2), I can calculate emf theoretically and by measuring potential difference I can check whether the two are equal or not.
But, I cannot derive a relation between the two mathematically.
To get the answer of this question, I refer to the recommended thread in post #14.

Last edited: Jan 18, 2018
20. Jan 18, 2018

### cnh1995

That's tricky!
It will depend on the orientation of the voltmeter. You will have to solve the KVL equation ∫closed loopE.dl=dΦ/dt for any loop containing the voltmeter in it.