# EMF produced across one side of a coil

• songoku
In summary: You calculated the emf produced by the whole coil - but you are only asked for the emf produced by side AC. Can you see why you got an answer double the correct answer?Max e.m.f produced by side AC = NBAω = 32 x 0.074 x 4.8 x 10-2 x 2.4 x 10-2 x 2π x 9 = 0.15 VIn summary, the emf produced by side AC of the coil is 0.15 V.
songoku
Homework Statement
Please see below
Relevant Equations
ε = - N . ΔΦ / Δt

I want to ask about part c (iii). This is what I did:

max e.m.f produced = NBAω = 32 x 0.074 x 4.8 x 10-2 x 2.4 x 10-2 x 2π x 9 = 0.15 V

But the answer is 0.077 V so my answer is off by factor of two.

Is my mistake related to "emf produced across the side of AC of the coil"?

Thanks

songoku said:
max e.m.f produced = NBAω = 32 x 0.074 x 4.8 x 10-2 x 2.4 x 10-2 x 2π x 9 = 0.15 V
You might like to consider the emf (including the emf's 'direction') produced by each side of the complete coil as it rotates.

songoku
Steve4Physics said:
You might like to consider the emf (including the emf's 'direction') produced by each side of the complete coil as it rotates.
By direction of emf, do you mean the direction of induced current?

If yes, then the direction of induced current is anticlockwise (viewed from above) so the current on side AC is from C to A and current on side BD is from B to D

songoku said:
By direction of emf, do you mean the direction of induced current?
Yes. I meant the direction from positive to negative. 'Polarity' would be a better word. And this gives the current direction.

songoku said:
If yes, then the direction of induced current is anticlockwise (viewed from above) so the current on side AC is from C to A and current on side BD is from B to D
Agreed. Note that sides AC and BD are each producing their own emf (a bit like in part C(II) of the original question). These emf sources are in series, so the total emf is their sum.

Sides AB and CD produce no emf - do you know why?

You calculated the emf produced by the whole coil - but you are only asked for the emf produced by side AC. Can you see why you got an answer double the correct answer?

songoku
Steve4Physics said:
Sides AB and CD produce no emf - do you know why?
Because point A and B are at same potential and also point C and D are at same potential?

Steve4Physics said:
You calculated the emf produced by the whole coil - but you are only asked for the emf produced by side AC. Can you see why you got an answer double the correct answer?
Yes, I should divide my answer by 2

songoku said:
Because point A and B are at same potential and also point C and D are at same potential?
These potentials are equal as you say - but that's because AB and CDproduce no emf. It is not the reason that AB and CD produce no emf.

The potentials at, say, A and C are different because an emf is induced in AC.
The potentials at, say, A and B are the same because no emf is induced in AB.

But why should emf be induced in AC but not in AB?

songoku said:
Yes, I should divide my answer by 2
Yes. Note that it's because AC and BD produce equal emfs and AB and CD produce no emfs.

songoku
Steve4Physics said:
These potentials are equal as you say - but that's because AB and CDproduce no emf. It is not the reason that AB and CD produce no emf.

The potentials at, say, A and C are different because an emf is induced in AC.
The potentials at, say, A and B are the same because no emf is induced in AB.

But why should emf be induced in AC but not in AB?
I think because AB and CD are parallel to the magnetic field so when they rotate they don't cut the magnetic field?

songoku said:
By direction of emf, do you mean the direction of induced current?
An unnecessary sidenote, but I find that a strange question.
They are indeed the same here, but I always think that the emf is induced and the current IF ANY is caused by that emf. Were the coil to be incomplete - with a break in AB or CD, say - then the induced emf would be the same, but there would be no current.
When calculating the current in a motor, the current will normally be in the opposite direction to the induced emf, because that current is driven by an external emf greater than the induced back-emf.

songoku
songoku said:
I think because AB and CD are parallel to the magnetic field so when they rotate they don't cut the magnetic field?
AB and CD are not always parallel to the magnetic field. But, as they rotate, they do not cut through the magnetic field in the direction required to produce an emf.

songoku
There's a contractio on adjecto already in the title of this thead. An EMF is defined as an integral along a closed loop,
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\dot{\Phi}_{\vec{B}},$$
where ##A## is a surface enclosed by its boundary ##\partial A##, with the velocity field ##\vec{v}## along this boundary, and
$$\Phi_{\vec{B}}=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}$$
is the magnetic flux through this area.

songoku
songoku said:
I think because AB and CD are parallel to the magnetic field so when they rotate they don't cut the magnetic field?
Steve4Physics said:
AB and CD are not always parallel to the magnetic field. But, as they rotate, they do not cut through the magnetic field in the direction required to produce an emf.
Another way of looking at it is this. AB and CD rotate in planes in which the magnetic field lies entirely. One might consider a motional emf across the diameter of the rod, but that would not generate a current around the closed loop.

songoku and Steve4Physics
vanhees71 said:
There's a contractio on adjecto already in the title of this thead.
Should that be 'contradictio in adjecto'?

vanhees71 said:
An EMF is defined as an integral along a closed loop,
At an introductory level it’s common to think about emfs induced in moving straight conductors. E.g. see part C-II of the original question in Post #1.

The coil’s total emf is then the sum of the emfs from the sides. So (in the context of the question) the title doesn't seem too bad.

songoku
Thank you very much for the help and explanation Steve4Physics, Merlin3189, vanhees71, kuruman

Steve4Physics and kuruman
Steve4Physics said:
Should that be 'contradictio in adjecto'?
yes, of course.
Steve4Physics said:
At an introductory level it’s common to think about emfs induced in moving straight conductors. E.g. see part C-II of the original question in Post #1.
That explains the confusion of students at the introductory level about Faraday's Law of induction.
Steve4Physics said:
The coil’s total emf is then the sum of the emfs from the sides. So (in the context of the question) the title doesn't seem too bad.
You have to take a line integral along a closed path in Faraday's Law. Anything else doesn't make mathematical sense!

## What is EMF?

EMF stands for electromagnetic force, which is a measure of the strength of the electric and magnetic fields produced by an electric current.

## How is EMF produced in a coil?

EMF is produced in a coil when there is a change in the magnetic field passing through the coil. This can be caused by a changing electric current or by moving the coil through a magnetic field.

## What is the relationship between the EMF and the number of turns in a coil?

The EMF produced across one side of a coil is directly proportional to the number of turns in the coil. This means that the more turns in the coil, the higher the EMF will be.

## How does the shape of a coil affect the EMF produced?

The shape of a coil can affect the EMF produced in two ways. First, a longer coil will produce a higher EMF than a shorter coil with the same number of turns. Second, the orientation of the coil in relation to the magnetic field can also affect the EMF produced.

## Can the EMF produced in a coil be increased?

Yes, the EMF produced in a coil can be increased by increasing the strength of the magnetic field passing through the coil, increasing the number of turns in the coil, or increasing the speed at which the coil moves through the magnetic field.

### Similar threads

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
984
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
12K