EMF induced in coil by magnetic field

In summary, Homework Equations state that the magnetic flux changes with time due to the current and the distance from the wire, and faraday's law is used to calculate the flux. The current induced in the loop obeys Faraday's Law, which depends on the changes in the total flux through the loop.
  • #1
iRaid
559
8

Homework Statement


attachment.php?attachmentid=63835&stc=1&d=1384141892.png



Homework Equations





The Attempt at a Solution


Well I tried this (I thought it'd work, but it's not the right answer and I'm not sure how to do it now) Anyway this is my attempt:

[tex]\Phi = \mu_{o}\frac{NIA}{\ell} \\ \varepsilon=N\frac{d\Phi}{dt}[/tex]
So then
[tex] \varepsilon=-N(I_{max} \omega cos(\omega t+\phi)\frac{\mu_{o}N(Lw)}{w+h} \implies \varepsilon=(-100)(5)(cos(200\pi t+\phi))(200\pi)\frac{\mu_{o}(100)(.2)(.05)}{.1}[/tex]

(which ended up giving me the wrong answer)
 

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  • #2
This is an exercize in faraday's Law - but you have to understand the equations.
The magnetic flux changes with time because the current changes - but it also changes with the distance from the wire. ##d\phi = B(r)dA##
 
  • #3
Simon Bridge said:
This is an exercize in faraday's Law - but you have to understand the equations.
The magnetic flux changes with time because the current changes - but it also changes with the distance from the wire. ##d\phi = B(r)dA##

I'm not understanding why it's $$d\phi = B(r)dA$$ specifically dA, the area wouldn't be changing, only the distance. The B is the magnetic field, but what is this value in this case? And finally, where do you get the r from?

(Basically I don't understand that equation)
 
  • #4
##d\phi## is the amount of magnetic flux though the area element ##dA## at a distance ##r## from the wire, in the plane of the coil.

The magnetic field due to a long straight wire depends on the distance from the wire as well as the current.
 
  • #5
Simon Bridge said:
##d\phi## is the amount of magnetic flux though the area element ##dA## at a distance ##r## from the wire, in the plane of the coil.

The magnetic field due to a long straight wire depends on the distance from the wire as well as the current.

Ok that makes sense, but what do I do with it? If I were to integrate both sides it would just end up ## \phi=BrA##

Also, wouldn't r therefore be changing too as the current moves around the loop?

Edit: So ##B=\frac{\mu_{o}I}{2\pi r}## Substituting for the above equation gives ## \phi=\frac{\mu_{o}IrA}{2 \pi r}## I think the r's will cancel so then: ##\phi=\frac{\mu_{o}IA}{2\pi}##. I don't know what to do with this now lol
 
Last edited:
  • #6
I think I got it...$$d\Phi_{B}=\vec{B} \cdot d\vec{A}=\frac{\mu_{o}IL}{2\pi r}dr$$ Then: $$\Phi_{B}=\int_{h}^{h+w}\frac{\mu_{o}IL}{2\pi r}dr=\frac{\mu_{o}IL}{2\pi}ln(\frac{h+w}{h})$$ Then I can just plug it into: $$\varepsilon=-\frac{d\Phi_{B}}{dt}$$

Correct?
 
  • #7
iRaid said:
Ok that makes sense, but what do I do with it? If I were to integrate both sides it would just end up ## \phi=BrA##
No it doesn't. Because dA is made up of a component of dr ... you need to integrate over the length and width of the loop.

i.e. if you divide the area into strips length L and width dr then the area element at distance r is dA=Ldr ... Now you have ##d\phi = B(r)Ldr## ... notice that B is a function of r?
Have you never constructed integrals like this before?

Also, wouldn't r therefore be changing too as the current moves around the loop?
The current induced in the loop obeys Faraday's Law. It depends on the changes in the total flux through the loop.
 
  • #8
Simon Bridge said:
No it doesn't. Because dA is made up of a component of dr ... you need to integrate over the length and width of the loop.

i.e. if you divide the area into strips length L and width dr then the area element at distance r is dA=Ldr ... Now you have ##d\phi = B(r)Ldr## ... notice that B is a function of r?
Have you never constructed integrals like this before?


The current induced in the loop obeys Faraday's Law. It depends on the changes in the total flux through the loop.

Yes I see, I was just confused mainly because of B as a function of r I thought you ment B multiplied by r, but I think I got the correct way of doing it now (above)

Thanks.
 
  • #9
Yeah: You posted while I was typing ... hang on...
 
  • #10
OK - that looks good. Well done :)
It was just a matter of getting your head in the right place huh?

Formatting notes: the natural logarithm is \ln in LaTeX.
In general, standard function are formatted by putting a backslash in front of their name ;)
You can make the brackets go around big arguments by using \left( \right) instead of ( ).

So - yeah: $$\varepsilon = \frac{\mu_0LI_{max}}{2\pi}\ln\left(\frac{h+w}{h}\right)\frac{d}{dt}\sin( \omega t +\phi )$$
 
  • #11
Simon Bridge said:
OK - that looks good. Well done :)
It was just a matter of getting your head in the right place huh?

Formatting notes: the natural logarithm is \ln in LaTeX.
In general, standard function are formatted by putting a backslash in front of their name ;)
You can make the brackets go around big arguments by using \left( \right) instead of ( ).

So - yeah: $$\varepsilon = \frac{\mu_0LI_{max}}{2\pi}\ln\left(\frac{h+w}{h}\right)\frac{d}{dt}\sin( \omega t +\phi )$$

I just had to slow down and analyze what I was doing and I saw an example similar to this in the book that made it clear.
Yeah that's what I got and the derivative is just ωcos(ωt+ϕ) then plug in numbers
 
  • #12
Of course the acid test is if this gives the answer that was expected.
If you feel keen you can try looking at the relationship in terms of the reactance of the coil.
 
  • #13
Simon Bridge said:
Of course the acid test is if this gives the answer that was expected.
If you feel keen you can try looking at the relationship in terms of the reactance of the coil.

Can you explain this further?
 
  • #14
Explain which bit?

If the treatment does not get the marks, then it means we misunderstood the question (or the model answer is wrong.)

A coil of wire is an electrical component called an inductor - it is a reactive load so it has a reactance. (Look up "back emf".)

The coil's presence close to the wire means there is an additional load on whatever is making the current in the wire... which is why you cannot just run a coil next to the power lines and get free electricity.

But you may prefer to wait until it comes up in the course.
 
  • #15
Simon Bridge said:
Explain which bit?

If the treatment does not get the marks, then it means we misunderstood the question (or the model answer is wrong.)

A coil of wire is an electrical component called an inductor - it is a reactive load so it has a reactance. (Look up "back emf".)

The coil's presence close to the wire means there is an additional load on whatever is making the current in the wire... which is why you cannot just run a coil next to the power lines and get free electricity.

But you may prefer to wait until it comes up in the course.

I don't believe that is part of my E&M physics course. We just had an exam on the magnetism part and now we're starting on waves and optics (more modern stuff). I already took an electrical engineering course and we did use inductors quite a bit, but we never talked about back emf other than in lab.
 
  • #16
Well ... that's why I said, "if you feel keen" ;)
It can help to make connections between different parts of the work you've done.
But that's OK - you're done with this problem now.
 

1. How is EMF induced in a coil by a magnetic field?

EMF (electromotive force) is induced in a coil when the coil is exposed to a changing magnetic field. This changing magnetic field creates a flow of electrons within the coil, resulting in the production of EMF.

2. What factors affect the strength of the induced EMF in a coil?

The strength of the induced EMF in a coil is affected by the rate of change of the magnetic field, the number of turns in the coil, and the strength of the magnetic field. Additionally, the material of the coil and the resistance of the circuit can also play a role in the strength of the induced EMF.

3. How is the direction of the induced EMF determined?

The direction of the induced EMF is determined by the direction of the changing magnetic field and the direction of the current flow in the coil. This is known as Lenz's Law, which states that the direction of the induced EMF is always such that it opposes the change in the magnetic field that caused it.

4. What is Faraday's Law and how does it relate to EMF induced in a coil?

Faraday's Law states that the magnitude of the induced EMF in a coil is directly proportional to the rate of change of the magnetic field. In other words, the faster the magnetic field changes, the stronger the induced EMF will be in the coil.

5. What are some real-world applications of EMF induced in a coil by a magnetic field?

EMF induced in a coil by a magnetic field is used in various technologies such as generators, transformers, and electric motors. It is also used in devices such as metal detectors, magnetic levitation trains, and induction cooktops. Additionally, EMF induced in a coil is essential for wireless charging and communication technologies such as RFID (radio-frequency identification) and wireless power transfer.

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