# EMF induced in coil by magnetic field

## 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:

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

(which ended up giving me the wrong answer)

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Simon Bridge
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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##

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)

Simon Bridge
Homework Helper
##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.

##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:
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?

Simon Bridge
Homework Helper
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.

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.

Simon Bridge
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Yeah: You posted while I was typing ... hang on...

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

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

Simon Bridge
Homework Helper
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.

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?

Simon Bridge
Homework Helper
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.

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.

Simon Bridge