Faraday's Law of Induction

  • #1
ayans2495
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Homework Statement:
The initial peak EMF and frequency in the generator (AC) are 1220 V and 50 Hz respectively. The magnetic field is halved, the area of the coil is tripled and the number of turns increases from 50 to 250. What is the new EMF and frequency?
Relevant Equations:
Faraday's law.
1654826980254.png
 

Answers and Replies

  • #2
kuruman
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Is there a question you wish to ask?
 
  • #3
ayans2495
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Is there a question you wish to ask?
It is in the homework statement
 
  • #4
ayans2495
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Is there a question you wish to ask?
Sorry.
The initial peak EMF and frequency in the generator (AC) are 1220 V and 50 Hz respectively. The magnetic field is halved, the area of the coil is tripled and the number of turns increases from 50 to 250. What is the new EMF and frequency?
 
  • #5
kuruman
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PF rules do not allow answering homework questions directly. We can help you get to the answer. Are you stuck somewhere? If so where?

I think you need a better relevant equation than "Faraday's law" which is not even an equation. How about the emf generator equation? What does that look like? If you don't remember, look it up.
 
  • #6
ayans2495
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PF rules do not allow answering homework questions directly. We can help you get to the answer. Are you stuck somewhere? If so where?

I think you need a better relevant equation than "Faraday's law" which is not even an equation. How about the emf generator equation? What does that look like? If you don't remember, look it up.
This was a question on a test I previously had. Faraday's law: Є= (NxBxA)/T. Where N is the number of turns, B is the magnetic field strength and T is the period. This is the formula I was given on the test. As you can see above in my attempt of the solution, I got stuck when I had an equation in terms of the final emf with two unknowns.
 
  • #7
kuruman
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What reason do you have to believe that the period changes? The problem says that N, B and A change but says nothing about changing T. You may assume that T stays the same as before in which case you have only one unknown.
 
  • #8
Steve4Physics
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@ayans2495, some additional advice. If you can understand/use proportionality, you only need 1 or 2 lines of working – not a whole page!

E.g. ##x = \frac {ab}{cd}##.
What happens to ##x## when:
##a## increases by a factor 3 (i.e. triples) and
##b## increases by a factor 4 and
##c## increases by a factor 6 and
##d## is unchanged?

Answer: ##x## changes by a factor ##\frac {3 \times 4}{6 \times 1} = 2##, (i.e. ##x## doubles).

P.S. I hope sufficient acts of kindness (as indicated on your Post #1 attachment) were performed!
 
  • #9
ayans2495
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@ayans2495, some additional advice. If you can understand/use proportionality, you only need 1 or 2 lines of working – not a whole page!

E.g. ##x = \frac {ab}{cd}##.
What happens to ##x## when:
##a## increases by a factor 3 (i.e. triples) and
##b## increases by a factor 4 and
##c## increases by a factor 6 and
##d## is unchanged?

Answer: ##x## changes by a factor ##\frac {3 \times 4}{6 \times 1} = 2##, (i.e. ##x## doubles).

P.S. I hope sufficient acts of kindness (as indicated on your Post #1 attachment) were performed!
That makes a lot of sense. Thank you for helping me see it, the period remains unchanged because it no where states in the question that it was changed. So the new emf, when rounded up, is 6473 emf.
 
  • #10
Steve4Physics
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That makes a lot of sense. Thank you for helping me see it, the period remains unchanged because it no where states in the question that it was changed. So the new emf, when rounded up, is 6473 emf.
I don't see how you get 6473. Can you post your working?
Also, you have forgotten units (lose 1 mark!) and have the wrong number of significant figures (lose 1 mark!).
 
  • #11
ayans2495
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I don't see how you get 6473. Can you post your working?
Also, you have forgotten units (lose 1 mark!) and have the wrong number of significant figures (lose 1 mark!).
The formula I acquired in my solutions suggests I merely have to divide 129.45 by the period (0.02 s because frequency is 50 Hz). Sorry about the unit, I certainly included it in my test. Also, it's not a requirement that I round to sig figs in my test but I'll be sure to do so from now in this forum.
 
  • #12
ayans2495
58
2
@ayans2495, some additional advice. If you can understand/use proportionality, you only need 1 or 2 lines of working – not a whole page!

E.g. ##x = \frac {ab}{cd}##.
What happens to ##x## when:
##a## increases by a factor 3 (i.e. triples) and
##b## increases by a factor 4 and
##c## increases by a factor 6 and
##d## is unchanged?

Answer: ##x## changes by a factor ##\frac {3 \times 4}{6 \times 1} = 2##, (i.e. ##x## doubles).

P.S. I hope sufficient acts of kindness (as indicated on your Post #1 attachment) were performed!
Oh I see what you're referring to when you say "acts of kindness" :) ! I felt compelled to work on the problem so I took the closest book I could find and tried my best.
 
  • #13
Steve4Physics
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The formula I acquired in my solutions suggests I merely have to divide 129.45 by the period (0.02 s because frequency is 50 Hz). Sorry about the unit, I certainly included it in my test. Also, it's not a requirement that I round to sig figs in my test but I'll be sure to do so from now in this forum.
You haven't provided your working so I've no idea where 129.45 (what units?) comes from, or why you are dividing it by the period.

EDIT - apologies. I see the figure 129.45 comes from your Post #1 attachment.

Try answering this:

##E_0 = \frac {NBA}{T}## isn’t strictly correct for an AC generator but it will do here.

B changes by a factor ½.
A changes by a factor 3.
N changes by a factor (250/50 =) 5.
T changes by a factor 1 (unchanged).

Q1. By what factor does##E_0## change?
Q2. ##E_0## = 1220V initially. So what is its final value?
Q3. Also, what is the final frequency (which is the other part of the original question)?
 
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  • #14
Steve4Physics
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You haven't provided your working so I've no idea where 129.45 (what units?) comes from, or why you are dividing it by the period. Try answering this:
Apologies - I see the '129.45' comes from your Post #1 attachment.
 
  • #15
Steve4Physics
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That makes a lot of sense. Thank you for helping me see it, the period remains unchanged because it no where states in the question that it was changed. So the new emf, when rounded up, is 6473 emf.
Aha! 6473(V) is the rms value! You have forgotten to convert it back to a peak value. This will then give the correct answer (ignoring the rounding errors which have accumulated).

But there is no need t convert peak to rms then back again. And remember, the whole problem can be done in one line of simple arithmetic if you use proportionality.
 
  • #16
ayans2495
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Aha! 6473(V) is the rms value! You have forgotten to convert it back to a peak value. This will then give the correct answer (ignoring the rounding errors which have accumulated).

But there is no need t convert peak to rms then back again. And remember, the whole problem can be done in one line of simple arithmetic if you use proportionality.
You're right! However, the question never asked for the peak value, it just said "new voltage". Is it implicit that I provide the peak value? As for the frequency, it would remain the same as we assumed the period to remain constant, thus 50 Hz.
 
  • #17
kuruman
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You're right! However, the question never asked for the peak value, it just said "new voltage". Is it implicit that I provide the peak value? As for the frequency, it would remain the same as we assumed the period to remain constant, thus 50 Hz.
It is implicit because you have already been given the peak value. If they wanted the rms as the new value, they would have said so.
 
  • #18
ayans2495
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It is implicit because you have already been given the peak value. If they wanted the rms as the new value, they would have said so.
Yes, you're right. I have one more question: what prompted you to assume that the period remains the same? Why couldn't the EMF remain constant and the period change by some factor?
 
  • #19
Delta2
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In a prototype generator, the frequency of the EMF equals the mechanical frequency (the mechanical rotation rate divided by ##2\pi##, that is ##\frac{\omega}{2\pi})## of the generator. The problems mentions nothing about changing that so we can assume that it remains the same.
 
  • #20
kuruman
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Yes, you're right. I have one more question: what prompted you to assume that the period remains the same? Why couldn't the EMF remain constant and the period change by some factor?
Because the problem is asking for the new EMF.
 
  • #21
ayans2495
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Because the problem is asking for the new EMF.
Forgive my English if I am wrong, but doesn't the adjective "new" apply to both nouns, namely emf AND frequency?
 
  • #22
kuruman
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Note that the full question is "What is the new EMF and frequency?" If two items were supposed to be new, the plural "are" should have been used as in "What are the new EMF and frequency?" In short,

What is the new EMF and frequency ⇒ new EMF + frequency
What are the new EMF and frequency ⇒ new (EMF + frequency) = new EMF + new frequency
 

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