New EMF and Frequency: Understanding Faraday's Law of Induction

In summary: 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.In summary, the conversation was about a generator problem in which the initial peak EMF and frequency were given. The question was to find the new EMF and frequency after changes were made to the magnetic field, area of coil, and number of turns. The conversation focused on using Faraday's law and the emf generator equation to solve the problem. The expert advised using proportionality and only providing a few lines of working instead of a whole page. They also reminded the student to include units and use the correct number of significant figures in their calculations
  • #1
ayans2495
58
2
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
 
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  • #2
Is there a question you wish to ask?
 
  • #3
kuruman said:
Is there a question you wish to ask?
It is in the homework statement
 
  • #4
kuruman said:
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
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
kuruman said:
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
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.
 
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  • #8
@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
Steve4Physics said:
@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
ayans2495 said:
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
Steve4Physics said:
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
Steve4Physics said:
@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
ayans2495 said:
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)?
 
Last edited:
  • #14
Steve4Physics said:
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
ayans2495 said:
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
Steve4Physics said:
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
ayans2495 said:
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
kuruman said:
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
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
ayans2495 said:
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
kuruman said:
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
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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What is Faraday's Law of Induction?

Faraday's Law of Induction is a fundamental principle in electromagnetism that explains the relationship between a changing magnetic field and an induced electric current.

How does Faraday's Law of Induction work?

Faraday's Law states that when a conductor (such as a wire) is exposed to a changing magnetic field, an electric current will be induced in the conductor. The direction of the induced current is always such that it opposes the change in the magnetic field.

What is the significance of Faraday's Law of Induction?

Faraday's Law is important because it is the basis for many modern technologies, such as electric generators, transformers, and motors. It also helps us understand the relationship between electricity and magnetism.

What is the difference between Faraday's Law of Induction and Lenz's Law?

Faraday's Law of Induction describes the relationship between a changing magnetic field and an induced electric current, while Lenz's Law states that the direction of the induced current will always oppose the change in the magnetic field that caused it.

How is Faraday's Law of Induction applied in real life?

Faraday's Law of Induction is applied in many practical applications, such as power generation, electric motors, and wireless charging. It is also used in devices like transformers and induction cooktops.

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