Find the max induced EMF in the loop by the changing loop

In summary: With that in mind, you will find that your answer will be correct. But if you really want to do it correctly, use the exponential form of the emf instead of the cosine form.In summary, The maximum induced EMF in the loop by the changing current is given by E(max)=(μ)*(n)*(N)*(A) and the magnitude of the EMF at t=0.8 seconds is given by E=(μ)*(n)*(N)*(A)*(i)*(alpha). It is important to remember that alpha is not the time constant, but 1/alpha is. Using the exponential form of the EMF is recommended for accuracy in calculations.
  • #1
SpringWater
27
0

Homework Statement


I have attached the two questions. the first has a diagram.

1. Find the MAX Induced EMF (in volts) in the loop by the changing current?

2. What is the magnitude of the EMF induced in the loop by the changing current at t=.8 (seconds) answer in Volts?


Homework Equations



All relevant equations are attached

The Attempt at a Solution



I believe I understand most of how to find the correct solution. However I am having trouble with deciding what is E(max) (1st question)? I am assuming I include E(max)=(μ)*(n)*(N)*(A) but i am not sure if i include I(initial)?

Any help will be greatly appreciated.
 

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  • #2
SpringWater said:

Homework Statement


I have attached the two questions. the first has a diagram.

1. Find the MAX Induced EMF (in volts) in the loop by the changing current?

2. What is the magnitude of the EMF induced in the loop by the changing current at t=.8 (seconds) answer in Volts?


Homework Equations



All relevant equations are attached

The Attempt at a Solution



I believe I understand most of how to find the correct solution. However I am having trouble with deciding what is E(max) (1st question)? I am assuming I include E(max)=(μ)*(n)*(N)*(A) but i am not sure if i include I(initial)?

If all else fails, revert to dimensional analysis!
Letting E = emf, would E = (μ)*(n)*(N)*(A) be dimensionally correct?
Would E = (μ)*(n)*(N)*(A)*(i) be correct?
How about E = (μ)*(n)*(N)*(A)*(i)* (alpha)?
Hint: one of them is, the others are not.

PS otherwise what you did was fine.
 
  • #3
rude man said:
If all else fails, revert to dimensional analysis!
Letting E = emf, would E = (μ)*(n)*(N)*(A) be dimensionally correct?
Would E = (μ)*(n)*(N)*(A)*(i) be correct?
How about E = (μ)*(n)*(N)*(A)*(i)* (alpha)?
Hint: one of them is, the others are not.

PS otherwise what you did was fine.

Thank you for the reply, I greatly appreciate it!

so here is what i have. A Volt units can be changed into a lot of different units. so

μ=(N) / (Amp)^(2)

n=turns / (meter)

N= Turns

I(initial)=Amp

Area=(meter)^(2)

alpha is (1) / (seconds) so then..

(Newton)*(Turns)*(Turns)*(meter^(2))*(Amp)
(Amp)^(2)*(meter)*(second)

then finally (N)*(Turn(in))*(Turn(out))*(Meter) / (Amp)*(second)

i am not sure how to eliminate turns?
 
  • #4
N is dimensionless. But watch out: n has dimensions of 1/L.

You slipped up somewhere. i needs to be in the numerator, obviously.

Make life easier for yourself: you know from your textbook that for a solenoid B = μ0*i*n and you should know emf = -N*d(phi)/dt = -N*A*dB/dt in this case. So emf must look like
emf = A*B*T-1 = A*μ0*i*n*T-1.

In other words, no need to break everything down into fundamental units.
 
  • #5
rude man said:
N is dimensionless. But watch out: n has dimensions of 1/L.

You slipped up somewhere. i needs to be in the numerator, obviously.

Make life easier for yourself: you know from your textbook that for a solenoid B = μ0*i*n and you should know emf = -N*d(phi)/dt = -N*A*dB/dt in this case. So emf must look like
emf = A*B*T-1 = A*μ0*i*n*T-1.

In other words, no need to break everything down into fundamental units.

okay, I assumed that T=time constant tau or in my case alpha. the answer was correct. thank you for the help
 
  • #6
SpringWater said:
okay, I assumed that T=time constant tau or in my case alpha. the answer was correct. thank you for the help

Alpha is not the tme constant. 1/alpha is. Alpha has units of T^(-1) as it must since you need a di/dt term in your emf.
 

1. What is induced EMF?

Induced EMF (electromotive force) is the voltage generated in a conducting loop due to a changing magnetic flux through the loop. It follows Faraday's Law of Induction, which states that the induced EMF is proportional to the rate of change of magnetic flux.

2. How is the max induced EMF calculated in a loop?

The max induced EMF in a loop can be calculated by taking the derivative of the magnetic flux with respect to time and multiplying it by the number of turns in the loop. This is represented by the equation: E = -N(dΦ/dt) where E is the induced EMF, N is the number of turns, and dΦ/dt is the rate of change of magnetic flux.

3. What factors affect the max induced EMF in a loop?

The max induced EMF in a loop is affected by the rate of change of magnetic flux, the number of turns in the loop, and the magnetic permeability of the material the loop is made of. It is also affected by the speed at which the loop is moving through the magnetic field and the strength of the magnetic field itself.

4. How does changing the orientation of the loop affect the induced EMF?

The induced EMF in a loop is dependent on the amount of magnetic flux passing through the loop. Changing the orientation of the loop can change the amount of flux passing through, thus affecting the induced EMF. For example, if the loop is parallel to the magnetic field, the induced EMF will be at its maximum, while if the loop is perpendicular to the field, the induced EMF will be zero.

5. How is induced EMF used in practical applications?

Induced EMF has many practical applications, including generating electricity in power plants, wireless charging of devices, and induction heating. It is also used in devices such as transformers and generators to transfer energy from one circuit to another.

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