How Does Induced EMF Vary Inside a Solenoid?

In summary: B}{dt}=N*(\frac{dA}{dt}+\frac{I}{dt})which gives \frac{d \vec B}{dt}=N*(\frac{dA}{dt}+\frac{I}{dt})where N=turns per lengthdA=area of the loopI=current through the loop
  • #1
physics24sam
10
0

Homework Statement


The current in a solenoid (you may treat it as a long solenoid of length 2.0 m, turns, and radius 0.5 m) is decreasing at a rate of 1.7 A/s.
What is the EMF at a point .35m inside the solenoid?

Homework Equations



  1. B=N(mu_0*I)/(2a)
    • where B=magnetic field, N=turns per length, I=current, and a=radius
  2. Epsilon=d(phi)/dt
    • where Epsilon=EMF, phi=magnetic flux, and t=time
  3. Phi=int(B dot dA)
    • where A=cross sectional area and you take the dot product of the two vectors
  4. A=pi*r^2
    • where r=new radius (.35m inside the solenoid)

The Attempt at a Solution


[/B]
  • I have tried saying that Epsilon = -N(BA/t) and then substituting (mu_0*I)/2a in for B and pi*r^2 in for A so that Epsilon= -N*((mu_0*I)/2a)*(pi*r^2)*(1/t) and then pulling out I/t as my decreasing current over time.
  • I have also tried when r=a and they are both .5m and when they are both .15m
  • I have also tried finding the flux at some random current (11A) and then finding the flux after one second due to the decreasing current and then plugging those values into my Epsilon=d(phi)/dt equation
 
Physics news on Phys.org
  • #2
Hello physics24sam,

physics24sam said:

Homework Statement


The current in a solenoid (you may treat it as a long solenoid of length 2.0 m, turns, and radius 0.5 m) is decreasing at a rate of 1.7 A/s.
What is the EMF at a point .35m inside the solenoid?

Homework Equations



  1. B=N(mu_0*I)/(2a)
    • where B=magnetic field, N=turns per length, I=current, and a=radius
  2. Epsilon=d(phi)/dt
    • where Epsilon=EMF, phi=magnetic flux, and t=time
  3. Phi=int(B dot dA)
    • where A=cross sectional area and you take the dot product of the two vectors
  4. A=pi*r^2
    • where r=new radius (.35m inside the solenoid)

The Attempt at a Solution


[/B]
  • I have tried saying that Epsilon = -N(BA/t) and then substituting (mu_0*I)/2a in for B and pi*r^2 in for A so that Epsilon= -N*((mu_0*I)/2a)*(pi*r^2)*(1/t) and then pulling out I/t as my decreasing current over time.
  • I have also tried when r=a and they are both .5m and when they are both .15m
  • I have also tried finding the flux at some random current (11A) and then finding the flux after one second due to the decreasing current and then plugging those values into my Epsilon=d(phi)/dt equation
I suspect that you haven't received a reply as of yet because the problem statement was not typed in correctly, word for word. [Edit: or if the problem statement was typed in correctly, it is ambiguous.]

(a) It seems that you left out how many turns are in the solenoid.

(b) What do you mean by ".35m inside the solenoid"? Do you mean that there is a wire loop in the center of the solenoid (coaxial with the solenoid) that has a radius of 0.35 m? Or are you asking about the emf of some sort of coil or something that lies 0.35 m from the edge of the solenoid? It is not clear from the original problem statement.

(c) From you response it seems to me that you have tried a number of approaches that did not work. Do you know what the correct answer is? (And how do you know that your attempted solutions were not correct?)

(d) Please show more detail in your attempted solutions. Perhaps we can help point out where something went wrong.
 
Last edited:
  • #3
Sorry about that...

a) there are 1.1x10^4 turns in the solenoid

b) In regards to the .35m, I'm not sure. That is how it is worded in the problem. All it says is .35m inside which I assumed to mean .35m radially inward.

c) My professor posted an answer key, so I do know the right answer. He did not post a solution though, so I still have to work the problem out. The answer is 4.5x10^-3 V
 
  • #4
I would have thought it to came to 71 volt?
 
  • #5
physics24sam said:
Sorry about that...

a) there are 1.1x10^4 turns in the solenoid

b) In regards to the .35m, I'm not sure. That is how it is worded in the problem. All it says is .35m inside which I assumed to mean .35m radially inward.

c) My professor posted an answer key, so I do know the right answer. He did not post a solution though, so I still have to work the problem out. The answer is 4.5x10^-3 V

I feel sorry to hear that your instructor worded it that way. The emf induced by the current in the solenoid, simply does not exist at a "point" inside the solenoid. You need to define a path in order to compute or measure the induced emf.

But I've worked the problem backwards to determine what your instructor likely meant:

Approximate the induced emf on a circular loop of radius 0.35 m, inside the solenoid, which shares the same axis as the solenoid. (You may assume the loop is near the center of the solenoid.)​

Start by calculating [itex] \frac{d \vec B}{dt} [/itex] inside the solenoid. Then you can find the rate of change of flux through the 0.35 m loop. It should be easy to get the emf using Lenz's law [Edit: or Faraday's law].
 
Last edited:
  • #6
My interpretation would be: determine the voltage across all 2m of solenoid, then calculate proportionally how much you'd measure across 0.35m of it.

How does that gel with the required answer? :cool:
 
  • #7
NascentOxygen said:
My interpretation would be: determine the voltage across all 2m of solenoid, then calculate proportionally how much you'd measure across 0.35m of it.

How does that gel with the required answer? :cool:
That's an idea for the interpretation of the problem: Treat the solenoid as an inductor. Calculate itself inductance, and then calculate the voltage at a tap located 0.35 m from the end, knowing the rate of change of the current. ([itex] v = L \frac{di}{dt} [/itex]).

But that approach/interpretation doesn't match the answer. :(
 

What is induced EMF in a solenoid?

Induced EMF (electromotive force) in a solenoid refers to the voltage generated in a coil of wire when there is a change in the magnetic field passing through the coil.

How is induced EMF in a solenoid calculated?

The induced EMF in a solenoid can be calculated using the formula E = -N * ΔΦ/Δt, where E is the induced EMF, N is the number of turns in the coil, and ΔΦ/Δt is the rate of change of magnetic flux through the coil.

What factors affect the magnitude of induced EMF in a solenoid?

The magnitude of induced EMF in a solenoid is affected by the number of turns in the coil, the strength of the magnetic field, the rate of change of the magnetic field, and the resistance of the coil.

What is the direction of induced EMF in a solenoid?

The direction of induced EMF in a solenoid is determined by the right-hand rule, which states that if you point your thumb in the direction of the magnetic field, and your fingers in the direction of the current, then the direction of your palm will indicate the direction of the induced EMF.

What is the practical application of induced EMF in a solenoid?

Induced EMF in a solenoid has many practical applications, such as in generators, transformers, and electric motors. It is also used in devices such as metal detectors and magnetic levitation trains.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
144
  • Introductory Physics Homework Help
2
Replies
41
Views
4K
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
513
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
907
  • Introductory Physics Homework Help
Replies
3
Views
164
  • Introductory Physics Homework Help
Replies
3
Views
968
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top