Calulate emf unsing electrodynamics

In summary: Here, let me get you started. It gives you the position functiony=asin(wt)so you know that the radius of the loop is a. You know that at time t=0 that y=a*0=0. That then shows the loop, with the radius of a, perpendicular to the magnetic field. So, \int\int B \cdot da = BAat that point in time. Now hopefully you can see the picture of how your loop is just varying how much flux goes through the loop based on your sine function. It's rotating about the z-axis. Once you've got
  • #1
leonardo123
9
0

Homework Statement


There is a Static magnetic field,with its intensity B. Assume a coil moving
in the magnetic field, with its position function y=asinwt. then calculate the voltage of U



Homework Equations





The Attempt at a Solution

 

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  • #2
Have you tried doing this? Do you know Faraday's law?
 
  • #3
Matterwave said:
Have you tried doing this? Do you know Faraday's law?

hey, it's not that easy. the movement of the coil change the magnetic field intensity. the thing is how can you calculate the changing magnetic intensity and use Maxwell's equation to solve the problem
 
  • #4
Hey, from our standpoint it is that easy! You're right, the movement of the coil changes the flux going through it. So figure out how quickly the flux changes, and you've got your answer. You'll have to show us specifically where your work stops if you want more help than that.
 
  • #5
Mindscrape said:
Hey, from our standpoint it is that easy! You're right, the movement of the coil changes the flux going through it. So figure out how quickly the flux changes, and you've got your answer. You'll have to show us specifically where your work stops if you want more help than that.

but how can you figure out how quick the flux changes? i mean coil is a loop.
assume the resistance of the coil is R, and the radius is r.
could you please show me the fuction of votage(electromotive force) in the coil?
 
  • #6
You can figure out how quick the flux changes by how fast the area is changing. Do you follow? Don't assume the radius, you're given the radius.
 
  • #7
Mindscrape said:
You can figure out how quick the flux changes by how fast the area is changing. Do you follow? Don't assume the radius, you're given the radius.

but the coil is a loop, my answer is V=0.
is that right?
 
  • #8
No, that's not right. Here, let me get you started. It gives you the position function

y=asin(wt)

so you know that the radius of the loop is a. You know that at time t=0 that y=a*0=0. That then shows the loop, with the radius of a, perpendicular to the magnetic field. So,

[tex]\int\int B \cdot da = BA[/tex]

at that point in time. Now hopefully you can see the picture of how your loop is just varying how much flux goes through the loop based on your sine function. It's rotating about the z-axis. Once you've got the flux, it's cake to get the voltage.
 
Last edited:
  • #9
i am sorry,but here is a mistake.
actually the radius of the coil is fixed, which is equal to r. the velocity of the coil is v=asinwt.
sorry for that.now still the same?
 
  • #10
Mindscrape said:
No, that's not right. Here, let me get you started. It gives you the position function

y=asin(wt)

so you know that the radius of the loop is a. You know that at time t=0 that y=a*0=0. That then shows the loop, with the radius of a, perpendicular to the magnetic field. So,

[tex]\int\int B \cdot da = BA[/tex]

at that point in time. Now hopefully you can see the picture of how your loop is just varying how much flux goes through the loop based on your sine function. It's rotating about the z-axis. Once you've got the flux, it's cake to get the voltage.

i am sorry,but here is a mistake.
actually the radius of the coil is fixed, which is equal to r. the velocity of the coil is v=asinwt.
sorry for that.now still the same?
 
  • #11
Ahh, okay, that makes a little more sense with the picture drawn then. Still kind of the same. If v=asinwt, then you have to integrate v w.r.t. time to get position. You still have to relate the flux to the rotation of the loop, and keep in mind that

[tex]\int\int B \cdot da = BAcos(\theta)[/tex]
 
  • #12
Mindscrape said:
Ahh, okay, that makes a little more sense with the picture drawn then. Still kind of the same. If v=asinwt, then you have to integrate v w.r.t. time to get position. You still have to relate the flux to the rotation of the loop, and keep in mind that

[tex]\int\int B \cdot da = BAcos(\theta)[/tex]

i am really greatful for your help. but here the coil doesn't rotate in the magnetic field. it just moves up and down, with the speed v=asinwt
 
  • #13
Are you sure about that? If all it does is move up and down in a magnetic field that takes up the entire yz-plane then there is no change in flux and no EMF. Why would the problem give you a sinusoidal position that is translational instead of rotational? That is either wrong or it's a really stupid problem.
 
  • #14
Mindscrape said:
Are you sure about that? If all it does is move up and down in a magnetic field that takes up the entire yz-plane then there is no change in flux and no EMF. Why would the problem give you a sinusoidal position that is translational instead of rotational? That is either wrong or it's a really stupid problem.

yes, i am sure.
from my point of view, the votage should be zero.but our professor told me the movement of the coil change the density of the magnetic field. and there will be a electromotive force in the ring.
what you think?
 
  • #15
Well, he could mean a couple of things. He could be talking about a material that develops eddy currents, which are super difficult to calculate. Or if the magnetic field itself has a density associated with it, then you'll get a change in flux from the loop traveling through those different magnitudes of magnetic field. Think about the actual statement of B dot da, it means the projection of B onto a differential area element. If different amounts of B are projected onto da over the course of da moving, then you get a changing flux that produces an EMF. However, all that being said, you did not list a density of magnetic field in your problem description.
 
  • #16
Mindscrape said:
Well, he could mean a couple of things. He could be talking about a material that develops eddy currents, which are super difficult to calculate. Or if the magnetic field itself has a density associated with it, then you'll get a change in flux from the loop traveling through those different magnitudes of magnetic field. However, you did not list a density of magnetic field in your problem description.

yes, i think i have to consider about the eddy current. but so far, i still didi not find any solution. i don't know where i can get started.
it is really hard
 
  • #17
There isn't a solution if you don't know the magnetic field density. You're going to have to give us a better problem description before we can go any further.
 

Related to Calulate emf unsing electrodynamics

1. How can I calculate the electromotive force (emf) using electrodynamics?

The emf can be calculated using the equation: emf = -N(dΦ/dt), where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux through the coil.

2. What is the significance of calculating emf using electrodynamics?

Calculating emf using electrodynamics allows us to understand the relationship between magnetic fields and induced currents, and how they are affected by factors such as coil size and magnetic flux.

3. Can emf be calculated for non-constant magnetic fields?

Yes, emf can be calculated for non-constant magnetic fields by using the equation: emf = -N(dΦ/dt), where dΦ/dt is the instantaneous rate of change of magnetic flux.

4. How does the direction of the induced current relate to the direction of the emf?

The direction of the induced current is always such that it opposes the change in magnetic flux that caused it. This is known as Lenz's Law. Therefore, the direction of the induced current is always opposite to the direction of the emf.

5. Are there any practical applications of calculating emf using electrodynamics?

Yes, calculating emf using electrodynamics is crucial in understanding and designing devices such as generators, transformers, and motors. It is also used in various industries, including power generation, telecommunications, and transportation.

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