Induced current in a metal ring

AI Thread Summary
A metal ring with a radius of 10cm and resistance of 0.1Ω moves through a magnetic field of 0.4T at 5m/s, inducing an electrical current. The discussion revolves around calculating the induced voltage using Faraday's law of induction and determining the direction and value of the current as the ring passes through different positions in the magnetic field. Participants suggest various approaches, including using effective length and motional emf formulas, while debating the need for calculus versus simpler methods. The induced emf is noted to vary sinusoidally as the ring enters and exits the field, with some confusion about the implications of the ring being a closed loop. Ultimately, the problem requires a combination of trigonometry and calculus to accurately describe the induced emf over time.
LogarithmLuke
Messages
82
Reaction score
3

Homework Statement


A metal ring with a radius of 10cm and resistance 0.1Ω passes through a delimited magnetic field, look at the photo posted below. The ring has a velocity of 5m/s, and the magnetic field has the field strength 0.4T. When the ring passes through the magnetic field, at some points an electrical current is induced in the ring.

Decide the direction and value of the current as the ring passes through.

Homework Equations


7aqxde5.jpg


Faradays law of induction: ε=-ΔΦ/Δt

The Attempt at a Solution


I tried to find the induced voltage at spot number two, by using the formula: ε=(-π*0.1m*0.4t)/x
I wasnt really able to figure out the time, since no distances except the radius of the ring are given.
 
Last edited by a moderator:
Physics news on Phys.org
Pictorial hint:
upload_2017-2-6_15-2-28.png


The yellow area represents the area over which there is flux at the present instant of time. During a small time interval Δt, an additional area represented by the blue strip will enter the region of B field.
 
  • Like
Likes cnh1995
TSny said:
Pictorial hint:
View attachment 112655

The yellow area represents the area over which there is flux at the present instant of time. During a small time interval Δt, an additional area represented by the blue strip will enter the region of B field.

What I am having trouble with is finding the increase in area in respect to time.
 
The increase in area is the area ΔA of the blue strip. Express the area of this rectangle in terms of the radius of the ring, the speed of the ring and Δt.
 
  • Like
Likes cnh1995
LogarithmLuke said:
What I am having trouble with is finding the increase in area in respect to time.
You have to find the function A(t) area under magnetic field at time t to calculate the flux. It looks that you have to do some integration to get A(t) but also after that some derivation (Faraday) so maybe they cancel each other, think about that.
 
  • Like
Likes cnh1995
LogarithmLuke said:
I tried to find the induced voltage at spot number two, by using the formula: ε=(-π*0.1m*0.4t)/x
I wasnt really able to figure out the time, since no distances except the radius of the ring are given.
You can solve it using TSny's suggestion. It will involve some differentiation and transformation of variables.

Another approach is to use the effective length of the arc in the magnetic field. Effective length of the arc in the magnetic field will be simply the straight line distance between its two end points. Once you write that length as a function of given parameters, you can directly use the motional emf formula E=B*v*Leffective.
 
cnh1995 said:
You can solve it using TSny's suggestion. It will involve some differentiation and transformation of variables.
No calculus or variable transformation is needed. Position "2" of the ring is special and finding the area of the blue rectangle is easy.
 
  • Like
Likes cnh1995
TSny said:
No calculus or variable transformation is needed. Position "2" of the ring is special and finding the area of the blue rectangle is easy.
LogarithmLuke said:
Decide the direction and value of the current as the ring passes through.
I was talking about the general expression for emf E(t), not just for position 2.

Or is it possible to find E(t) just by using the speciality of position 2? I'll see if I can do that..
Thanks!
 
cnh1995 said:
I was talking about the general expression for emf E(t), not just for position 2.
OK, I see. I might have misinterpreted the problem. I was assuming that you only had to find the induced emf at the 5 specific positions shown.
 
  • #10
TSny said:
OK, I see. I might have misinterpreted the problem. I was assuming that you only had to find the induced emf at the 5 specific positions shown.

The positions marked are basically general positions, so position 2 basically goes from where the ring is at in positon 2 in the picture, till it gets to position 3. I am excpected to find a min value, max value as well as an average value for the voltage in all positions.
 
  • #11
LogarithmLuke said:
The positions marked are basically general positions, so position 2 basically goes from where the ring is at in positon 2 in the picture, till it gets to position 3. I am excpected to find a min value, max value as well as an average value for the voltage in all positions.
cnh1995 said:
Another approach is to use the effective length of the arc in the magnetic field. Effective length of the arc in the magnetic field will be simply the straight line distance between its two end points. Once you write that length as a function of given parameters, you can directly use the motional emf formula E=B*v*Leffective.
Can you express this effective length in terms of radius of the ring? Join the two ends of the arc in the magnetic field and express that length in terms of known quantities or variables.
Hint: Rectangular to polar conversion..
 
  • #12
LogarithmLuke said:
The positions marked are basically general positions, so position 2 basically goes from where the ring is at in positon 2 in the picture, till it gets to position 3. I am excpected to find a min value, max value as well as an average value for the voltage in all positions.
OK. Sorry for the misinterpretation.
 
  • #13
The area of the ring starts from zero and increases to maximum area , then constant for some time after that the area decreases to zero
time to increase the area from zero to max = 20/5000= 0.004s or 4ms ,
 
  • #14
malemdk said:
The area of the ring starts from zero and increases to maximum area , then constant for some time after that the area decreases to zero
time to increase the area from zero to max = 20/5000= 0.004s or 4ms ,

Should it not be 0.2m/5m/s=40ms ?
 
  • #15
LogarithmLuke said:
Should it not be 0.2m/5m/s=40ms ?
The equation for emf comes out to be sinusoidal.
 
  • #16
LogarithmLuke said:
Should it not be 0.2m/5m/s=40ms ?
Yes, that correct
 
  • #17
cnh1995 said:
The equation for emf comes out to be sinusoidal.
Since the ring intersects magnetic fluxes constantly for some time -depending on the width of fluxes -we cannot consider it as sinusoidal function
 
  • #18
malemdk said:
Since the ring intersects magnetic fluxes constantly for some time -depending on the width of fluxes -we cannot consider it as sinusoidal function
I mean when the ring is entering the field region, the emf is sinusoidal. Once it is completely inside the field, there is no emf for some time. When it is leaving the region, there is again a sinusoidal emf.
 
  • #19
cnh1995 said:
I mean when the ring is entering the field region, the emf is sinusoidal. Once it is completely inside the field, there is no emf for some time. When it is leaving the region, there is again a sinusoidal emf.
When completely inside the field the emf will not be zero, since it is moving not stationary
 
  • #20
malemdk said:
When completely inside the field the emf will not be zero, since it is moving not stationary
Though it is moving, the rate of change of flux through it is zero when it is completely inside the region. So there would be no emf in the ring once it is completely in the field.

Thinking in terms of motional emf, the effective length of the ring becomes zero when the ring is completely inside the field region, making the emf (BvLeffective) zero.
 
  • #21
Thanks for highlighting, yes indeed the induced emf would be sinusoidal
 
  • #22
malemdk said:
Thanks for highlighting, yes indeed the induced emf would be sinusoidal
I don't think so. How do you get that?
 
  • #23
As the ring enters (or leaves) the field, I get that the emf vs time graph is part of an ellipse.
 
  • #24
haruspex said:
I don't think so. How do you get that?
Isn't the effective length of the arc in the magnetic field varying sinusoidally as the ring enters the region?
Am I missing something?
 
  • #25
cnh1995 said:
Isn't the effective length of the arc in the magnetic field varying sinusoidally as the ring enters the region?
Am I missing something?
Are you assuming that the angle subtended by the effective length (from the center of the ring) increases proportional to time?
 
  • #26
TSny said:
As the ring enters (or leaves) the field, I get that the emf vs time graph is part of an ellipse.
Effective length of the arc in the magnetic field would be the length of the segment joining the ends of the arc.

We can express that length as Leffective=2rsinθ, where 2θ is the angle subtended by the arc in the B-field.
Now, as the loop enters the region, doesn't θ vary from 0 to 180°? This should make Leffective vary sinusoidally, from 0(completely outside) to Lmax(half in) to 0 again (completely inside).

TSny said:
Are you assuming that the angle subtended by the effective length (from the center of the ring) increases proportional to time?
Yes.
I used rdθ=dx, which gives dx/dt=v=rdθ/dt.
Is that a mistake?
 
  • #27
When the ring is entirely inside the magnetic field the left and right side of the ring exposed to the same amount of flux and since the ring is closed no emf can be generated
 
  • #28
x = r cos(θ/2) where x is distance of center of ring from edge of field region and θ is subtended angle of effective length.
 
  • Like
Likes cnh1995
  • #29
haruspex said:
I don't think so. How do you get that?
When the ring is entirely inside the magnetic field the left and right side of the ring exposed to the same amount of flux and since the ring is closed no emf can be generated
Area exposed to the magnetic field is sinusoidal.
 
  • #30
malemdk said:
Area exposed to the magnetic field is sinusoidal
It may very well vary as the sine of some angle, but that angle is not changing uniformly with time.
Consider the rate of change of area as the ring moves at constant linear speed.
(I concur that it is half an ellipse, the top half on the way in and the bottom half on the way out, or vice versa.)
 
  • Like
Likes cnh1995
  • #31
TSny said:
x = r cos(θ/2) where x is distance of center of ring from edge of field region and θ is subtended angle of effective length.
Ah..I see! I had written this equation earlier, don't know why I didn't contonue with that.
I'll work out the math later.

So Leffective=2rsinθ.
Which means motional emf=2Blvrsinθ.
This equation is sinusoidal w.r.t. the angle θ, but since θ doesn't vary linearly with time, the equation is not sinusoidal w.r.t. time. Is that correct?
 
Last edited:
  • #32
cnh1995 said:
I'll work out the math later.
Yes, the equation is indeed of an ellipse.
Thanks @TSny, @haruspex for catching the mistake.
 
Last edited:
  • #33
The area exposed to magnetic field is much more than simple Sine function
 
  • #34
malemdk said:
The area exposed to magnetic field is much more than simple Sine function
Of course, but it's the rate of change of area that generates the emf.
 
  • #35
What I mean is that rate of change area is not simple function of Sine
 
  • #36
malemdk said:
What I mean is that rate of change area is not simple function of Sine
"Function of Sine" is meaningless. Do you mean it is not the sine function of the subtended angle?
The rate of change of area is the linear speed multiplied by chord length. The chord length is 2r sin(θ)
 
  • #37
That's what try to mean,
But I have doubt, since the ring is closed loop how could emf develop?
Since no high or low potential point
 
  • #38
malemdk said:
That's what try to mean,
But I have doubt, since the ring is closed loop how could emf develop?
Since no high or low potential point
The induced electric field in the ring is non-conservative. Therefore, concept of potential is not applicable here.

Here, ∫closed loopE⋅dl=dΦ/dt. In electrostatic case (circuits with battery) this integral for any loop is zero. The concept of potential is applicable in electrostatic field.

Here, the E field everywhere in the ring will be same (and non-zero), given by the above formula.
 
  • #39
You mean that the current keep on flowing?

Yes that makes sense
 
  • #40
malemdk said:
You mean that the current keep on flowing?
Yes.
 
  • #41
Has anyone found a way to solve it yet though? I am quite certain you don't need to use calculus to solve this, but I am sure there are multiple valid methods one can use.
 
  • #42
LogarithmLuke said:
Has anyone found a way to solve it yet though? I am quite certain you don't need to use calculus to solve this, but I am sure there are multiple valid methods one can use.
This problem needs trigonometry and some calculus (chain rule of differentiation).
The emf graph w.r.t. time is a part of an ellipse.
Read #23, #25, #28, #30 and #31..
 
  • #43
t
cnh1995 said:
This problem needs trigonometry and some calculus (chain rule of differentiation).
The emf graph w.r.t. time is a part of an ellipse.
Read #23, #25, #28, #30 and #31..

That would probably work, but it seems like that would be more complicated than this really has to be. All the other problems in the textbook on induction have required far less complex methods than what you're describing. What i was thinking was to just find the average and max value for the emf in all the general positions. The problem doesen't really require us to represent all of the emf values induced with a graph, although I am sure its possible to solve the problem that way.
 
  • #44
LogarithmLuke said:
The problem doesen't really require us to represent all of the emf values induced with a graph, although I am sure its possible to solve the problem that way.
Then I guess they want you to find the voltage and current only at the 5 positions shown. Why else would they draw those 5 positions instead of just saying that the ring passes through the field with a constant velocity of 5m/s?

You can compute the emf at the given positions without calculus and any variable transformation. See TSny's pictorial hint in #2.
 
  • #45
LogarithmLuke said:
The problem doesen't really require us to represent all of the emf values induced with a graph, although I am sure its possible to solve the problem that way.
I think it would be helpful if you quoted the statement of the problem exactly as it was given to you. In the first post, you mention having to get the magnitude and direction of the induced current. Then in a later post, you mention having to get max, min, and average values of emf.
 
  • #46
TSny said:
I think it would be helpful if you quoted the statement of the problem exactly as it was given to you. In the first post, you mention having to get the magnitude and direction of the induced current. Then in a later post, you mention having to get max, min, and average values of emf.

I wrote the whole problem statement i my first post. The problem is that they did not specify whether they wanted the current as a function of the time or just spesific values. However in the answer section for the problems in the very back of the textbook they used a max value as well as an average value as an answer to this specific problem.
 
  • #47
So i was able to find the average value of the current for current for position 2. I found that the emf must be (((0.4T*pi*0.1m^2)/2)/20ms)/0.1Ω

This gave me 3.1A which apparently is the average current. However i do not really understand why that gave me the average current. Isnt the emf constant when part of the ring is in the magnetic field, then 0 when all of the ring is inside? Additionally, how do i go about finding the max value for the current?
 
  • #48
LogarithmLuke said:
Isnt the emf constant when part of the ring is in the magnetic field,
The emf is never constant (except when it is 0).

Are you asked to find the average emf from position 1 to position 5?
 
  • #49
cnh1995 said:
The emf is never constant (except when it is 0).

Are you asked to find the average emf from position 1 to position 5?

Why isn't it constant when the speed is constant and the rate of change in area is constant?

The problem doesent specify, but based on the answers section were supposed to find the average and max values for each of the positions. However, positions 1,3 and 5 all have emf values of 0 (why is this?), so in reality only for positions 2 and 4.

Like i said the entire problem statement is in the first post, however the creators of the physics textbook were quite unclear in the way they formulated the problem. I am sure there is still a lot to learn through working with this though.
 
  • #50
LogarithmLuke said:
Why isn't it constant when the speed is constant and the rate of change in area is constant?
Area is not changing at a constant rate.
cnh1995 said:
The emf graph w.r.t. time is a part of an ellipse.
Read #23, #25, #28, #30 and #31..
 
Back
Top