Induced current in a metal ring

[LogarithmLuke]

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

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.

 

[TSny]

Pictorial hint:

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.

 

[LogarithmLuke]

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.

 

[TSny]

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.

 

[Diegor]

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.

 

[cnh1995]

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*L_effective.

 

[TSny]

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.

 

[cnh1995]

No calculus or variable transformation is needed. Position "2" of the ring is special and finding the area of the blue rectangle is easy.

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!

 

[TSny]

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.

 

[LogarithmLuke]

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.

 

[cnh1995]

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.

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..

 

[TSny]

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.

 

[malemdk]

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 ,

 

[LogarithmLuke]

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 ?

 

[cnh1995]

Should it not be 0.2m/5m/s=40ms ?

The equation for emf comes out to be sinusoidal.

 

[malemdk]

Should it not be 0.2m/5m/s=40ms ?

Yes, that correct

 

[malemdk]

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

 

[cnh1995]

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.

 

[malemdk]

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

 

[cnh1995]

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 (BvL_effective) zero.

 

[malemdk]

Thanks for highlighting, yes indeed the induced emf would be sinusoidal

 

[haruspex]

Thanks for highlighting, yes indeed the induced emf would be sinusoidal

I don't think so. How do you get that?

 

[TSny]

As the ring enters (or leaves) the field, I get that the emf vs time graph is part of an ellipse.

 

[cnh1995]

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?

 

[TSny]

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?

 

[cnh1995]

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 L_effective=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 L_effective vary sinusoidally, from 0(completely outside) to Lmax(half in) to 0 again (completely inside).

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?

 

[malemdk]

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

 

[TSny]

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.

 

[malemdk]

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.

 

[haruspex]

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.)

 

[cnh1995]

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 L_effective=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?

 

[cnh1995]

I'll work out the math later.

Yes, the equation is indeed of an ellipse.
Thanks @TSny, @haruspex for catching the mistake.

 

[malemdk]

The area exposed to magnetic field is much more than simple Sine function

 

[haruspex]

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.

 

[malemdk]

What I mean is that rate of change area is not simple function of Sine