How does the induced voltage change when pulling a coil out of a magnetic field?

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The discussion centers on calculating the induced voltage when a coil is pulled out of a magnetic field and when it rotates within the field. For part A, the induced voltage is zero since the coil's plane is parallel to the magnetic field lines. In part B, the participants explore two interpretations: calculating an average induced voltage based on the change in magnetic flux over time or deriving a time-dependent function for the induced voltage as the coil exits the field. The conversation also touches on part C, where the coil's rotation introduces a varying induced voltage, modeled as a sine function due to the changing angle between the coil and the magnetic field. Ultimately, the induced voltage is expressed as a function of time, reflecting the dynamic nature of the coil's motion within the magnetic field.
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A loop of area 4 cm2 has its plane parallel to the field lines of the magnetic field, B = 0.6 T, as shown in the figure. The loop is pulled in the opposite direction of the field with a constant velocity of v = 6 m/s.

http://img375.imageshack.us/img375/4139/picbs3.th.jpg

A. What is the induced voltage?
B. Use a coil of 10 loops. Now the loop is turned so that it’s plane is perpendicular to the B field and then the coil is pulled out of the field. What is the induced voltage?

I know how to do part A, it's just zero. But I have tried many formulas with part B but cannot get an answer, because there is always some variable that is unknown. Can someone help me?
 
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Do you know Faraday's law? induced voltage equals change in flux over time
 
cryptoguy said:
Do you know Faraday's law? induced voltage equals change in flux over time

yah. but time is not given.
 
It seems to me that part B can be interpreted in one of two ways:

either

1) they want an average induced voltage, in which case you would just find the amount of magnetic flux through the loop initially (just before it leaves the field), figure out how long it takes the loop to completely clear the field (introductory physics problem fields have sharp edges), and find the average rate of change in the flux during that interval (you will come up with a number in volts);

or

2) you are to find a function of time for the induced voltage, which will be zero just before the loop begins to exit the field, zero again once it completely leaves, and something related to how the area of the circle still inside the field varies with time; this will require a little geometric consideration. (My personal suspicion is that this is what they want; you have enough information to construct the function.)
 
dynamicsolo said:
It seems to me that part B can be interpreted in one of two ways:

either

1) they want an average induced voltage, in which case you would just find the amount of magnetic flux through the loop initially (just before it leaves the field), figure out how long it takes the loop to completely clear the field (introductory physics problem fields have sharp edges), and find the average rate of change in the flux during that interval (you will come up with a number in volts);

or

2) you are to find a function of time for the induced voltage, which will be zero just before the loop begins to exit the field, zero again once it completely leaves, and something related to how the area of the circle still inside the field varies with time; this will require a little geometric consideration. (My personal suspicion is that this is what they want; you have enough information to construct the function.)

how would you find the time? I'm a bit confused.
 
If you are to find an average value for the voltage, keep in mind that the loop is circular (according to the diagram), you are given an area for it, and you know the speed at which it is moving. From this, you can find the time the loop requires to escape the field.

If you are to construct a function, this time will tell you when the induced voltage goes back to zero. Call time t = 0 the moment the loop begins to leave the field. The loop moves at a constant speed, so you will need to figure out how much of the circle is still inside the field as it progresses across the "edge". At the time calculated for the loop to be moved out, there is no longer any flux through the loop.
 
dynamicsolo said:
If you are to find an average value for the voltage, keep in mind that the loop is circular (according to the diagram), you are given an area for it, and you know the speed at which it is moving.

Okay. So I am thinking that I take the 4 cm^2 then using the area of a circle: A=pir^2 I will figure out the radius. Then times that by 2 to get the diameter. After that convert cm to m. Then divide that by 6 to get the time?
 
i actually thought its more like #1 where you can find the time by dividing the diameter of the circle (which can be figured from the area) by 6 (which is the speed of the loop as it exits the field).
 
pooka said:
Okay. So I am thinking that I take the 4 cm^2 then using the area of a circle: A=pir^2 I will figure out the radius. Then times that by 2 to get the diameter. After that convert cm to m. Then divide that by 6 to get the time?

Sounds right and you should have found a rather short time interval, which was...?
 
  • #10
dynamicsolo said:
Sounds right and you should have found a rather short time interval, which was...?

I got 0.0037 s.
 
  • #11
cryptoguy said:
i actually thought its more like #1 where you can find the time by dividing the diameter of the circle (which can be figured from the area) by 6 (which is the speed of the loop as it exits the field).

That's true, though the induced voltage in the loop is going to be variable during that interval. I don't know what level of course this problem is intended for, but in a calculus-based course, you would be expected to work out the flux variation as the loop leaves the field and find an induced voltage function. If this isn't that kind of course, they may just want an average value...
 
  • #12
Okay so there's also part C to this problem:
If you use B = 0.6 T and the area of the coil = 4 cm^2, with the coil of 10 loops, but now the loop is rotating in the B field at a speed of 4 revolutions per second, what is the induced voltage as it rotates through one revolution?

For part C, do you just use the diameter and divide is by 4?
 
  • #13
pooka said:
I got 0.0037 s.

So the flux through the loop will go from its value when the loop is completely inside the field down to zero in that interval. That would let you find the average rate of change in flux. (Don't forget that the loop has 10 turns in it when you go to find the induced voltage...)
 
  • #14
pooka said:
Okay so there's also part C to this problem:
If you use B = 0.6 T and the area of the coil = 4 cm^2, with the coil of 10 loops, but now the loop is rotating in the B field at a speed of 4 revolutions per second, what is the induced voltage as it rotates through one revolution?

For part C, do you just use the diameter and divide is by 4?

The fact that part C asks for this is making me think they do want a function in part B. This problem is three-dimensional now: the coil is spinning around inside the field, so it is presenting a constantly varying face to the field lines. If you've been given the definition for flux as a dot product, you know it's flux = B · A. Here, neither B nor A is changing, but the angle between the field and the line perpendicular to the "face" of the loop is changing continually (at a constant rate). Is your physics course calculus-based?
 
  • #15
dynamicsolo said:
The fact that part C asks for this is making me think they do want a function in part B. This problem is three-dimensional now: the coil is spinning around inside the field, so it is presenting a constantly varying face to the field lines. If you've been given the definition for flux as a dot product, you know it's flux = B · A. Here, neither B nor A is changing, but the angle between the field and the line perpendicular to the "face" of the loop is changing continually (at a constant rate). Is your physics course calculus-based?

No it's not calculus based.
 
  • #16
Have you been given the definition of flux as

\phi_{B} = BA sin\theta ?
 
  • #17
dynamicsolo said:
Have you been given the definition of flux as

\phi_{B} = BA sin\theta?
<br /> <br /> sry i can't read the equation you just wrote
 
  • #18
So if the coil is rotating at 4 revs per second, what is the time needed for the flux to go from maximum to 0? keep in mind, one revolution of the coil will have 2 positions of max flux
 
  • #19
cryptoguy said:
So if the coil is rotating at 4 revs per second, what is the time needed for the flux to go from maximum to 0? keep in mind, one revolution of the coil will have 2 positions of max flux

Does this mean I divide the diameter by 8 then?
 
  • #20
Diameter doesn't enter into this part. The coil is spinning about a diameter, so the coil is presenting a constantly varying area to the field.

I'll try explaining this without calculus. At time t = 0, say, the coil is face-on to the field, so the flux is BA. As it turns, the angle the "face" of the loop makes to the field rotates. A quarter-turn later, it is sideways to the field and the flux is zero, as you said for part A. As the coil continues to turn, the area presented to the field increases again, reaching a maximum again at half-a-turn. The next half-a-turn repeats this (in reverse). So the flux through the coil varies like a sine or cosine function. If you think about where the change in flux is greatest, that will be around the one-quarter and three-quarter marks in the rotation. At the start and the halfway point, the area presented to the field is changing very little. So we can say that the rate of flux change is proportional to

sin(kt),

k being the rate at which the coil spins. For 4 revs per second, this will be

k = (4 revs)·(2 \pi radians/rev) / (1/4 second period)<br /> = 32 \pi radians/second.

You can also see that the flux change will be faster if the coil rotates faster. So the rate of flux change, which gives, the induced voltage, will be

k · N · B · A · sin (kt) ;

that is, the induced voltage varies like a sine function. (This is related to the way alternating current is made.) You have all the values for these factors, so you will have the function of time for the induced voltage. (And a check of the units will show that this does come out in volts.)
 
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  • #21
dynamicsolo said:
Diameter doesn't enter into this part. The coil is spinning about a diameter, so the coil is presenting a constantly varying area to the field.

I'll try explaining this without calculus. At time t = 0, say, the coil is face-on to the field, so the flux is BA. As it turns, the angle the "face" of the loop makes to the field rotates. A quarter-turn later, it is sideways to the field and the flux is zero, as you said for part A. As the coil continues to turn, the area presented to the field increases again, reaching a maximum again at half-a-turn. The next half-a-turn repeats this (in reverse). So the flux through the coil varies like a sine or cosine function. If you think about where the change in flux is greatest, that will be around the one-quarter and three-quarter marks in the rotation. At the start and the halfway point, the area presented to the field is changing very little. So we can say that the rate of flux change is proportional to

sin(kt),

k being the rate at which the coil spins. For 4 revs per second, this will be

k = (4 revs)·(2 \pi radians/rev) / (1/4 second period)<br /> = 32 \pi radians/second.

You can also see that the flux change will be faster if the coil rotates faster. So the rate of flux change, which gives, the induced voltage, will be

k · N · B · A · sin (kt) ;

that is, the induced voltage varies like a sine function. (This is related to the way alternating current is made.) You have all the values for these factors, so you will have the function of time for the induced voltage. (And a check of the units will show that this does come out in volts.)

I think I am beginning to understand, so V=kNBA sin (kt)
so plugging all this in V=(32)(10)(0.6)(4*10^-4)sin(32t)

but what is t? Is it one, because if it is like a sine wave, one revolution is from 0 to 2pi? or is it 2pi?
 
  • #22
The voltage in this part (and in part B, really...) are functions of time: the voltage in these situations will not have a constant value, but will depend on what time it is. So the result for part C is expressed as

V=(8\pi)(10)(0.6)(4*10^{-4})sin(8\pi t) volts.

(And I wrote that wrong earlier -- shouldn't do replies at 2 AM -- the value for k will be (4 rev/sec)(2\pi radians/rev) = 8\pi radians/sec.)

For the time, one revolution would require the time, say, from t = 0 to t = 1/4 sec., so the argument in the sine function (the number in parentheses) goes from 8(pi)·0 = 0 to 8(pi)·(1/4) = 2(pi) . The induced voltage (and current) will vary continually, following a sine curve-type graph.
 
  • #23
thank you so much for helping! =)
 
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