# Homework Help: What is the induced voltage?

1. Jul 5, 2008

### pooka

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 [Broken]

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?

Last edited by a moderator: May 3, 2017
2. Jul 6, 2008

### cryptoguy

Do you know Faraday's law? induced voltage equals change in flux over time

3. Jul 6, 2008

### pooka

yah. but time is not given.

4. Jul 6, 2008

### dynamicsolo

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.)

5. Jul 6, 2008

### pooka

how would you find the time? i'm a bit confused.

6. Jul 6, 2008

### dynamicsolo

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.

7. Jul 6, 2008

### pooka

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?

8. Jul 6, 2008

### cryptoguy

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).

9. Jul 6, 2008

### dynamicsolo

Sounds right and you should have found a rather short time interval, which was...?

10. Jul 6, 2008

### pooka

I got 0.0037 s.

11. Jul 6, 2008

### dynamicsolo

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. Jul 6, 2008

### pooka

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. Jul 6, 2008

### dynamicsolo

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. Jul 6, 2008

### dynamicsolo

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. Jul 6, 2008

### pooka

No it's not calculus based.

16. Jul 6, 2008

### dynamicsolo

Have you been given the definition of flux as

$$\phi_{B} = BA sin\theta$$ ?

17. Jul 6, 2008

### pooka

sry i cant read the equation you just wrote

18. Jul 6, 2008

### cryptoguy

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. Jul 6, 2008

### pooka

Does this mean I divide the diameter by 8 then?

20. Jul 6, 2008

### dynamicsolo

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

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

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.)

Last edited: Jul 6, 2008
21. Jul 6, 2008

### pooka

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. Jul 7, 2008

### dynamicsolo

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. Jul 7, 2008

### pooka

thank you so much for helping!! =)