# Homework Help: Faraday's Law - very tricky question

1. Jun 7, 2013

### assaftolko

1. The problem statement, all variables and given/known data
A straight infinitly long wire with a current of I upwards is stationary. A rectangular loop with length a and height b is moved to the right with constant speed of v across the wire in the plane of the paper. At t=0 the right edge of the loop is exactly at the same location of the wire.

2. Relevant equations
What is the flux through the loop as a function of the time?
What is the induced emf as a function of the time?

3. The attempt at a solution
Well I have 2 main problems here:
1. At any moment up untill the time that all the loop has passed to the right of the wire I'm facing the fact that some part of the loop (the part that is to the right of the wire) is under the influence of a magnetic field in the direction which is into the paper, and another part (the part that's to the left of the wire) is under the influence of a magnetic field in the direction which is out of the paper - how can I deal with this problem in terms of my integration?
2. If I want to switch the integral from dA to dr (where r is the distance from the wire) - I have again one part of the loop that is coming towards the wire while another part of the loop is moving away from the wire (At least until the whole loop passes through the wire) - How can I deal with this fact when I try to write the integration limits?

This is a bonus question and isn't mandatory but it pisses me off!

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2. Jun 7, 2013

### tiny-tim

hi assaftolko!

choose a direction (eg, into the paper), then flux the same way is positive and flux the opposite way is negative

so just subtract one from the other!

3. Jun 7, 2013

### assaftolko

I know but I don't get how to really do the calculus here... because the two fluxes are present in the same time... I'm quite lost

4. Jun 7, 2013

### tiny-tim

ok, you calculate flux L(t) through the left-hand part at time t, and R(t) through the right-hand part

then your total flux at time t is F(t) = L(t) - R(t), and that's what you integrate

5. Jun 7, 2013

### assaftolko

Ok I tried something tell me if so far so good: I said that dA is in the out of the page direction:

For the flux in the left area:

∅L(t) = ∫$\vec{B}$*$\vec{dA}$ = ∫μ0I/2∏r*dA*cos0 = μ0I/2∏∫$\frac{1}{r}$dA

But dA=b*dr and r itself is a function of time, and at t=0 we get (with respect to the left edge of the loop) that r=a, and the final location of the left edge when we deal with the left side flux is r=0 and so:

∅L(t)=μ0Ib/2∏∫$^{0}_{a}$$\frac{1}{r}$*-dr

-dr and not dr because r is getting smaller as the loop moves towards the wire.

How am I so far?

6. Jun 7, 2013

### tiny-tim

at each time t, you have to integrate across the whole loop

your "r" is the distance from the wire

if the wire is at position w(t) at time t,

then you have positive flux on the left from r = 0 to r = …

and negative flux on the right from r = 0 to r = … ?

(btw, once you've found what the limits are, there's a fairly obvious way of simplifying the integral)

7. Jun 7, 2013

### assaftolko

well I guess you meant to write r(t) and not w(t) - I think that if r=0 is the location of the wire then r=a-vt is the location at t=0, and for the right part we go from r=0 to r=vt?

8. Jun 7, 2013

### tiny-tim

no, you're confusing r(t), which is a function of t and is the position of the wire is at time t

with r(x,t), the distance of a general point x (0 ≤ x ≤ a) from the wire

since your flux formula already uses r for the second one, i've used w(t) for the first one, and r(x,t) = |x - w(t)| for the second one

(now, to find the flux at time t, integrate over x, with t constant)

9. Jun 7, 2013

### assaftolko

How is r(t) the position of the wire at time t? the position of the wire is constant... it's the loop that moves. I still don;t get what you're trying to tell me :(

10. Jun 7, 2013

### tiny-tim

what difference does it make, which one moves?!!

you have to measure r (in the formula) from the position of the wire, and it goes up to the two sides

11. Jun 7, 2013

### assaftolko

Ok but aren't the limits I've written in my before last post correct? from r=0 to r=vt to the right and from r=0 to r=a-vt to the left (or maybe r=vt-a to the left because this position is at the negative side of the r axis?)?

12. Jun 7, 2013

### tiny-tim

ah, that's not what you originally wrote

yes that's correct now

ok, now do the integration!
no, it's very confusing if you let r be negative … it makes it too easy to make mistakes

it's best to keep r positive, even if that means writing two integrals instead of one

13. Jun 7, 2013

### assaftolko

Ok but what I don't get is this - if some of my integration limits are 0 - then we'll end up at some point with ln(0) which is a big no-no... also - I still don't get why it's not correct to say that for the right part we start from r=a (I'm looking at the left edge of the loop with respect to the wire) at t=0 and r=0 at t=the moment the left edge of the loop passes through the wire, and similarly to say that for the right part we start from r=0 (I'm now looking at the right edge of the loop with respect to the wire) and end up at r=a at t=the moment the left edge of the loop passes through the wire. Why is it critical to represent r as a function of t?

14. Jun 7, 2013

### tiny-tim

true

but the smaller part will cancel out with some of the larger part, so you won't need to bother with 0
sorry, i don't understand this

the flux is a function of t

i'm not even sure what you mean by "r"

the r in your basic formula is the distance frrom the wire to a general point inside the loop, and that distance changes with t

15. Jun 7, 2013

### assaftolko

Yes it changes with time - but if I integrate over dr, and I look at a fixed point in the loop - like it's left edge, then I can say that at t=0: r=a and because the loop moves to the right, r is getting (for the left edge of the loop) smaller and smaller until the left edge of the loop passes through the wire (and then r=0) and from this point on there isn't any area of the loop that is left to the wire, and so there is no flux from here on to the left of the wire. Why isn;t this representing the total flux through the area to the left of the wire?

16. Jun 7, 2013

### tiny-tim

yes, but
i don't understand

17. Jun 7, 2013

### assaftolko

I want to write an expression that represents the flux from the area to the left of the wire (the wire being this rectangular area's right edge) and the total flux from the area to the right of the wire - why aren't I doing just that when I integrate dr from r=a to r=0 (let's just focus on the left flux for now)?

18. Jun 7, 2013

### assaftolko

Ok I think I got it - if the loop wasn't moving I think that my integration limits would be correct...

19. Jun 7, 2013

### tiny-tim

but r doesn't go from a to 0 (except at t = 0)

20. Jun 7, 2013

### assaftolko

Yep exactly - if the loop wasn't moving my limits are correct I think

21. Jun 7, 2013

### assaftolko

Well something still bothers me - if I integrate for the right part from 0 to vt, and we know that the position of the right and left edges of the loop changes according to time - won't we get that for a moment in which the left edge of the loop is a distance 2m from the wire (for example), we integrate also over an area which the loop isn't present? I mean from 0 to 2m we don't have any part of the loop but we still integrate from 0 to vt... If we had a similar problem where the problem starts when the loop is a distance c to the right of the wire (and still moves with constant speed v to the right) - it's quite clear the integration limits would be c+vt (bottom) and c+vt+a (upper) - in this way we represent that for every time we want to integrate over the edges of the loop we have to consider that these 2 edges (the left and the right) change position according to the moment we've chosen to look at the loop. But if we get back to our problem - how can we say we're doing this by choosing the lower integration limit to be 0? It's quite clear that after some time the loop will be completely to the right of the wire and will be seperated from the wire by some growing distance, but we want to integrate over the loop for every moment in time... so if I take this moment in time I can't say that the bottom limit of integration for the loop is r=0...

Last edited: Jun 8, 2013
22. Jun 8, 2013

### tiny-tim

yes, that's correct, if the loop is entirely to the right, then the flux has the same sign through the entire loop, and there's only one integral, from r = vt to r = vt+a

(i thought you were only worried about when the flux changed sign? )

23. Jun 8, 2013

### assaftolko

That's true also :)
But I thought from their solution next to the question that a single expression represents the flux at any given time, and now I understand it only represents the flux from t=0 to t=the moment the entire loop passes to the right of the wire.

I'll write my calculations here briefly and I'll be happy if you'll give me your feedback. Thanks a lot!

24. Jun 8, 2013

### assaftolko

Let dA be positive in the $\otimes$ direction:

1. from t=0 to t=a/v the loop moves through the wire to the right - let ∅$_{L}$(t) be the flux from the left (bounded by the loop's left edge and the wire) and let ∅$_{R}$(t) be the flux from the right (bounded by the wire and the loop's right edge):

∅$_{L}$(t)=∫$\vec{B}$$\bullet$$\vec{dA}$ = ∫$\frac{μ_{0}I}{2∏r}$dAcos180 = -$\frac{μ0I}{2∏}$∫$\frac{1}{r}$*b*-dr = $\frac{μ0Ib}{2∏}$∫$^{d}_{a-vt}$$\frac{1}{r}$dr = $\frac{μ0Ib}{2∏}$*ln($\frac{d}{a-vt}$)

where d is an infitisimaly small distance left to the wire. We write -dr and not dr because r is getting smaller for the flux to the left of the wire, as the loop moves to the right.

∅ $_{R}$(t) = $\frac{μ0Ib}{2∏}$∫$^{vt}_{d}$$\frac{1}{r}$*dr = $\frac{μ0Ib}{2∏}$ln($\frac{vt}{d}$)

This time the angle between B and dA is 0, so cos0=1 and dr is positive becuase r increases to the right of the wire as the loop passes through it. d is the same infitisimaly small distance from the wire, this time to the right of the wire.

∅$_{L}$+∅$_{R}$ = $\frac{μ0Ib}{2∏}$*[ln(d)-ln(a-vt)+ln(vt)-ln(d)] = $\frac{μ0Ib}{2∏}$*ln($\frac{vt}{a-vt}$)

That's the same answer that's written in the question, only by a minus. Is this because of my choice of dA? Because if I would have chosen it to be positive in the other direction I'm quite sure I would get the answer as it's given in the question...

2. from t=a/v and on the whole loop is to the right of the wire and so:

∅(t) = ∫$\vec{B}$$\bullet$$\vec{dA}$ = ∫$^{r(t-t0)+a}_{r(t-t0)}$$\frac{μ0Ib}{2∏r'}$*dr'

where t0 is the time it takes the whole loop to pass through the wire, which means t0=a/v and we get:

∅(t) = $\frac{μ0Ib}{2∏}$∫$^{vt}_{vt-a}$$\frac{1}{r}$*dr = $\frac{μ0Ib}{2∏}$*ln($\frac{vt}{vt-a}$)