# Homework Help: Faraday lenz excercise

1. Jun 11, 2009

### Dell

in the following question i am given an infinite straight wire with a current changing according to i=2t, next to the wire is a conducting square with a side of a=5cm at t=0 the square is at a distance of b=5cm from the wire and the frame is moving away from the frame at v=2m/s,
the frame has a resistance of R=2 Ω

http://lh6.ggpht.com/_H4Iz7SmBrbk/SjDX68M6InI/AAAAAAAABEg/thgSA-myiaI/Untitled.jpg [Broken]

what is the induced current in the frame at t=0.5s??

my problem here is that i would usually have a constant B field so i would find the flux, but now my field changes according to R, when R=b+v*t. and my current changes according to t, so can i do one single integral of t??

dΦ=AdB, = a2*$$\frac{μ*2t}{2π(b+v*t)}$$dt

but the problem now is that that integral would only take into account the field at the left hand side of the square,

so i thought maybe a double integral

∫dt∫a2*$$\frac{μ*2t}{2π(r)}$$dr where r goes from (b+v*t) to (a+b+v*t)

=a2μ/π * ∫tdt∫dr/r

= a2μ/π ∫t*ln($$\frac{(a+b+v*t) }{(b+v*t)}$$dt from 0 to 0.5

does this seem okay or am i totally off,

then once i find the flux at t=0.5

dΦ/dt=-I*R

I=-dΦ/(Rdt)

Last edited by a moderator: May 4, 2017
2. Jun 11, 2009

### Cyosis

I am confused about the problem statement, i=2t makes little sense because the units don't match. Could you explain this?

3. Jun 11, 2009

### Dell

every 0.5 second the current gets 1A bigger , i(t=0)=0, i(t=1)=2....

4. Jun 11, 2009

### Cyosis

Okay.

To obtain the flux you should not integrate over t. To obtain the flux we need to know dA. As you said yourself the magnetic field at a time t varies with a distance perpendicular to the wire, but not parallel to the wire. Therefore you can express dA=adr, with r the distance away from the wire.

The flux then becomes $\phi_B=\int B a dr$. You just need to find the correct boundaries and expression for B now.

5. Jun 11, 2009

### Dell

i tried this, didnt work though

dA=a*dr
B=μI/(2πr)

dΦ = BdA = aμI/(2πr)dr, ---> r=v*t ; dt=vdr

dΦ = aμI/(2vπr)dt ---> I=2*t

dΦ = aμt/(vπr)dt

dΦ/dt=aμt/(vπr) = -ε= -IinducedR

Iinduced=-aμt/(Rvπr) = -aμ/(Rv2π)

where have i gone wrong here?

i know you said not to integrate with 't' but i need to for dΦ/dt

6. Jun 11, 2009

### Cyosis

If r=vt, then dr=v dt for constant v, therefore dt=dr/v.

I don't think this approach will be very fruitful however. Could you calculate the magnetic flux at a time t=t_0?

7. Jun 11, 2009

### Dell

thanks for the help, i got it, onlt thing is my answer has an opposite sign to the correct one, but from what i understand the sign is not that important if you state the direction,,, this is what i wrote,

how will i know what direction the curren flows, i need to know if the flux is increasing or decreasing, and since the current is increasing AND the distance is increasing, (1 increases flux one decreases flux)

8. Jun 11, 2009

### Cyosis

A sign difference means that you're going in the opposite direction. You could of course calculate the magnitude of the emf and then explain in what direction it flows. What expression did you find for the flux?

9. Jun 11, 2009

### Dell

Φ=aμt/π*(ln(a+b+tv)-ln(b+tv))

my final answer came to
I=3.22997*10^-11 A and the answer is meant to be

I=-3.23*10^-11 A

10. Jun 11, 2009

### Cyosis

I just ran the numbers myself and your value is correct, however your sign is wrong. The emf is defined as $-\frac{d\phi}{dt}$. I have a feeling you only calculated $\frac{d\phi}{dt}$ and forgot to put the minus in front. You can see from the flux you calculated that the emf has to be negative as well, because dphi/dt>0. Have you been able to find the direction of the current?

Last edited: Jun 11, 2009
11. Jun 11, 2009

### Dell

Φ=aμt/π*(ln(a+b+tv)-ln(b+tv))

emf=-dΦ/dt = IR

I=-aμ/πR*{ln(a+b+tv)-ln(b+tv)+vt*[1/(a+b+tv)-1/(b+tv)]}

when i plug in the values i get 3.22997*10^-11 A

12. Jun 11, 2009

### Dell

if you say that the flux increases as t increases, then the current would have to be anti clockwise right??

13. Jun 11, 2009

### Cyosis

You can't say that based on the flux alone, you also need to know the direction of the magnetic field. I can't find that in the problem statement though, perhaps it's in the picture which isn't working for me.

14. Jun 11, 2009

### Dell

o, of course, the current is flowing upwards, therefore the field on the right hand side, where the square is is into the page, and my induced field must be opposite to that,,,

do you see anything wrong with my calculations that i cant get the minus??

15. Jun 11, 2009

### Cyosis

Are you using a pocket calculator or something? I copy pasted the formula for the emf you derived (which is the same one I derived) straight into mathematica and plugged in the given quantities. All I can think off is that you forgot a minus somewhere.

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