Solve Induction Slider Homework: Find AB Voltage & Current

[fayled]

Homework Statement

An induction device is set up as shown below. It has a U-shaped conductor of negligible resistance along which a conducting slider of resistance R moves in the x direction. The assembly is immersed in a uniform magnetic flux density B of magnitude B₀ that is directed out of the page. The slider is moved such that its position with respect to the end of the U-shaped conductor is given by x=L+x0coswt with x₀<L.

I need to derive the voltage across AB and the current flowing around the U.

Then I need to find the voltage across AB when the U has a resistance per unit length alpha.

Homework Equations

Faraday's law. V=IR.

The Attempt at a Solution

The induced emf is B₀dx₀wsinwt in both cases. This is induced in the sliding rod.

Now comes my confusion.

In the first case, I believe we have a voltage rise across the rod, and a voltage drop of equal magnitude across the rod due to the resistance. These cancel so V=0. The current that flows is B₀lx₀wsinwt/R.

In the second case, we now have a voltage rise across the rod, some voltage drop due to a resistance R, and then around the U conductor a voltage drop due to a resistance of R'=α[d+2(L+x₀coswt)]. Thus I believe the voltage drop across A and B should be the induced emf minus the the circuit current times the resistance R, with the circuit current the induced emf over total resistance. However the answer is in fact only the second component, i.e just the current times R.

Thanks for any help in advance.

 

[rude man]

First case: correct.

Second case: think of the rod as a battery with internal resistance R and shorted by a wire of resistance alpha times the length of the U. You will get the right answer.

 

[fayled]

First case: correct.

Second case: think of the rod as a battery with internal resistance R and shorted by a wire of resistance alpha times the length of the U. You will get the right answer.

Unfortunately that is what I was thinking of...

 

[rude man]

Unfortunately that is what I was thinking of...

You short a battery with a wire of length L (not the L of the illustration) and resistance per unit length α, so the wire resistance is αL, and the battery's internal resistance is R, well, what is the current? What is the voltage drop across the internal resistance?

 

[fayled]

You short a battery with a wire of length L (not the L of the illustration) and resistance per unit length α, so the wire resistance is αL, and thebattery's internal resistance is R, well, what is the current? Whatis the voltage drop across the internal resistance?

Current is V/(R+aL) I believe. Voltage drop is then that multiplied by R. However we need the voltage between A and B and surely this involves a voltage gain due to the induced emf aswell?

Alternatively this pd should be path independent - if I found the p.d across the u circuit resistance it would give a different result.

 

[rude man]

Current is V/(R+aL) I believe. Voltage drop is then that multiplied by R. However we need the voltage between A and B and surely this involves a voltage gain due to the induced emf aswell?

Alternatively this pd should be path independent - if I found the p.d across the u circuit resistance it would give a different result.

You wrote that the correct answer is emf - Ri. Where emf = Bvd. That is correct. And the p.d. across the U section is αLi with i = Bvd/(αL + R) which is exactly the same as the drop across the rod. There is no 'different result'.

 

[fayled]

You wrote that the correct answer is emf - Ri. Where emf = Bvd. That is correct. And the p.d. across the U section is αLi with i = Bvd/(αL + R) which is exactly the same as the drop across the rod. There is no 'different result'.

Nope I said it was the drop across the resistance only despite me thinking otherwise... So I guess the answer isn't right.

 

[rude man]

Nope I said it was the drop across the resistance only despite me thinking otherwise... So I guess the answer isn't right.

That's not what you said in your first pst. I quote:

" Thus I believe the voltage drop across A and B should be the induced emf minus the the circuit current times the resistance R. "

Which is exactly correct.

 

[fayled]

That's not what you said in your first pst. I quote:

" Thus I believe the voltage drop across A and B should be the induced emf minus the the circuit current times the resistance R. "

Which is exactly correct.

When I say answer I mean the solution to the problem, I realize that may have been confusing?

Thanks for your help anyway.

 

[rude man]

When I say answer I mean the solution to the problem, I realize that may have been confusing?

Thanks for your help anyway.

That is the answer to the problem. The problem asks for the voltage across the rod.

 

[fayled]

That is the answer to the problem. The problem asks for the voltage across the rod.

As in my personal answer disagrees with THE solution.

 

[rude man]

As in my personal answer disagrees with THE solution.

So what IS THE solution?

OK, if their solution is v = iR then they're wrong and you are right.

 

[rude man]

fayled, I guess you've given up on this but I wanted to try to clarify by looking at it a bit differently (but the answer is unchanged):

In part (a) the "U" is just a shorting bar. Since no emf is generated around it, the voltage across the sliding rod has to be the voltage across the "U" which has zero resistance and so is zero volts.

In part (b) the "U" is no longer a short. The current is now i = emf/(R+αL) with L = length of the "U". So the voltage V across the U is i x the resistance of the "U" or V = emf/(R+αL) x αL. So this also has to be the voltage across the bar: V = iαL, not iR.

Think about it: Say you start with zero "U" resistance and so the rod voltage is zero. Now increase the U resistance αL slightly. Then, does it make sense that the rod voltage all of a sudden jumps to iR when i has changed just a tiny bit, from emf/R to emf/(R+αL)? Wheras V = iαL makes sense.

Before I said V = emf - iR but that amounts to the same voltage, remembering that i went down from emf/R to emf/(R+αL).

P.S. my L is your 2L + d.