# Induction slider

1. May 23, 2014

### fayled

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

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2. May 23, 2014

### 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.

3. May 23, 2014

### fayled

Unfortunately that is what I was thinking of...

4. May 23, 2014

### rude man

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?

Last edited: May 23, 2014
5. May 23, 2014

### fayled

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.

6. May 23, 2014

### rude man

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'.

7. May 23, 2014

### fayled

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

8. May 23, 2014

### rude man

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.

9. May 23, 2014

### fayled

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

10. May 23, 2014

### rude man

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

11. May 23, 2014

### fayled

As in my personal answer disagrees with THE solution.

12. May 23, 2014

### rude man

So what IS THE solution?

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

13. May 24, 2014

### 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.

Last edited: May 24, 2014