# EMF induced in a circuit

1. May 20, 2013

### LilyY

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

See attachment.

2. Relevant equations

3. The attempt at a solution

For the first part: (B' and x' are B subscript 0 and x subscript 0 respectively.)

$\phi$=BA=B'xl=B'l(L+x'cos(ωt))
EMF=-d$\phi$/dt=-B'ωlx'sin(ωt)
Therefore V=-EMF=B'ωlx'sin(ωt)
So I=V/R=B'ωlx'sin(ωt)/R

For the second part the resistance is now R+α(2x+l)=R+α(2(L+x'cos(ωt))+l)
But I don't know how to get the required expression. I don't know any circuit theory other than V=IR, so maybe that's why I'm having trouble.

Thanks in advance for helping! :-)

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

2. Relevant equations

3. The attempt at a solution

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2. May 21, 2013

### BruceW

Good so far. I agree with I=V/R=B'ωlx'sin(ωt)/R although in your working, I think the sign is the wrong way around EMF=-dϕ/dt=-B'ωlx'sin(ωt) here, there is a negative, but you also just differentiated a cosine, so the negatives cancel, giving a positive. But you have used V=-EMF to give the positive, which is not a rule that I know of... Or maybe you are just saying that you want the magnitude of the voltage, and don't worry about the direction? yeah, to be honest, you can always work out the direction by Lenz' law, so it is not that important to remember the sign in the equation.

Anyway, you get the right answer. And for the second part, you have the correct idea. You have worked out the value of the resistance as a function of time, and you have the voltage as a function of time, so now just use your V=IR equation to get I. Yes, it is as easy as that :)

Edit: ah, once you get 'I', then you need to think a little bit about what will be the voltage between A and B. Once you have 'I', there is not much more calculation you need to do. just thinking really. p.s. welcome to physicsforums :)

Last edited: May 21, 2013
3. May 21, 2013

### LilyY

Hi Bruce, thanks so much for your help.

So I've found that the new resistance is R+α(2(L+x'cos(ωt))+l), and the EMF is -B'ωlx'sin(ωt). So I=V/R=-B'ωlx'sin(ωt)/(R+α(2(L+x'cos(ωt))+l)). This implies that V=IR=-B'ωlx'sin(ωt), which is the original EMF, not the answer given. Where am I going wrong?

4. May 21, 2013

### Yukoel

Hello LilyY,
I think you mean R'=R+α(2(L+x'cos(ωt))+l)(effective resistance of the whole circuit)and I=V/R'=-B'ωlx'sin(ωt)/(R+α(2(L+x'cos(ωt))+l)).
I suppose you are trying to measure potential difference between the ends of rod right?In other words you are trying to find the V across a single element R in the circuit.How will you get that ?

Regards
Yukoel

5. May 21, 2013

### LilyY

Hi Yukoel,

Thanks for your reply- I can now see that the potential difference across the slider is the expression given in the question. However, why wouldn't it also be correct to find V along the U shaped wire? This would give a different expression, i.e. R'-R would replace R in the numerator.

Now I'm a bit confused. I thought that potential difference was path independent? Sorry if this is a dumb question; I haven't studied much circuit theory.

Lily

6. May 21, 2013

### BruceW

You use the resistance of the loop to find 'I' (which you have done correctly). And the voltage across AB is not due to the emf of the entire circuit. You need to use a different resistance to get the voltage across AB. Hint: you can think of AB as being parallel to the slider.

Also, I think your original answer of B'wlx'sin(wt) was correct. It is just that I think you made two sign errors in your method. 1)-dϕ/dt=-B'ωlx'sin(ωt) was not correct, because when you differentiated cosine, this gives a minus. 2) V=-EMF I don't know why you did this. surely just V=EMF

7. May 21, 2013

### LilyY

Hi Bruce,

Thanks for your reply. Would you mind having a look at my post just above yours? I'm confused about whether or not V is path independent.

Lily :-)

8. May 21, 2013

### BruceW

ah, yeah that's a good question. there are two definitions of 'voltage' in circuit problems:
$$\int \vec{E} \cdot d \vec{l}$$
Which is the 'proper' way to define change in electrical potential. And this certainly is path independent. But then there is also the definition:
$$\int (\vec{E} + \vec{v} \wedge \vec{B}) \ \cdot d \vec{l}$$
This takes account of the magnetic field and v is the velocity of the wire. And this definition is path dependent. I think this definition is more commonly called emf. In the question, they ask "derive the voltage V appearing across AB" They want you to use the second definition. If you use the first definition, the answer is simply zero because there is no electric field in this problem. So really, if they used better terminology, they would have used the word emf instead of voltage. But these get used interchangeably, so it is one of those things you just need to be careful of.

edit: actually, the first definition is only path independent for situations with a magnetic field that does not change with time (which is what we have in this problem).

2nd edit: another way to think about the problem is that induction is happening. and generally in circuits, you can think of the sum of voltage changes around a closed loop as being equal to the induction in that loop. (but you need to be careful about the sign here). Anyway, you have the right answer.

Last edited: May 21, 2013