# The emf of a string in a circle - magnetic fields

1. Feb 5, 2012

### axcelenator

1. The problem statement, all variables and given/known data
there is a cylindrical area in xy plane with a magnetic field (into the plane) that changes with time like that: (dB/dt)=δ
the magnetic field outside the area is 0. the radius of the cylinder is a.
we pick a string in the circle with lentgh L which is smaller than the diameter(2a). let's call the points that connect the string - A and B.
I have to find the emf between those points.

its looks like that:

http://img714.imageshack.us/img714/7092/emfk.jpg [Broken]

the answer is -(Lδ/2)√[(a^2)-(L/2)^2]
please show me how can I get it. thanks

Last edited by a moderator: May 5, 2017
2. Feb 5, 2012

### axcelenator

I got it but can't understand something: if i take the inner loop that creates the triangle I'll get the answer. But, if i take the loop that goes from A to B via the arc of the circle the EMF will be other. why?(i understand that the Elec field is nonconservative)
thanks

3. Feb 5, 2012

### vela

Staff Emeritus
You need to show us your calculations.

4. Feb 5, 2012

### axcelenator

Hey Vela, the original question goes like that: there is a conducting rod between A and B and a resistor R connected to it: http://img692.imageshack.us/img692/2079/faraday.jpg [Broken]

Now. in the first question(when there is no resistor and no conductor) I'm asked to find only the EMF between the points.
I did it like that: i picked the loop that created the triangle and so: EMF=-(d/dt)∫ ∫B*dS=
-δ*(territory of the traingle)=-(Lδ/2)√[(a^2)-(L/2)^2]
(that is the right answer by the teacher)

In the second question the resistor connected to the conductor. What is the current via the resistor?? so I took now the area under the triangle and the new calculation of the flux, EMF and in the end the current gave me:
I=(-δ/R)*{[(a^2)(arcsin(L/2a)]-[(L/2)√[(a^2)-(L/2)^2]]}
(that is the right answer by the teacher)
Now the problem is like that: the teacher asks why can't you say that the current is not:
I=EMF/R=-(Lδ/2R)√[(a^2)-(L/2)^2] ??
Thanks

Last edited by a moderator: May 5, 2017
5. Feb 5, 2012

### vela

Staff Emeritus
That's a good question. What do you think? You're probably assuming there's no emf induced in the wires connecting the resistor to the rod. Is that true?

6. Feb 5, 2012

### axcelenator

emmmm that's nice.. (-:
I didn't assume like that but I thought:(in the first question) if i took a smaller triange with a smaller area so the flux is smaller too my EMF would be smaller.... -->so if i take a loop with an area that limits 0 I would get EMF≈0... LOL (-:
Where is my mistake sir?

7. Feb 5, 2012

### vela

Staff Emeritus
The EMF around the entire loop would be smaller, but in the first part, you're only interested in the EMF along the rod. That wouldn't change because you chose a different loop, but the total EMF around the loop would be affected by the EMF induced on the other parts of the loop, which would change if you modified the loop.

8. Feb 5, 2012

### axcelenator

i cant understand why the current of R is not the EMF/R - (emf of the triangle)

9. Feb 6, 2012

### axcelenator

Alright now I understand it: the emf of the triangle is the emf of the rod only becaude the Electric field on the radius is perpendicular to it and so E*dl = 0.

And in the second question the E*dl on the wires and the resistor is not 0 so I get another EMF across the loop!

10. Feb 6, 2012

### vela

Staff Emeritus
Right!