# EMF between the rim and centre of the disc

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1. Oct 19, 2016

### moenste

1. The problem statement, all variables and given/known data
A copper disc of radius 10 cm is situated in a uniform field of magnetic flux density 1.0 * 10-2 T with its plane perpendicular to the field.

The disc is rotated about an axis through its centre parallel to the field at 3.0 * 103 rev min-1. Calculate the EMF between the rim and centre of the disc.

2. The attempt at a solution
Attempt 1

E = B A N ω, where ω = 2 π / T, where T = 1 / f.

f = 103 rev min-1 / 60 = 16.7 rev s-1.

ω = 2 π / 16.7 = 104.9 rad s-1.

E = 10-2 * (0.1 * 0.1) * 104.9 = 0.01049 V.

Attempt 2
I also used another formula: E = f B π (r22 - r12) where r2 = radius of rim, r1 = radius of axle.

E = 16.7 * 10-2 * π * 0.12 = 5.2 * 10-3 V.

What am I missing?

2. Oct 19, 2016

### cnh1995

Are you familiar with differential equations and integration?

3. Oct 19, 2016

### moenste

Like x2 + 3 x + 7 = 2 x + 3? How is it applied here?

4. Oct 19, 2016

### cnh1995

Not exactly. But first think about the physics behind this.
Assume any radius of the disc to be a wire. Erase the rest of the disc. Now you'll have a conductor of length r=10cm rotating about a point (center of the disk) in a magnetic field B. What is the emf induced in this wire? What is the formula for motional emf?
You can see that you can consider the disk as infinite number of such conductors. But the voltage induced in a single conductor will be same as the voltage between center and rim of the disk. All the points on the rim will be equipotential. So all you need to do is compute the voltage induced in the single conductor. What is the formula for motional emf when a conductor of length l is moving in a magnetic field B with velocity v?

Last edited: Oct 19, 2016
5. Oct 19, 2016

### moenste

E = B L v?

6. Oct 19, 2016

### cnh1995

Right. Is v constant along the length of the conductor (radius of the disk)?

7. Oct 19, 2016

### moenste

Doesn't it depend on the problem? I mean if it says "the velocity is X" then it's constant.

Update:
E = B L v = B r f = 10-2 * 0.1 * 16.7 = 0.0167 V. That simple?

8. Oct 19, 2016

### cnh1995

You have angular velocity. But you need linear velocity for emf. Is the linear velocity constant along the radius?

9. Oct 19, 2016

### moenste

Like this?

10. Oct 19, 2016

### cnh1995

No.
E=Blv where v is constant along the length of the conductor. Here, since the conductor is revolving, v is not constant along its length but is a function of its distance from the center of the disk. That's where calculus comes into picture.

11. Oct 19, 2016

### moenste

We know radius and angular velocity. So we can find regular velocity: ω = v / r → v = ω r = 104.9 * 0.1 = 10.49 m s-1. Correct?

12. Oct 19, 2016

### cnh1995

That is the linear velocity at the rim.
You need to consider all the linear velocities along the radius. Consider an infinitesimally small length of the wire of length 'dr' at a distance r from the center of the disk. What is the emf induced in that infinitesimal portion of the wire?

13. Oct 19, 2016

### moenste

At the edge of the disc?

Don't know how to find it. I only have this formula in my book that has radius in it: E = f B π (r22 - r12) where r2 = radius of rim, r1 = radius of axle.

14. Oct 19, 2016

### cnh1995

Yes.
What is the linear velocity of an infinitesimal element of the wire of length 'dr' at a distance r from the center? You know ω is constant throughout.

15. Oct 19, 2016

### moenste

v = ?
Lwire = dr
r = ?

I don't see any relationship.

Maybe a = v2 / r → v = √v r?

16. Oct 19, 2016

### cnh1995

Good.
Since the length of the wire is infinitesimally small, you can treat it as a point (dr→0). So, what is the linear velocity of the infinitesimal length element at a distance r from the center of the disk? How will you express v as a function of its distance 'r' from the center? Hint: Angular velocity is constant along the length.

17. Oct 19, 2016

### moenste

v = ?
L = dr = 0
Isn't L = r? Isn't the length equal to radius?

v = ω r = 0?

18. Oct 19, 2016

### cnh1995

v=rω , dr→0 but dr≠0. This is one of the basic concepts in calculus. Look up 'limit of a function'.
So, you have velocity of the length element v=r*ω and length of the element=dr. What is the emf induced in this infinitesimal length element?

Last edited: Oct 19, 2016
19. Oct 19, 2016

### moenste

E = B L v and since dr tends to zero but doesn't approaches it, therefore E also tends to zero?

20. Oct 19, 2016

### lychette

another approach is to use induced emf = rate of change of flux linkage = area swept out per second x B
Area swept out by a radius of 0.1m = π x 0.12... 50 rotations per second and B= 1 x 10-2....can you complete this calculation?

In your original post I think you have mis- read the revs....3 x 1000 not 1000