EMF between the rim and centre of the disc

[moenste]

Homework Statement

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.

Answer: 16 mV.

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?

 

[cnh1995]

Homework Statement

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.

Answer: 16 mV.

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?

Are you familiar with differential equations and integration?

 

[moenste]

Are you familiar with differential equations and integration?

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

 

[cnh1995]

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

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?

 

[moenste]

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?

E = B L v?

 

[cnh1995]

E = B L v?

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

 

[moenste]

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

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?

 

[cnh1995]

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

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

 

[moenste]

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

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

Like this?

 

[cnh1995]

Like this?

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.

 

[moenste]

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.

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?

 

[cnh1995]

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?

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?

 

[moenste]

That is the linear velocity at the rim.

At the edge of the disc?

You need to take 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 portion of the wire?

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.

 

[cnh1995]

At the edge of the disc?

Yes.

Don't know how to find it.

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.

 

[moenste]

What is the linear velocity of an infinitesimal element of the wire of length 'dr' at a distance r from the center?

v = ?
L_wire = dr
r = ?

I don't see any relationship.

Maybe a = v² / r → v = √v r?

 

[cnh1995]

Lwire = dr

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.

 

[moenste]

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.

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

v = ω r = 0?

 

[cnh1995]

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

v = ω r = 0?

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?

 

[moenste]

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?

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

 

[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

 

[cnh1995]

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

That would be a very easy way to do it. And @moenste, I am sure that's what you are expected to use. Since you have not studied calculus yet, I believe you should drop my method and go with this. Needless to say it gives the same result.Good luck!

 

[moenste]

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

You mean my second attempt is correct? (Aside from my 1000 instead of 3000 mistake.)

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.

E = f B π r² = [(3 * 10³) / 60] * 10^-2 * π * 0.1² = 0.0157 V or 16 mV.

 

[cnh1995]

E = f B π r2 = [(3 * 103) / 60] * 10-2 * π * 0.12 = 0.0157 V or 16 mV.

Right.
You can use calculus method in this way:
Velocity of the length element 'dr' at a distance r from the center=v(r)=r*ω.
So, infinitesimal emf induced in the length element
dE=B*dr*(rω)=Bωr*dr
Integrating this between r=0 to r=0.1 will give you the total emf induced along the radius.
E=∫Bωr*dr from r=0 to r=0.1
=B*ω*r²/2 from r=0 to r=0.1
=B*ω*0.1²/2.
Put the values of B and ω, you'll get the same answer.