# Homework Help: A spinning disc due to a magnetic field variation

1. Oct 11, 2013

### Gabriel Maia

Consider a charged wire with constant linear charge density λ. The wire has length 2πa and is attached to the edge of a disc with radius a. In the central region of the disc (a circular region of radius b<a) a constant magnetic field B is applied (perpendicular to the disc).

The magnetic field is then suddenly turned off. Obtain the torque due to the emf induced in the wire, in terms of ∂B/∂t.

From this result calculate the disc final angular momentum.

Now... when we suddently turn the magnetic field off a emf is induced in the wire. This emf is given by

ε=-πb^2∂B/∂t.​

It is the variation of the magnetic flux. Right? I suppose I could obtain the electric field associated to ε by taking its gradient. The B field is, however, say, in the z direction. How come it will generate a central force to rotate de disc?

Thank you.

2. Oct 11, 2013

### TSny

The electric field generated by the changing B field is not a conservative electric field. So, E is not the gradient of a potential. However, ε is related to the electric field via a line integral of E. Can you make this more precise by setting up the line integral? Think about the direction of the induced E field at the edge of the disc.

You do not need to generate a central force to rotate the disc. You need a torque. Can you see how the electric field at the edge of the disc will produce a torque? Again, you will need to be clear on the direction of E at the edge of the disc.

3. Oct 11, 2013

### Gabriel Maia

The Lenz Law tells us that an electric current will be generated in the wire in order to oppose the magnetic field variation. Therefore the electric field should be perpendicular to the radius of the disc. If the magnetic field is in the z direction and the disc lay on the xy plane, the electric field (and consequently the electric force) should also be on the xy plane. The force will be, as I said, perpendicular to the radius of the disc and therefore the torque will be in the z direction as the magnetic field.

I'm thinking about the expression for the torque, but is this vectorial analysis ok?

4. Oct 11, 2013

### TSny

Yes, the electric field will be perpendicular to the radius and tangent to the disc. You will not need to worry about the induced current in the wire. Just worry about the force that the electric field exerts on the linear charge density of the wire.

5. Oct 11, 2013

### Gabriel Maia

Let's see...

the emf is given by

ε=∫Edl

where dl is an element of the wire. It has the same direction of E. This is a line integral so we have that

ε=2πaE

where a is the radius of the disk. Now... ε is also wrote as minus the time variation of the magnetic flux. We arrive then at

ε=-πb^2∂B/∂t

where b is the radius of the circular area where the magnetic field is applied. The electric field is then

E=-b^2/2a*∂B/∂t

The total charge of the wire is q=2πaλ. The electric force is therefore

F=-λπb^2∂B/∂t

and the torque, at last should be

T=-λπab^2∂B/∂t

I believe that's it... is it?

6. Oct 11, 2013

### TSny

I believe your answer is correct. But, the derivation might be questioned a little. You wrote that the electric force on the wire is F=-λπb^2∂B/∂t. But, in fact, the total electric force is zero. Think about the direction of the force on each element of arc length of the wire and the result of adding all of the force elements (as vectors).

A better way is to consider the torque $\vec{d\tau}$ on each element of arc length. What is the direction of $\vec{d\tau}$? Are the directions of $\vec{d\tau}$ the same for all arc length elements? Express $d\tau$ in terms of $a$, $\lambda$, $E$, and $dl$. Then integrate $d\tau$ over all the arc length elements. Your expression should involve $\oint{E\;dl}$ which you can relate directly to ε.

Last edited: Oct 11, 2013
7. Oct 12, 2013

### Gabriel Maia

I see... my derivation is really flawed. Let me rethink it.

the torque of an element dl of the wire is

dT=a × dF

This electric force is

dF=dqE

where, we know, dq=λdl. We know the direction of the torque so we can write it as

dT=aλdlE

where dl = ad$\varphi$ . Integrating all over the wire gives us

T=2$\pi$a$^{2}$λE

Now let's focus on the electric field E. We know that its line integral is equal to minus the time derivative of the magnetic flux

$\oint$Edl=-$\pi$b$^{2}$$\frac{∂B}{∂t}$

E, however, varies with time, not with space. Therefore we obtain

E2$\pi$a=-$\pi$b$^{2}$$\frac{∂B}{∂t}$

E=-$\frac{b^{2}}{2a}$$\frac{∂B}{∂t}$

and then... finally

T=-$\pi$$\lambda$ab$^{2}$$\frac{∂B}{∂t}$

and a new aspect of the problem... the angular moment of the disc is

L=-$\pi$$\lambda$ab$^{2}$$\frac{∂^{2}B}{∂t^{2}}$

right?

But the angular moment is

L=I$\omega$

where I is the moment of inertia of the wire-disk configuration and $\omega$ is its angular velocity. The speed of the spinning disk is then

v=-$\pi$$\lambda$$\frac{a^{2}b^{2}}{I}$$\frac{∂^{2}B}{∂t^{2}}$

Last edited: Oct 12, 2013
8. Oct 12, 2013

### Gabriel Maia

Thank you.

9. Oct 12, 2013

### TSny

Your derivation of the torque now looks good! If you want, you can make a bit more "elegant" by noting that the integral over the torque is $\int{d\tau} = \oint{a\lambda E\;dl} = a\lambda\oint{E\;dl}=a\lambda\varepsilon$. So, you don't need to actually find an expression for E separately. But your derivation is good.

Now, your relationship between angular momentum and torque appears to be backwards. Net torque equals rate of change of angular momentum. You appear to have assumed that angular momentum is rate of change of torque.

Last edited: Oct 12, 2013
10. Oct 12, 2013

### Gabriel Maia

My bad. You're right. So... I'm trying to find the magnetic field this spinning disk will produce at the axis around which it's spinning.

We have found the torque to be

T=-$\pi$λab$^{2}$$\frac{∂B}{∂t}$

and the torque is related to the angular moment by

T=$\frac{∂L}{∂T}$

and therefore L is

L=-$\pi$$\lambda$ab$^{2}$B

We can write L as

L=I$\omega$

where I is the moment of inertia and $\omega$=v/a is the angular velocity. v can then be written as

v=-$\pi\lambda$a$^{2}$b$^{2}$$\frac{B}{I}$

We know that the electric current is

i=$\frac{∂q}{∂t}$=$\frac{λ∂l}{∂t}$=λv

which gives us

i=-$\pi\lambda^{2}$a$^{2}$b$^{2}$$\frac{B}{I}$

I then have planned to use this current in the very known result of the magnetic field due to a current in a circular circuit

B$_{ind}$=μ$_{0}$i$\frac{a^{2}}{2r^{3}}$

Is my current derivation ok?

11. Oct 12, 2013

### TSny

That looks right to me. Good work.

12. Oct 12, 2013

### Gabriel Maia

Thank you. And thank you very much for your time. You helped me a lot.