Ratio of electric field to applied magnetic field

[Rijad Hadzic]

Homework Statement

A metal rod of mass M and length L is pivoted about a hinge at point O as shown in Figure P32.80 (I have attached this to this post)

The axis of rotation passes through O into the page. Find the ratio of the maximum electric field inside the rod to the applied magnetic field when the rod is rotated with angular speed ω. Assume the speed of the rod is determined by the linear speed of its center of mass, and its mass is uniformly distributed.

Homework Equations

electric field in a translating rod, E = vB

The Attempt at a Solution

Ok since e = vB, v = ω, I have e = ωB

It wants to find the ratio of the maximum electric field and the applied magnetic field.

so I have e/B = ω

But this is problem 80 in my textbook. Seems a bit too simple. Is there a trick here?

 

[kuruman]

First off the motional emf is not given by E = vB. You are probably thinking of a rod translating, not rotating, in a magnetic field, but this equation is incorrect even for that. Secondly, what is v? The velocity changes as one moves from the pivot to the end of the rod. Think of all those charge carriers in the rod and what kind of forces act on them as the rod spins. Also, v is never equal to ω.

 

[Rijad Hadzic]

First off the motional emf is not given by E = vB. You are probably thinking of a rod translating, not rotating, in a magnetic field, but this equation is incorrect even for that. Secondly, what is v? The velocity changes as one moves from the pivot to the end of the rod. Think of all those charge carriers in the rod and what kind of forces act on them as the rod spins. Also, v is never equal to ω.

Sorry, yes I meant E = vB for a translating rod.

Why is the equation wrong though? My book is telling me that the max electric field (at equilibrium) is given by equation E = vB

V is the velocity of the rod. I thought in the question "Assume the speed of the rod is determined by the linear speed of its center of mass, and its mass is uniformly distributed." impies that we can ignore that the velocity changes from pivot to end??

On second thought I'm going to give the question a little more thought. I do see now that the magnetic force acting on the positive protons will not be the same when rotating and translating.

 

[Rijad Hadzic]

I also found this:
http://physics.bu.edu/~duffy/sc545_notes02/motional_emf.html

Which states E = vB similarly to how my book does it.

 

[Rijad Hadzic]

So linear speed v = 1/2 L * angular velocity,
in my book it says

E = vB

In the case of transnational motion. Even in rotational motion, I could see the force pushing the positive charged particles to the end opposite the hinge, thus giving E.

My doubt comes from karuman's post, because all of this is implying that the equation E = vB is correct...

So I'm thinking just plug in E = 1/2 L * avel then E/B = ( 1/2 L * avel) / B

Does this seem right??

 

[kuruman]

My doubt comes from karuman's post, because all of this is implying that the equation E = vB is correct...

Yes, E = vB is the correct expression for the electric field in the case of a translating rod.

So I'm thinking just plug in E = 1/2 L * avel then E/B = ( 1/2 L * avel) / B

What is avel? What is its relation to the angular speed ω?

Does this seem right??

It seems right with the correct expression for avel, but that's not enough. Other than what's written in the sources you consulted, do you understand how the answer is put together? If you were asked on a test to "derive" the expression, would you be able to do it as opposed to just writing it down because you have memorized it?

 

[Rijad Hadzic]

Yes, E = vB is the correct expression for the electric field in the case of a translating rod.

What is avel? What is its relation to the angular speed ω?

It seems right with the correct expression for avel, but that's not enough. Other than what's written in the sources you consulted, do you understand how the answer is put together? If you were asked on a test to "derive" the expression, would you be able to do it as opposed to just writing it down because you have memorized it?

avel = angular velocity, which is the speed of the middle of the rod.

I'm not sure what you mean to derive the expression though?

I understand why E = vB in the case of a translating rod. Actually, I don't. I understand why the magnetic force pushes the protons to one side of the rod, but I don't understand why electrons would be on the other side instead of following the protons. I don't get why a dipole is formed. I thought electrons were attracted to protons so when the rod starts moving and protons and pushed in one direction, wouldn't the electrons be attracted to them? Why is a dipole assumed?

Also by derive do you mean start from scratch in the case of a translating rod and apply it to the case of a rotating rod? Also is my expression correct given that avel = angular velocity?

 

[kuruman]

avel = angular velocity, which is the speed of the middle of the rod.

You are presenting a contradicting picture. Angular velocity is has units of rad/s. Speed has units of meters/s. I repeat my previous question. What is avel in relation to ω?

I understand why the magnetic force pushes the protons to one side of the rod, but I don't understand why electrons would be on the other side instead of following the protons.

Protons are not free to move in the rod, only electrons are. What is the direction of the magnetic force on the negatively charged electrons?

I don't get why a dipole is formed.

The magnetic force pushes the electrons to one end of the rod. The excess electrons at that end leave a deficit of electrons at the other end which means that the other end is no longer neutral but has net positive charge. This creates an electric field inside the rod that opposes the magnetic force. Eventually, the force due to this electric field is as large as the magnetic force at which point the migration of additional electrons stops. The total charge on the rod is, of course, zero.

Also by derive do you mean start from scratch in the case of a translating rod and apply it to the case of a rotating rod?

That's exactly what I mean.

Also is my expression correct given that avel = angular velocity?

It is correct, but I am not convinced that you understand why. If that doesn't bother you, so be it.

 

[Rijad Hadzic]

You are presenting a contradicting picture. Angular velocity is has units of rad/s. Speed has units of meters/s. I repeat my previous question. What is avel in relation to ω?

Protons are not free to move in the rod, only electrons are. What is the direction of the magnetic force on the negatively charged electrons?

The magnetic force pushes the electrons to one end of the rod. The excess electrons at that end leave a deficit of electrons at the other end which means that the other end is no longer neutral but has net positive charge. This creates an electric field inside the rod that opposes the magnetic force. Eventually, the force due to this electric field is as large as the magnetic force at which point the migration of additional electrons stops. The total charge on the rod is, of course, zero.

That's exactly what I mean.

It is correct, but I am not convinced that you understand why. If that doesn't bother you, so be it.

Hey thanks for being patient with me.

So I guess I need to brush up on my rotational dynamics...

but from what I read just recently..:

I need to find E, which = vB

this applied to a circle of radius r, but in my problem I am using 1/2 L as r...

I am given ω, which is in rad/s, I need to use this to find an expression for the arc length in meters divided by time. I will call the arc length y

ω=θ/s, and v (linear speed) = y/t

θ*r = y, ---> θ = y/r

so ω = y/rs

so ωr = y/s

thus linear speed = ωr

in my problem r = 1/2 L

so E = (ωLB)/2 and E/B = ωL/2

So in my post # 5, I was actually wrong.

Not only am I suppose to use angular speed (not velocity) but I forgot to cancel the B out as well.As for the proton part, that is also my fault. Really stupid mistake in writing that. I forgot the book is implying that the protons are "imaginary," but the force on the electrons is going to be opposite in direction to the force of the imaginary protons.

I understand now why the electrons are being pushed to one end now, the protons aren't following because its not the protons that are "free," its the protons. I wish my book stated this a little better although I'm sure they did... was this the point that you were trying to get across? If so I think I understand a bit better now..

and it does bother me lol, I want to understand. I also want to pass the course though lol, so just getting the answer is important as well :s, again thanks for all your time man..

so is my new answer of E/B = ωL/2 correct now though?

 

[kuruman]

Your new answer is correct. Good luck with your course.