Faraday's Law - induced emf of a rotating loop

[Feodalherren]

Homework Statement

Consider the loop in the figure below. What is the maximum induced emf in each of the following cases if A = 600 cm2, ω = 31.0 rad/s, and B = 0.490 T?

Rotating about x,y,z?

Homework Equations

Faraday's law

The Attempt at a Solution

This should be relatively easy since the B is constant but I can't seem to figure it out.

$ε=(.490)\frac{d}{dt} \int dA cos(ωt)$
$ε=(.490)(.06)\frac{d}{dt} cos(ωt)$
$ε=-ω(.490)(.06) sin(ωt)$

Now then, do I take the time derivative again and set it equal to zero to find max/min?

Oh yeah an by inspection I can tell that the change in flux is zero as it spins around the X axis and thus emf=0 in that case.

ps. what's the command fora a new row in itex?

 

[collinsmark]

Homework Statement

Consider the loop in the figure below. What is the maximum induced emf in each of the following cases if A = 600 cm2, ω = 31.0 rad/s, and B = 0.490 T?

Rotating about x,y,z?

Homework Equations

Faraday's law

The Attempt at a Solution

This should be relatively easy since the B is constant but I can't seem to figure it out.

$ε=(.490)\frac{d}{dt} \int dA cos(ωt)$
$ε=(.490)(.06)\frac{d}{dt} cos(ωt)$
$ε=-ω(.490)(.06) sin(ωt)$

That's pretty-much correct so far. (It's arguably correct as-is, depending on how you look at it.)

By "arguably", I'm just making note of the negative sign in $\varepsilon = - \frac{d \Phi}{dt}$

But since the initial phase angle and $\varepsilon$ direction convention are not given in the problem statement, your answers are good enough as they are, I suppose.

Now then, do I take the time derivative again and set it equal to zero to find max/min?

You could. There's nothing stopping you from doing so.

But the only part of the answer that varies with time is $\sin(\omega t)$ and that simply varies between 1 and -1. It should be pretty obvious what the maximum $\varepsilon$ is by inspection.

Oh yeah an by inspection I can tell that the change in flux is zero as it spins around the X axis and thus emf=0 in that case.

Eek! No! [Edit: it's spinning around the y-axis, btw. See below. (Yes, if the loop were to instead be spinning along the x-axis, you are correct and the emf would be zero.)]

Don't forget that the area is a vector, with a direction normal to the surface. In this case, the magnitude of the area is constant and the magnetic field is constant, but the direction of the area changes (with respect to $\vec B$)!

So here, in this problem with constant magnetic field and constant area magnitude, we have $\Phi = \vec B \cdot \vec A$. B and A each have constant magnitudes, but their vector dot product changes with time because the direction of $\vec A$ changes with time with respect to $\vec B$. So no, the change in flux is not always zero. The correct answer is definitely not zero. Go back to your equations above; you were on the right track there.

[Edit: But yes, you are correct that if the loop spins around the x-axis, the emf would be zero in that case. But in the figure, as I interpret it, the loop is spinning on the y-axis. The answer is not zero if the loop spins on any axis except the x-axis.]

 

[Feodalherren]

That's pretty-much correct so far. (It's arguably correct as-is, depending on how you look at it.)

By "arguably", I'm just making note of the negative sign in $\varepsilon = - \frac{d \Phi}{dt}$

But since the initial phase angle and $\varepsilon$ direction convention are not given in the problem statement, your answers are good enough as they are, I suppose.

I never use the sign in Faraday's law. I suppose I should have clarified that but I just take the absolute value and use Lenz's law for direction.

You could. There's nothing stopping you from doing so.

It didn't seem to work. I will re-try it when I get home but for now I didn't get the correct answer.

Eek! No! [Edit: it's spinning around the y-axis, btw. See below. (Yes, if the loop were to instead be spinning along the x-axis, you are correct and the emf would be zero.)]

Actually it's asking for when it's spinning around X, Y and Z. The picture only shows one case.

 

[Feodalherren]

It worked perfectly, I did something weird with my math. Thanks.

 

[BvU]

And the character for a new line is \\

 gives ##\#\###a\\b\\c]##\#\### : it inserts..

gives $a\\b\\c$ : it inserts a three line box, so it really does a "new row" job...