What is the Maximum Induced emf of a Rotating Loop?

[mbrmbrg]

Homework Statement

A rectangular loop of wire has area A. It is placed perpendicular to a uniform magnetic field B and then spun around one of its sides at frequency f. The maximum induced emf is:

• BAf
• BAf
• 4(pi)BAf
• 2(pi)BAf
• 2BAf

Note: No, it's not my mistake that BAf appears as an option twice.
The correct answer (verified by WebAssign) is 2(pi)BAf.

Homework Equations

[tex]\varepsilon=-\frac{d\phi_B}{dt}[/tex]

[tex]d\phi_B=\vec{B}\cdot \vec{A}[/tex]

The Attempt at a Solution

Looking for maximum emf, so the dot product B*A simplifies to BA.
The crux of the problem will be finding out how A varies with time.
Frequency is given, and A will be very closely related to frequency.

And then I didn't know where to go from there, so I guessed, and got lucky.

Help with a systematic approach to this problem, please?

 

[Hootenanny]

Can you write A in terms of an angle? Think of the standard wave function...

 

[mbrmbrg]

Basic wave function: [tex]y(x,t)=y_m\sin{(kx-\omega t)}[/tex]

• y(x,t) will be the area perpindicular to the magnetic field at a given time
• y_m=amplitude=actual area of the loop=A
• [tex]kx-\omega t[/tex] is the angle that the loop makes with the magnetic field... I think...

I feel like an idiot
I've also been doing nothing but physics for a couple of days, please excuse me.
:the sort of emoticon that looks all normal then spontaneously explodes, sending tiny bits of itself all over the screen, but it explodes in a sphere, of course, because space is isotropic:

 

[Hootenanny]

Basic wave function: [tex]y(x,t)=y_m\sin{(kx-\omega t)}[/tex]

• y(x,t) will be the area perpindicular to the magnetic field at a given time
• y_m=amplitude=actual area of the loop=A
• [tex]kx-\omega t[/tex] is the angle that the loop makes with the magnetic field... I think...

I feel like an idiot
I've also been doing nothing but physics for a couple of days, please excuse me.
:the sort of emoticon that looks all normal then spontaneously explodes, sending tiny bits of itself all over the screen, but it explodes in a sphere, of course, because space is isotropic:

I know the feeling, I've just done a QM exam today . But don't despair, your very close. Perhaps my hint of the wave function was a little unfair. Since you've already posted your homework I don't mind showing you this derivation. Okay, so you know Gauss' law;

[tex]\Phi_B = \int \vec{B}\cdot d\vec{A}[/tex]

Now, recall the definition of the dot product, namely [itex]\vec{A}\cdot\vec{B} = |A||B|\cos\theta[/itex] thus for a coil of area A and N loops we can write;

[tex]\Phi_B = \int \vec{B}\cdot d\vec{A} = BAN\cos\theta[/tex]

Now, from Faraday's law;

[tex]\xi = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\left( BAN\cos\theta \right)[/tex]

Recall that [itex]\omega = \theta/t \Rightarrow \theta = \omega t[/itex]; hence,

[tex]\xi = -\frac{d}{dt}\left( BAN\cos\omega t \right)[/tex]

[tex]\xi = BAN\omega\sin\omega t[/tex]

Since [itex]\omega = 2\pi f[/itex] we arrive at our desired result;

[tex]\xi = 2BAN\pi f\sin(2\pi f t)[/tex]

I hope this helps. Don't be to hard on yourself, if you've been doing nothing but physics lately, your probably very tired and very bored of physics. Try forgetting about physics for a few hours and doing something different, you'll probably find you'll come back refreshed

 

[mbrmbrg]

No, it wasn't unfair at all. If anything, it was a leetle too fair

Thank you! Thank you! Thank you!
I followed what you said, but deriving it myself would have been slow and tortuous work.

THAT WEARY FEELING
by Piet Hein

Do you know that weary feeling
when your mind is strangely strangled
and your head is like a ball of wool
that's very, very tangled;
and the tempo of your thinking
must be lenient and mild,
as though you were explaining
to a very little child.

 

[Hootenanny]

Thank you! Thank you! Thank you!
I followed what you said, but deriving it myself would have been slow and tortuous work.

It was my pleasure. Don't worry about the derivation, its just practise