Induced EMF — generator coil area calculation

[Celso]

Homework Statement

What is the necessary area for a generator that produces an emf of $\mathcal{E} = 150V$ when it spins at a ratio of 60 revolutions per second, in a magnetic field of $B = 0.5 T$?

Homework Equations

$\oint_{c} E \cdot dl = \mathcal{E} = -\frac{d}{dt}\iint_{s} B \cdot dS$

The Attempt at a Solution

First I tried solving for the magnetic flux $\iint_{s}B \cdot dS=\int_{\theta{1}}^{\theta{2}}BAcos(\theta)d\theta$
$B$ and $A$ are constant, so $\iint_{s} B \cdot dS = BA \int_{\theta{1}}^{\theta{2}} cos(\theta)d\theta$
Now here is where I'm in doubt, please correct me if you spot a mistake, it spins 60 times a sec and since these engines switch the field or the current so the emf won't change to the other direction every half revolutions, one must consider the flux being always positive:
$60BA \int_{0}^{\frac{\pi}{2}}2cos(\theta)d\theta$
($2cos(\theta)$ represents a full revolution, considering only the positive domain of the function).
Which leads to $\frac{d}{dt}120BA = \mathcal{E}$
I'm stuck here

 

[Charles Link]

The magnetic flux is given by $\Phi=BA \cos(\omega t)$ where $\omega =2 \pi f$. $\\$ The EMF $\mathcal{E}=-\frac{d \Phi}{dt}$. See if that is helpful.

 

[ZapperZ]

Homework Statement

What is the necessary area for a generator that produces an emf of $\mathcal{E} = 150V$ when it spins at a ratio of 60 revolutions per second, in a magnetic field of $B = 0.5 T$?

Homework Equations

$\oint_{c} E \cdot dl = \mathcal{E} = -\frac{d}{dt}\iint_{s} B \cdot dS$

The Attempt at a Solution

First I tried solving for the magnetic flux $\iint_{s}B \cdot dS=\int_{\theta{1}}^{\theta{2}}BAcos(\theta)d\theta$
$B$ and $A$ are constant, so $\iint_{s} B \cdot dS = BA \int_{\theta{1}}^{\theta{2}} cos(\theta)d\theta$
Now here is where I'm in doubt, please correct me if you spot a mistake, it spins 60 times a sec and since these engines switch the field or the current so the emf won't change to the other direction every half revolutions, one must consider the flux being always positive:
$60BA \int_{0}^{\frac{\pi}{2}}2cos(\theta)d\theta$
($2cos(\theta)$ represents a full revolution, considering only the positive domain of the function).
Which leads to $\frac{d}{dt}120BA = \mathcal{E}$
I'm stuck here

You operated the integral on something that you shouldn't.

The starting equation in your "Relevant equations" is the correct starting point. But you need to be aware that this is the surface integral bounded by the closed loop of current. In this case, this is a constant. You already had done this integral when you came up with "A" for the area, so you should have dropped the integral after that.

What you are left with is
<br /> \mathscr{E}=-\frac{d}{dt}BA cos \theta

Since B and A are constants, the derivative only operates on cosθ.

You should see that what you have now is identical to the derivative of the magnetic flux.

Zz.

 

[Delta2]

The Attempt at a Solution

First I tried solving for the magnetic flux $\iint_{s}B \cdot dS=\int_{\theta{1}}^{\theta{2}}BAcos(\theta)d\theta$

This equation is wrong. You switch from integration with respect to $dS$ to integration with respect to $d\theta$ which is the core of your mistake and it does not give the correct flux. The correct equation would be $\Phi=\iint_{s}B \cdot dS=\iint_{s} B\cos\theta dS$ where $\theta=\omega t$ is the angle that the magnetic field vector makes with the normal to the surface. This angle is time varying (not constant with respect to time), but it is constant with respect to the variable of integration $dS$ (it does not change as we move at various points of the surface) and also constant with the latter is the magnetic field vector so both can be taken out of integral and we ll have $\Phi=B\cos\theta\iint_SdS$. I believe it is easy to understand $\iint_SdS=A$ . So at the very end we get
$$\Phi=BA\cos\theta=BA\cos(\omega t)$$ where $$\omega=2\pi60=120\pi$$
and $$\mathcal{E}=-\frac{d\Phi}{dt}$$

Now here is where I'm in doubt, please correct me if you spot a mistake, it spins 60 times a sec and since these engines switch the field or the current so the emf won't change to the other direction every half revolutions, one must consider the flux being always positive:
$60BA \int_{0}^{\frac{\pi}{2}}2cos(\theta)d\theta$
($2cos(\theta)$ represents a full revolution, considering only the positive domain of the function).
Which leads to $\frac{d}{dt}120BA = \mathcal{E}$
I'm stuck here

if the generator has commutator that keeps the polarity constant this doesn't affect the EMF as you say but it would rather be:
$$\mathcal{E}=| \frac{d\Phi}{dt}|$$ where |x| denotes the absolute value of x.

 

[rude man]

Depends on how many windings there are in the armature. I presume you mean 1.
Also, the emf is time-varying so it needs to be specified whether the sought emf is average or peak or rms or ...

BTW there is a simple formula for average emf which is not a function of angle or field configuration or anything else except frequency and maximum flux (and N if N>1).