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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: dekoyl on November 09, 2009, 05:44:05 pm

Title: Half frequency = halved EMF?
Post by: dekoyl on November 09, 2009, 05:44:05 pm: I can't see a formula that shows the fact that if you half the frequency of a spinning coil, the induced EMF will be halved as well.
I can kinda see why, but I can't see the link via formulae (or maybe I'm just so rusty I forgot a formula).
Title: Re: Half frequency = halved EMF?
Post by: appianway on November 09, 2009, 05:46:22 pm: It does.

Induced EMF is inversely proportional to the time taken, and the time taken for 1/4 of a revolution is 1/4 of the period. The period is calculated by 1/f, so hence halving the frequency halves the EMF.
Title: Re: Half frequency = halved EMF?
Post by: /0 on November 09, 2009, 05:46:48 pm: Normally you're asked for the emf of a quarter turn. If the whole period is T, then

$emf = \frac{\Delta\Phi}{\Delta t} = \frac{\Delta \Phi}{\frac{1}{4}T} = \frac{\Delta \Phi}{\frac{1}{4f}} = 4f\Delta\Phi$
Title: Re: Half frequency = halved EMF?
Post by: dekoyl on November 09, 2009, 05:48:38 pm: Oh I see =\ Thanks /0 and appianway
Title: Re: Half frequency = halved EMF?
Post by: /0 on November 09, 2009, 06:53:45 pm: You can also use this formula to answer such a question, however it is not in the course.

The flux through the coil at any time is $\Phi = NBA\cos{\theta}$ , where $\theta$ is the angle between the magnetic field and a vector perpendicular to the face of the coil.

If the coil rotates at a frequency of $f$ with period $T$ , the coil will move through an angle of $2\pi$ in time $T$ , so the angular velocity is $\frac{d\theta}{dt}=\frac{2\pi}{T}=\omega$

Therefore, in time $t$ , the coil will rotate through $\theta = \frac{d\theta}{dt}t=\omega t$

Thus, we may say that $\Phi = NBA\cos (\omega t})$ . (we assume the coil starts perpendicular to field, but this is no problem)

Finally,

$V = -\frac{d\Phi}{dt}= -\left(- NBA\omega\sin{(\omega t)}\right)$

$\therefore \boxed{V=NBA\omega\sin{(\omega t)}}$

Now, $NBA\omega=NBA\left(\frac{2\pi}{T}\right) = NBA2\pi f$ decides the amplitude of the graph, so by decreasing $f$ you decrease the amplitude.

The maximum emf is given by $NBA2\pi f$ .