# Difference between Vav and Vrms

## Homework Statement

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I encountered this problem during my exam. A coil of 50 turns, each with area 0.02m2 is rotated in a uniform magnetic field of magnitude 0.4T. The coil is rotated at a frequency of 60Hz. Given the average power generated is 50W, what is the resistance of the wire?

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E = dΦb/dt
P = E2/R

## The Attempt at a Solution

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E = dNBA/dt = 50*0.4*0.02/(1/60*1/4) = 96V

(1/60 * 1/4 is the time taken when magnetic flux through the coil is max to zero)

R = 962/50 = 184Ω

However, the correct answer is 227Ω. As the prof had another class to teach, i could not ask him about it. My friend who got the answer said to use another formula: E = ωBAsinωt because Vav is different from Vrms.

• Delta2

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haruspex
Homework Helper
Gold Member
because Vav is different from Vrms.
Quite so. In fact, you did not really calculate Vav. You only considered the change in area over one quarter cycle. Had you done the same calculation over a full cycle you would have got zero.

The energy in each interval dt is V2dt/R, so that is what needs to be integrated.

The energy in each interval dt is V2dt/R, so that is what needs to be integrated.
if i calculate the change in flux over 1/4 cycle and divide by 1/4 of the period, dont i get Vav?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
if i calculate the change in flux over 1/4 cycle and divide by 1/4 of the period, dont i get Vav?
No. As already stated, Vav = 0.

haruspex
Homework Helper
Gold Member
if i calculate the change in flux over 1/4 cycle and divide by 1/4 of the period, dont i get Vav?
You do get some kind of average voltage, but it is not of any great interest. The value will depend on which quarter cycle. E.g. try the quarter cycle from -π/8 to +3π/8. And it will tell you nothing about the average power.

No. As already stated, Vav = 0.
Vav over the whole cycle is 0, that i understand. Then what exactly have i calculated?

You do get some kind of average voltage, but it is not of any great interest.
Because V is a sinusoidal function, each of the 4 quarters have the same area right? So if i integrate each quarter individually i should be able to get V over the whole cycle?

haruspex
Yes, but for half the cycle V is negative. You could take the absolute value, but as I posted it is not very meaningful. The energy over the whole cycle is $\int_0^T \frac{V(t)^2}R.dt$, not $\frac{(\int_0^T |V(t)|.dt)^2}{TR}$,
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