Electric Flux Equation for Rotating Loop?

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The discussion revolves around deriving the electric flux equation for a rotating circular loop in a uniform electric field. Initially, the electric flux through the loop is zero, which requires careful consideration of the angle between the area vector and the electric field. The correct equation for electric flux is determined to be Eπr²sin(wt), as this accounts for the initial condition where the electric field is perpendicular to the loop's surface at t=0. The confusion arises from the incorrect definition of the angle, where the angle should be π/2 at t=0, leading to a maximum flux if defined incorrectly. Ultimately, the understanding is clarified that the sine function correctly represents the time-dependent flux due to its phase shift compared to the cosine function.
Rijad Hadzic
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Homework Statement


A circular loop with radius r is rotating with constant angular velocity w in a uniform electric field with magnitude E. The axis of rotation is perpendicular to the electric field direction and is along the diameter of the loop.Initially, the electric flux through the loop is 0. Write an equation for the electric flux through the loop as a function of time in terms of r, E, and w.

Homework Equations

The Attempt at a Solution


So I use flux = EAcos(x)

A = pir^2

Epir^2 and cos x is going to vary with time, and this is where I should put w, angular velocity.

so flux = Epir^2cos(wt)

but my book is telling me its Epir^rsin(wt)

Why would it be sin? That's going against the flux equation..
 
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At time ##t=0## the flux is zero but your equation gives
$$E\pi r^{2}\text{cos}(\omega\cdot 0)=E\pi r^{2}\neq 0$$
This is because your angle ##\theta## which you call ##x## is defined incorrectly. At time zero, ##\theta=\pi/2##, so how would you correct this?
 
NFuller said:
At time ##t=0## the flux is zero but your equation gives
$$E\pi r^{2}\text{cos}(\omega\cdot 0)=E\pi r^{2}\neq 0$$
This is because your angle ##\theta## which you call ##x## is defined incorrectly. At time zero, ##\theta=\pi/2##, so how would you correct this?

I see. So starting at t = 0, my equation would give me the flux if Area vector and E are pointing in the same direction, which would be the maximum flux, but the problem is saying at t= 0, E is perpendicular to the surface, so flux should be 0..

so using (E)(3.14...)(r)^2sin(wt), my inital flux would be 0, and this is the correct function.

Is my understanding right now?
 
Yes, because sin is shifted over from cos by ##\pi/2##.
 
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