Electric Flux Equation for Rotating Loop?

In summary, the equation for the electric flux through a circular loop rotating with constant angular velocity in a uniform electric field is given by Eπr^2sin(wt) where E is the magnitude of the electric field, r is the radius of the loop, and w is the angular velocity. This is because the angle between the electric field and the area vector changes over time, with the initial flux being zero when the electric field is perpendicular to the surface.
  • #1
Rijad Hadzic
321
20

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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  • #2
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?
 
  • #3
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?
 
  • #4
Yes, because sin is shifted over from cos by ##\pi/2##.
 

1. What is the equation for electric flux?

The equation for electric flux is Φ = E * A * cos(θ), where Φ represents the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

2. How is electric flux related to electric field?

Electric flux is directly proportional to the electric field. This means that as the strength of the electric field increases, the electric flux also increases. Similarly, as the electric field decreases, the electric flux decreases.

3. What is the significance of the angle in the equation for electric flux?

The angle (θ) in the equation for electric flux represents the angle between the electric field and the surface normal. This angle is important because it determines the amount of electric field passing through the surface and contributes to the overall value of electric flux.

4. Can the electric flux be negative?

Yes, the electric flux can be negative. This occurs when the electric field and the surface normal are in opposite directions, resulting in a negative value for cos(θ). This means that the electric flux is directed in the opposite direction of the electric field.

5. How is the unit of electric flux measured?

The SI unit for electric flux is volt-meters (V∙m). This is because the electric field has the unit of volts per meter (V/m) and the area has the unit of square meters (m²), resulting in the unit of V∙m for electric flux.

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