Induced Voltage as a function of time.

[ElijahRockers]

1. The problem.
A circular loop of radius 4cm has a plane which is perpendicular to the electric field B. (The area vector is +k, so is the field vector.) The strength of the field varies with time according to the equation B=4/(t^3). Give an expression for the voltage as a function of time.

2. Homework Equations .
OK, I'm using V_induced = -d/dt(B_flux), which is Faraday's law, and B_flux = Integral(B.dA)

The Attempt at a Solution

I'm not great at this calculus stuff, but when trying to find the field flux, int(B.dA), I can assume that dA is equal to A, and that it's constant, since the area is not changing right? Well A = .04m^2, so I get B_flux = .04*int(4/(t^3))dt, right? Are my limits of integration from 0 to DeltaT? Either way, I get an error because the integral will leave me with t in the denominator, and since the initial limit is 0, it is undefined.

Anyway, after I do find an expression for the B_flux, i can take the derivative of it to find d/dt(B_flux) and that should be the Induced voltage as a function of time... right?

Thanks in advance.

 

[gneill]

In order to find the flux you want to integrate the field over the area of the loop. This has nothing to do with time; time is a constant at any given instant, and you're interested in finding the flux through the loop at any given instant in time.

The equation that you quoted, B_flux = Integral(B.dA), is summing over differential elements of flux B.dA . It seems that the magnetic field B is a constant over the area of the loop, since it doesn't have any geometrical terms related to x, y, or z. So the magnetic field is uniform over the loop at any given instant in time, so we can pull it out of the integration over the loop area.

\Phi = \int B \cdot dA = B \int dA = BA = 4 t^{-3} \pi r^2

Where you have to worry about the time is when you differentiate this flux with respect to time in order to determine the induced voltage. Since we're not concerned about the polarity of the induced voltage,

|\varepsilon| = \left| \frac{d \Phi}{dt} \right|

You should be able to carry on from here.

 

[ElijahRockers]

You should be able to carry on from here.

Thank you very much! I think I got it. Wish I had been a member of this forum several semesters ago, haha!