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Energy Dissipated in Loop (Magnetic Field)-- Please Help!
What is the energy dissipated as a function of time in a circular loop of N turns of wire having a radius of r and a resistance of R if the plane of the loop is perpendicular to a magnetic field given by
B(t)=B0e-t/tau
P=I^2*R=EMF^2/R
EMF=-d/dt(flux)=-dB/dt*NA
I first wanted to calculate the EMF, and since the B field varies, I found the derivative of the B field, times the number of turns times the area:
EMF=NA*(t/tau)Be^(-t/tau)
Then squaring this term and multiplying it by 1/R will equal the time-derivative of the energy dissipated
dE/dt = (NA)^2*(t/tau)^2*(B^2)e^(-2t/tau)
Bringing dt over and integrating from 0 to t I find (and substituting pi*r^2 for A):
E=(N*pi*r^2)^2*(B^2)(tau^2-e^(-2t/tau))(2t^2+2t*tau + tau^2)/(4R*tau)
This is not the correct answer, though.
Any help? Please!
Homework Statement
What is the energy dissipated as a function of time in a circular loop of N turns of wire having a radius of r and a resistance of R if the plane of the loop is perpendicular to a magnetic field given by
B(t)=B0e-t/tau
Homework Equations
P=I^2*R=EMF^2/R
EMF=-d/dt(flux)=-dB/dt*NA
The Attempt at a Solution
I first wanted to calculate the EMF, and since the B field varies, I found the derivative of the B field, times the number of turns times the area:
EMF=NA*(t/tau)Be^(-t/tau)
Then squaring this term and multiplying it by 1/R will equal the time-derivative of the energy dissipated
dE/dt = (NA)^2*(t/tau)^2*(B^2)e^(-2t/tau)
Bringing dt over and integrating from 0 to t I find (and substituting pi*r^2 for A):
E=(N*pi*r^2)^2*(B^2)(tau^2-e^(-2t/tau))(2t^2+2t*tau + tau^2)/(4R*tau)
This is not the correct answer, though.
Any help? Please!