A square loop 24.0cm on a side has a resistance of 5.20 ohms. It is initially in a 0.665 T magnetic field, with it's plane perpendicular to B, but it is removed from the field in 40.0 ms. Calculate the electric energy dissipated in this process.
The answer in the back of the book is 7.05 x 10^-3 J. I can get through part of the problem but I'm getting confused at how to use the information given to get an answer in joules. Can someone help?
The answer in the back of the book is 7.05 x 10^-3 J. I can get through part of the problem but I'm getting confused at how to use the information given to get an answer in joules. Can someone help?
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The induced voltage V is given by:
V = -d(BA)/dt where B = magnetic field strength and A = area perpendicular to B
Now you know the BA -> 0 in 40 ms so you can approximate d(BA)/dt by
d(BA)/dt = BA/t where t = 40 ms.
Now the current generated is I = V/R where R= resistance of loop
I = V/R = -BA/(tR)
and power P = VI = (BA)^2/(t^2R)
FInally energy = power times time = E = Pt = (BA)^2/(tR) = (0.665T*[0.24m]^2)^2/(0.04s*5.2 Ohms)
E = 7.05x 10^-3 J
V = -d(BA)/dt where B = magnetic field strength and A = area perpendicular to B
Now you know the BA -> 0 in 40 ms so you can approximate d(BA)/dt by
d(BA)/dt = BA/t where t = 40 ms.
Now the current generated is I = V/R where R= resistance of loop
I = V/R = -BA/(tR)
and power P = VI = (BA)^2/(t^2R)
FInally energy = power times time = E = Pt = (BA)^2/(tR) = (0.665T*[0.24m]^2)^2/(0.04s*5.2 Ohms)
E = 7.05x 10^-3 J