Question: a piece of copper wire is formed into a single circular loop of radius 12 cm. a magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.60 T in time of 0.45 s. the wire has a resistance per unit length of 3.3 x 10^-2 ohms/m. what is the avg. electrical energy dissipated in the resistance of the wire?
My Approach: I found the area using the radius...A = .0144 m^2 and then found L using the radius....L = .75 m
Then, I used this equation to determine emf... Emf = (BA - initial BA)/ change(time)
Emf = (.60)(.0144) - 0/ 0.45 = .0192 V
Next, I found resistance by multiplying the L to ohms/L value:
R = 3.3 x 10-2 ohms/m (.75m) = .025 ohms
Then I found current using V = IR:
I = .0192V/.025 ohm = .768A
Then I plugged to P = IV (to find power):
P = IV = (.768)(.0192) = .0147 J/s
Then I multipilied by time to get the energy:
.0147 J/s (.45 s) = .0066J ---- THAT'S THE ANSWER!
Again, just rectify my approach and correct what is wrong!
2007-02-25
13:16:37
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2 answers
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asked by
Jimmy
3