Faraday's law and loop of wire of resistance

[Dominguez Scaramanga]

Hello there, I'm new to this place, I thought these forums looked like a wonderful source of knowledge, so purhaps some of you could be so kind as to help with the following problem...? (here's hoping I'm in the right forum for a start...)

sorry about the length of it, but i did draw a picture! http://groups.msn.com/_Secure/0QADfAkYTL2tjEDZUv*E*lQ54rKER9RkgjSMAd6dd2kQv4rwGnfWenVWaLADu*HOLB85dAhlAI!P2SHdFPJ!VG04XQLHUrfTpzozfAgBwORc/8c.jpg?dc=4675485221241691806
(damn, sorry bout the smudge marks, I must have used two different whites :blush: )

...Consider the rectangular loop of wire of resistance R shown above, being pulled with velocity v perpendicular through a uniform magnetic field B (which is coming out of the screen).

now, I have worked out that current I will be induced when the circuit enters, and leaves the B-field, due to that being the times when there is a rate of change of flux - \varepsilon=-\frac{d\Phi_{B}}{dt} and the fact thatI=\frac{\varepsilon}{R}.Using this I found that the induced current is
I=\frac{Bav}{R}.

(first question, is this correct?)

now, I have to;
"Determine the total external power required to pull the loop through the region of the magnetic field. show that this is equivilent to the electrical power dissapated in the wire".

This is the part I am stuck on...

P=I^2R

but I am not sure as to how I could work out the power needed to pull the circuit through the field, and equate it to the above. any suggestions?

any help would be greatly appreciated

 

[pervect]

but I am not sure as to how I could work out the power needed to pull the circuit through the field, and equate it to the above. any suggestions?

any help would be greatly appreciated

Well, it seems to me they want you to calculate the force that is induced on the loop of wire, and calculate

\vec{force} \cdot \vec{velocity} = (\vec{force} \cdot \vec{distance}) / time

 

[Dominguez Scaramanga]

thankyou very much

pitty I've sent the paper off now hehe, ah well, thanks all the same!