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Hi, thanks for this. Just about the F=qvB acting on the rod where v is the horizontal velocity of the rod (and charge carriers) to the right. Can you explain why we do not consider the vertical component of velocity of the charge carriers in the qvB expression. I mean there is a current (due to the emf) already set up due to the changing flux. So why ignore this vertical component of the charge carriers? More broadly, is there a F=qvB on the other sections of the loop due to the velocity along the wire of their charge carriers. I know it won't cause displacement of these sections but does such a force exist due to their velocities? This is something I don't understand about the extent of self-induction that you may be able to clarify.

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Thank you for this video

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Shoutout to Patches for making studying for finals less painful 😸

Many thanks for the videos, sir! Now we just need all of NYC... ;p

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The notes to follow along can be found at: drive.google.com/file/d/1-rJ4OrVAjQ8AmgXnBBXlAhzemzW5tlPv/view?usp=sharing

The notes to follow along can be found at: drive.google.com/file/d/1-rJ4OrVAjQ8AmgXnBBXlAhzemzW5tlPv/view?usp=sharing

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Hey I also thought of that idea

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I seem to have made a mistake in 1 c). I write P=IV, but then for some reason I say to multiply the current times the resistance and I write the current and the resistance in each case. But the answers I give are in fact the current times the voltage, so the answers are correct. (Thanks Niko).

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sec 2 lol

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