A long, straight wire as shown below carries a constant current I. A metal bar with length L is moving at a constant velocity V as shown. Point ‘a’ is a distance ‘d’ from the wire. a) Derive an expression for the EMF induced in the bar. b) Which point ‘a’ or ‘b’ is at a higher potential?
B=μI/2πr for a straight wire. EMF=Blv for motional EMF.
The Attempt at a Solution
For part a) I split up the bar into a bunch of small segments of length dr, a distance "r" away from the wire. so dB=μIdr/2πr. Integrating that from point (a) to point (a+b) I get B=(μI/2π)ln((a+b)/a). While I feel reasonably confident about that part, I thought about it a bit more, and I am not sure how multiplying the B-field, μI/2πr by the length, dr, really makes sense, am I forgetting something important? But assuming that is right, I just plug that into EMF=Blv, and get my answer.
Part b) I feel stuck on, only thing I could thing of was using the equation B=μI/2πr and plugging in and (a) and (a+b) for r. Since if I plug in (a) it will be a greater value, then the EMF will be greater there.