In Fig. E29.25 a conducting rod of length L = 30.0 cm moves...

In $\textbf{Fig. E29.25}$ a conducting rod of length $L =$ 30.0 cm moves in a magnetic field $\overrightarrow{B}$ of magnitude 0.450 T directed into the plane of the figure. The rod moves with speed $v =$ 5.00 m/s in the direction shown. (a) What is the potential difference between the ends of the rod? (b) Which point, $a$ or $b$, is at higher potential? (c) When the charges in the rod are in equilibrium, what are the magnitude and direction of the electric field within the rod? (d) When the charges in the rod are in equilibrium, which point, $a$ or $b$, has an excess of positive charge? (e) What is the potential difference across the rod if it moves (i) parallel to $ab$ and (ii) directly out of the page?

Transcript

00:01 Okay, so we're doing chapter 29, problem 25.

00:05 So in this problem, we have a rod here, length l, and they call this side a, this side b, and it's moving in this direction, that is v.

00:20 So this is all in a magnetic field pointing into the page.

00:28 The vector line, okay.

00:32 So it tells us that l is 30 centimeters and it says the magnitude of v is 0 .45 tesla and it's moving at 5 meters per second in the direction shown.

00:55 So part a asks what is the potential difference between the ends of the rod? so we know that a conductor moving in a magnetic field may have a potential difference.

01:07 Induced across it depending on how it's moving and we know that the induced emf from the equations is given as speed times magnitude of the b field times the length of the rod and times the sign of this angle phi.

01:26 Phi is the angle between the velocity vector and a magnetic field.

01:38 So this is the angle between these two vectors.

01:42 Okay.

01:43 So now we can just calculate this and we know that for this case, since v is in the plane of the screen here and the field is into the page, we know that phi for our case is 90 degrees.

02:01 So we're evaluating this.

02:04 We have 5 meters per second of speed times 0 .3 meters.

02:11 Four the length, 30 centimeters is 0 .3 meters, and then times 0 .45 tesla.

02:21 And then sign of 98 degrees is one.

02:23 So that is easy.

02:26 We put this in and we see that the induced emf magnitude is 0 .75 volts.

02:35 Awesome.

02:36 So for part b, we want to figure out which side has a higher potential.

02:40 So an easy way to do this is by figuring out which way the charges will be pushed.

02:46 So we know that the force that these from the magnetic field is given as qv cross b.

02:55 So using the right hand rule for a positive charge, v, you can just take v cross b.

03:04 V cross b gives the right hand rule of pushing the positive charge this way.

03:09 So that means the negative charges are pushed this way.

03:13 So since positive charges are pushed to b, then b is the higher voltage.

03:30 Easy enough because the voltage is always higher, or the higher potential, sorry, as a higher potential, i should say instead, because the higher potential voltage is always at the place where the positive charges accumulate...

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