A conducting rod is pulled along horizontal, frictionless,...

A conducting rod is pulled along horizontal, frictionless, conducting rails at a constant speed of 2.3m/s. The rails are spaced 82cm across. The entire system is in a uniform magnetic field of 0.018T that points into the screen. Assuming that the resistance of the rod is 0.40? and that the resistance of the rails is neegligible, determine the force that must be applied to the rod in order to maintain the constant speed.

Transcript

00:01 All right, hello.

00:01 In this question, we're given this setup.

00:03 We have a conducting rod moving across two conducting rails.

00:08 And so we have a full circuit being formed here.

00:10 And we're told that the rod is moving to the left at some speed v of 2 .3 meters per second.

00:15 And this whole thing is in a uniform magnetic field.

00:18 We're told the resistance of the rod and the length between the rods, or the rails, rather.

00:24 And we're asked, what force must be applied in order to maintain a constant speed? so we're going to have some applied force here.

00:31 We need to figure out what that is.

00:34 So in order to do this, we're first going to have to establish, if we are not applying a force to this, what would happen? so we have this conducting bar moving to the left in a magnetic field at some speed v.

00:49 And so because we have a change in flux, we're going to induce some voltage in the rail.

00:54 And i know we have a change in flux because we're changing the area of our enclosed part within our circuit here.

01:03 So we're going to have induced voltage.

01:06 We know that our flux, generally speaking, is going to be b times a.

01:10 We don't have more than one loop here, so we don't need an n.

01:12 And everything's nice and perpendicular, so we don't need to worry about any angles.

01:15 In this case, b is not changing.

01:18 B is a constant value of 0 .018 teslas.

01:21 So we have b.

01:23 We can pull that out.

01:24 And then our area, well, our area is going to be l, this end here, times whatever this distance here is.

01:31 I'm going to call that x.

01:33 And so we know that l's not changing, so we can pull that out as well.

01:37 And then we're going to just get dx dt.

01:39 So our induced voltage is going to be negative bl dx dt.

01:43 You might recognize this term here as b, our velocity, a change in distance over a change in time.

01:50 So in this case, our induced voltage is negative b times l times v.

01:55 We know that voltage also equals i times r.

02:00 That's ohm's law there.

02:02 So we can go ahead and figure out what our induced current is.

02:05 It's going to be negative blv over r, our resistance.

02:09 Cool, so we know that there is an induced current in this loop.

02:13 Well, what direction is it going to be? that's going to be important if we're looking at the force...

Question

Please give Ace some feedback