00:02
In this question we are going to find a number of things regarding the damped oscillation motion for this bar attached to a spring that lies on a u -shaped rail within a magnetic field.
00:22
So this is going to pull together some of the things we know about electromagnetic induction as well as damped oscillations from simple harmonic motion.
00:36
So we're given a we're instructed to go look at equation 14 .41 to get the equation that describes a damped oscillator.
00:51
So that equation is negative kx minus bv equals ma and this is for a single direction a single dimension.
01:13
And where this came from was we have a restoring spring force that doesn't look very legible.
01:33
We have a restoring spring force and then we have a frictional force.
01:42
It could be friction but we'll call it a damping force right.
01:46
We have some sort of resistive force that depends on the velocity at which the system is moving and then of course we have the ma that completes newton's second law.
02:06
So this equation is just an expression of newton's second law.
02:10
And to find the damping coefficient b for this given system what we need a force that is dependent on the velocity from all of our magnetic induction kinds of things.
02:44
And the first thing to note is i mean we could either you can apply faraday's law or use your relationship for motional emf.
02:56
You get the same result.
02:59
I'm going to go for the fast one which is the equation for motional emf.
03:14
So the emf induced in this bar will be equal to b times the width of the bar w multiplied by v the velocity at which this bar is going to move.
03:32
And i'm not going to worry about signs we're just going to go with magnitudes here.
03:37
Trust me if you wanted to start with faraday's law that the induced emf is equal to negative d dt of the flux you're going to get the same thing.
03:49
Okay cool so how do we calculate now the force on a length of current carrying wire in a magnetic field? well that's found with the current in the wire multiplied by its length which again in this case is w multiplied by the magnetic field strength.
04:20
And since they're already perpendicular we don't have to worry about the sine theta that moves in there.
04:24
Well, swell! we need something for i.
04:30
Well our current i is just going to be equal to the emf divided by the resistance of the bar.
04:38
So b w v divided by r.
04:43
So now we have our force is going to be equal to b w v over r times w times b.
04:57
And so we get b squared w squared over r multiplied by v...