00:01
So for this problem, we want to find the m f at t equals nine seconds.
00:07
So we know that the e m f is the change in magnetic flux over time.
00:13
And magnetic flux is defined as the b magnetic field times dotted with the surface theo area.
00:30
And because b is perpendicular to a yeah, we would have that.
00:41
The manic field times the area cosign data.
00:46
Well, actually, just leave us with b a.
00:49
Exactly.
00:51
Now.
00:53
Although this is our magnetic flux, we cannot necessarily use this shit because our area changes over time.
01:04
So in this problem, we have a wire and there is a byfield.
01:13
There's going out of the page, so we have be filled.
01:20
Going out of the page on our loop of wire is actually shrinking.
01:30
So imagine our wire is getting smaller and smaller with time.
01:47
So out here we have time.
01:52
The initial time, which zero and nine seconds.
01:58
T nine.
02:04
Our circle has are weir has shrunk.
02:10
So our initial conditions that were given from this problem is that the circumference is 165 centimeters.
02:23
So we know that the magnetic field, the unit of magnetic flux is tesla meters squared, so we would want to work with centimeters rather than meters.
02:44
I mean this down a bit.
02:48
So we're given that the circumference, which is two pi r, is equal to 165 centimeters, which is equal to 1.65 m, which is the units we wanna work with.
03:07
Now.
03:09
We would want to know the radius of this circle because we know that area of a circle is pi r squared.
03:18
So we'll probably the radius.
03:21
But we have to also keep in mind that this changes over time.
03:23
So we're gonna have to do more stuff with this.
03:26
So we will have our well label.
03:30
This, as the initial are so are not, is equal to 1.65 m over two pi, which gives us 0.2626 m on dso.
03:48
We know we have way know that we have a constant magnetic field, a constant magnetic field.
04:00
It is equal to 0.5 tesla's and are changing value.
04:12
So what? we're told that the circumference of this circle is decreasing by 12 m, 12 centimeters per second.
04:23
So the circumference see this right this, as are changing values we read are changing value value as, um point 12.12 m per second, which is just what they give us.
04:49
And we could write a negative sign to indicate that the circle is getting smaller, so negative 12 centimeters per second is what they gave us.
04:58
And we know that the relationship so this is the circumference shrinking.
05:04
So as we saw in our diagram eso this is two pi r.
05:11
So this contains our initial radius to buy our and this is some other our value that depends on time because each time each time a second goes by, we are losing our radius of our circle is getting smaller and smaller.
05:32
Right? so what we want to do is find is making equation to describe that change in our radius.
05:39
So we know that two pi r and that's right.
05:44
This as well believe it eso to fire is our circumference and our circumference is changing bye 12 m per 2nd 120.12 m per second and we would want to look at this in terms of time.
06:05
So if we want to see this as it changes with time so we end up with d.
06:13
R.
06:14
Because we know that the radius is changing with time and these air just constant so we can take this out.
06:21
So we have our radius that's changing with time.
06:27
And if we divide this to the other side so if you divide two pi on both sides, we are left with, um that give us that gives us a value of negative because we are decreasing negative 0.190986 meters per second.
06:57
So i'm just living in all those values in so that we can have a more accurate answer at the end.
07:04
And so now we have, uh, we have, ah, description of how our changes with time.
07:11
Now, the reason we needed this is because rmf, uh is the change in flux and the flux depends on the on area, which is also changing, and to describe this change.
07:26
So we know that the area is changing with time, so area is changing with time.
07:35
So if we're taking the derivative of d a with respect to time, uh, we know that our is also changing with time, so we would have to take d t to both sides of the equation pi r squared.
07:56
And here we apply the product rule because we have our square, which is our other variable.
08:02
So we have d a dt is equal to two pi.
08:12
So if we take the product rule of our square, we get to our.
08:18
So i dropped down to two pi so to from there and the pie and are so two pi r and we can see that this is the circumference, obviously, because the if we add up a group of, uh if we just let's let's say we have a group of circles or this, for example, if you want oh, add up.
08:44
Um, all of the circumference is of a circle we would get actually an area.
08:53
So if we know that the radius changed with, let's say five, for example, then we would go out.
09:06
Okay, 1234 five and the value will actually fill in the area.
09:19
Uh huh.
09:21
So what we're left here is the a over d t is equal to two pi r now we took the product rule of this with respect to time, and that's d r d.
09:32
T here.
09:34
That is that value.
09:35
So this value represents that change that i was mentioning of...