00:01
This question tells us that we have a coil of radius r, and it is in a uniform magnetic field oriented perpendicular to the coil's surface.
00:14
So this tells me that i don't have to worry about having cosine or sign because the field is already perpendicular to the surface, which means it will be parallel to the surface area vector.
00:28
And that simplifies our calculation for flux later in case that that's something we need.
00:34
And then it's asking us, what is the total electric charge that passes through the coil's wire loop if the magnetic field decreases at a constant rate to zero? so a lot of information there.
00:47
They're looking for the total charge that passes through, so essentially the change in charge.
00:54
And they tell us that b field will decrease.
00:59
At a constant rate to zero.
01:04
So b field is changing.
01:06
And then finally, they say that the resistance of the coils wire is r.
01:11
So how do we approach this question? well, if ultimately changing charge is what we're looking for, hopefully you'll remember that changing charge can be found by the current times the changing time.
01:25
Because by definition, current is how much charge that passes through, a cross -sectional area of a wire over time.
01:34
So rearranging that equation, you have delta q is equal to i times delta t.
01:39
Well, how do i find delta i then? hopefully we remember that once we have the resistance, we can use oms law to find current being emf divided by resistance because v equals i are.
01:57
And how then do i find emf? well, that's where we have to use faraday's law of induction.
02:04
And from there, we can find the induced emf.
02:10
So let's go ahead and solve for that.
02:13
Induced emf is equal to the change in magnetic flux over the changing time.
02:19
Again, i care about the magnitude of emf in this case because it seems like they want us to express a everything algebraically.
02:28
And of course, we're only looking at one coil, so there's no need for n here because it's just going to be one.
02:35
Now, the problem tells us it is b that is changing, so i am going to write that out.
02:45
Basically, i expanded phi into b times a, and i note that because b is already perpendicular to a...