00:01
Okay, this is question 2840, and we have a few different wires and different paths or different circles specifically are or loops are shown labeled a, b, c, and d.
00:15
And we want to figure out what the line integral is, or integral of b .dl for each path.
00:22
So here we need to use amper's law where the integral of b .dl is equal to mu not times the current that's, enclosed in that path.
00:33
So let's start with loop a.
00:38
So loop a here, we see that the current enclosed is zero.
00:44
So i enclosed here is going to be zero, which means that the line integral e .d .l is also going to equal zero.
00:55
So that's an easy one.
00:57
For part b, let's take a look at the current's enclosed.
01:04
Now for loop b, we see that it's enclosed by current i -1, but i -1 here is pointing into the page.
01:15
So we need to know which sign the current really should be.
01:20
So if we, let's see, we're told here that the line integral is going around the path in the counterclockwise direction.
01:31
So using the right -hand rule, when we're going around the path in a counter -clockwise direction, currents out of the page are positive and currents into the page are negative.
01:41
So here we have this current is going, i -1 is going into the page, so that's going to be negative here.
01:50
So our eye -enclosed is going to be negative 4 .0 amps.
01:55
So we want our b .dl in this case, to just be equal to mu not times our minus 4 .0 amps, and that's going to give us minus 5 .03 times 10 to the minus 6 tesla meters.
02:18
I'm going to set this unit here is tesla times meters because we have magnetic field in tesla and then another length, which is measured in meters.
02:28
Now for loop c, let's take a look.
02:33
We have two currents enclosed here now.
02:35
So we have i1 and i2...