WIRES GOT CROSSED Part 1 of 2 Suppose two identical very...

WIRES GOT CROSSED Part 1 of 2 Suppose two identical very long straight wires are held so that they cross each other at the origin as shown below, with wire 1 carrying a current of (0 A, 18 A, 0 A) and wire 2 carrying a current of (-17 A, -26 A, 0 A). y ?x X What is the magnetic field at (0.4 mm, 1.6 mm, 0 mm)? (Enter the components in units of mT) $\vec{B}_1$ = What is the magnetic field at (-0.4 mm, 1.6 mm, 0 mm)? (Enter the components in units of mT) $\vec{B}_2$ = What is the magnetic field at (0.4 mm, -1.6 mm, 0 mm)? (Enter the components in units of mT) $\vec{B}_3$ = What is the magnetic field at (-0.4 mm, -1.6 mm, 0 mm)? (Enter the components in units of mT) $\vec{B}_4$ =

Transcript

00:01 All right, so let's say we have two wires placed in the xy plane, and one of them is right here with the current coming out of the page.

00:07 The other one is here with the current going into the page.

00:10 This is i1.

00:12 This is i2.

00:14 And we want to look at the magnetic field that is basically completely in line of these charges as shown.

00:21 So like the distance, p has coordinates of 13 .1 and 9 .7 millimeters.

00:31 And so that tells you basically what the coordinates of the wires are as well.

00:35 I2 is at 13 .1 millimeters, and i2 is at 9 .7 millimeters up.

00:43 And we're told that i1 has a value of 11 .7 amps, and i2 has a value of 8 amps.

00:53 And we want to find the x component of the magnetic field at point p, and then the y component of the magnetic field point p so if we look at what the magnetic field of i2 looks like coming out of the page it's going to circle around in a counterclockwise fashion and then if we look at what i2 is going to do at the same point it's basically going clockwise like this so the x component of the magnetic field at y2 is going to come purely from i2 if we look at it it'll be bx is going in this direction because i2 only produces a vertical magnetic field at that point so bx is just going to be mu not times i2 over two pi times the distance to point p we'll just call it's d and it'll have a negative sign because it's pointing to the left at that point we can see from our direction so this is going to be like two times or negative two times 10 to the negative seventh henries per meter times of the current in i2 which is eight amps divided by the distance to the distance to this point, which is 0 .007 meters.

02:13 The 2 times 10 to the negative 7 comes from the fact that mu not divided by 2 pi.

02:20 It basically comes up to 2 times into the negative 7.

02:22 So we've got that divided by 0 .0097 with a negative sign.

02:27 So this would be negative 1 .65 times 10 to the negative 4th tesla.

02:35 And then b .y, using much the same logic, is going to be basically purely due to the other current...

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