An alpha particle charge 2e and an electron move in opposite...

An alpha particle (charge +2$e$) and an electron move in opposite directions from the same point, each with the speed of 2.50 $\times$ 10$^5$ m/s ($\textbf{Fig. E28.4}$). Find the magnitude and direction of the total magnetic field these charges produce at point $P$, which is 8.65 nm from each charge.

Transcript

00:01 Problem 28 .4, we have an awful particle that's this here, an electron that move in opposite directions from the same point.

00:09 And we have each with the same speed, and we're supposed to find a magnitude direction of the total magnetic field of these charges that they produce at point b, which is a certain distance way 8 .65 nanometers from each charge.

00:20 Okay, so we need to solve this using the equation for the b field of a moving point charge.

00:29 So that equation is b.

00:32 Equal to new not over four pi and we have q vector for velocity vector across r hat over r square.

00:49 Okay, so q, that's just the charge of each of them, v, that's the vector in, vector and for each them in its different directions, and r is just this distance given.

01:02 R hat is going to be the vector extending from their origin point out to the point p.

01:08 So we need to write down what this is going to be for each of them.

01:13 So i'm going to start off with the equation for the alpha particle.

01:18 So that's going to be mu not over 4 pi.

01:24 And it's going to be q is 2e.

01:27 So that's going to be electron charge here.

01:30 And then we have the v value for v, 2 .5 times times to the 5.

01:35 And we need to write down what the cross product.

01:41 That second part of that cross product is going to be the sign of the angle between them.

01:44 So that's going to be sign of 140 degrees.

01:52 And that's going to be over the distance r squared.

01:56 So that's the b field for the alpha particle.