 Now I'm ready for my students to finish up this calculation that we started in the part one video. You're trying to find the magnetic force on a charged particle moving through a magnetic field. And we use the information in the word problem to put our magnetic field and our velocity, the velocity being the one in red and the magnetic field being the one in blue, into vector notation. And now we're ready to do the cross product. Now this example problem was set up to give you some numbers that are really easy to work with. In general, when you go to do that cross product, you would have to multiply out all of these terms to find the i, the j, and the k components. And we can do that for this vector, but we've got lots of zeros in a lot of these different places. Now I also show in some of my videos in my reference information for the students how you can actually write that out in term of the matrix. And then following along you can do the forward i, the negative i, the forward j, the negative j, the forward k, and the negative k. Which for me is a lot easier than trying to follow all the little subscripts of the x, y's, and z's on these components. If I've got i, I've got zero times zero and then minus zero times three. For j, zero times zero, and then the minus j would be five times zero. The positive k five times three and the negative k zero times zero. Which is how I got all of these terms, and it does actually follow this equation. Now in this case, I've left the units off just for a little bit of clarity with the numbers. But if I were to go through and plug that all in, the v cross b part of this is going to give me zero for i hat, zero for j hat, and 15 tesla meter per second for the k hat part. Now a lot of students are so relieved that they've finally gotten through the cross product that they think they're done. But don't forget that you've got a charge in there as well. And the problem that my students are working on, it was a charge of two micro coulombs. So in this case, we have to multiply that two micro coulombs by what I found for the vector cross product. And I find it's much easier to do that multiplication after you've done the cross product, especially since the charge is normally a very small amount and it's going to introduce a lot of scientific notation. And in this case, it's really easy because the two times zero gives me zero. The two times zero gives me zero. And then I have to worry about the two micro coulombs times the 15 tesla meter per second. Which gives me three times ten to the minus fifth newtons in the k hat direction. So your final solution for the force tells you that you've got nothing left or right, nothing up or down. But you do have a force, although it's a small force, out of the board in that positive k hat direction.