00:03
In this question, we are given the motion of proton in the region of uniform magnetic field.
00:13
So the proton has a velocity of negative 2ihed plus 4j hat plus minus 6 k hat meters per second.
00:27
And then the magnetic field is 2ihat minus 9.
00:33
For j hat plus 8k hat, mini tesla.
00:40
You want to find a magnetic force acting on the proton, the angle between the velocity and the magnetic force and the angle between velocity and magnetic view.
00:51
So to do this question for part a, okay, we're using fb equals to qv cross b.
01:04
So we'll calculate b cross b first.
01:09
So v cross b using the given information.
01:13
We'll be using the determinants method to calculate the cross product.
01:20
So we have negative 2, 4, negative 6, and then 2, negative 4, and 8.
01:28
And remember that b is in the really tesla.
01:34
So we need to multiply 10 to the positive 3.
01:38
So you do this, you get 8 i.
01:43
Head plus 4 j head times 10 to the pound negative 3...