00:01
In the given problem here this is a magnetic field v bar and this is the direction of motion of a proton vertically up like this with the velocity v bar and the values are given as speed of the proton is v is equal to b .60 km per second or we can say 3 ,600 meter per second.
00:42
Magnitude of magnetic field is 0 .750 tesla and the angle between them is 55 .0 degree.
00:55
Angle theta is 55 .0 degree.
00:59
Now in the first part of the problem, we have to find the magnitude and direction of the magnetic lawrence force experienced by this proton.
01:10
So, magnetic lawrence force will be given by in vector form f equals to q into v.
01:24
Cross b.
01:25
So, first of all, using right -hand screw rule in this v -cross -b, we conclude that the direction of this force will be into the plane of paper.
01:52
Direction of force f will be into the plane of paper and its magnitude will be given by f equals to q v b sine theta plugging in all known values for the charge over proton this is 1 .6 into 10 dash bar minus 19 kulam for the speed this is 3600 meter per second then 0 .750 tesla magnetic field and sign 55 degree.
02:38
So it comes out to be 3 .54 into 10 dash per minus 16 newton direction we have already found into the plane of paper.
02:55
And this is the answer for the first part of this problem...