00:01
In this problem on the topic of circular motion, we are told that within a particle accelerator, a deuteron with a given mass will reach a final speed of 10 % at the speed of light, and this is whilst moving in a circular path of a given radius of 0 .48 meters.
00:18
We want to know the magnitude of the magnetic force that is required to maintain the deuteron in a circular path.
00:24
Now if we neglect relativistic effects, we know the magnitude of this force, f, must equal to m, a, c, for the centripetal acceleration, which we can write as m times v squared over r, where v is the speed of the deuteron and r is the radius of its orbit.
00:50
So we can write this as its mass, which is two atomic mass units, which is two times 1 .661 times 10 to the minus 27 kg times the square of its speed, which is 10 % the speed of light, which is to 0 .998 times 10 to the power 7 meters per second.
01:22
And that's squared...