00:01
Okay, in this question, we have a proton which is accelerated through an electric field into a magnetic field.
00:07
And first question is we're asked calculate the magnitude of the magnetic force acting on the proton in the field.
00:13
And for that, we're going to need to use the equation f is equal to b, q, v, where b is the magnetic field strength, q is the charge, and v is the velocity.
00:25
Now, we are given b and q, b is 0 .1 tesla.
00:30
Q is a proton, so it's 1 .6 times 10 to the minus 19 kuloms.
00:35
However, we don't have v.
00:38
We are told, however, that it is accelerated through a potential difference.
00:41
So we can use the equations e is equal to qv, and e is equal to half mv squared.
00:49
Now, just a bit of a note here.
00:51
I'll try and write voltage here, a sort of bolder and more capitalized, then lower case v is velocity.
00:58
So this is the electrical energy, gained by.
01:00
Charge particle moving through a potential difference and that will then be transferred to kinetic energy so we can say that qv is equal to half mv squared electric energy is equal to kinetic energy and therefore two qv over m all square rooted is equal to velocity again two different vs here so capitalize v for voltage and little v lower case v for velocity so substituting in all of our values the square root of 2 times 1 .6 times 10 to minus 19 multiplied by the voltage, the potential difference to 2 ,100, divided by the mass, which is 1 .67 times 10 to the minus 27.
01:49
And that would then give you the velocity, which is 634 ,346 meters per second, which we're just going to round to 634 ,000 meters per second, or 6 .34 times 10 to the 5 meters per second...