0:00
Hi there.
00:01
So for this problem, we are told for part a that a cosmic ray proton in the interstellar space has an energy that is equal to 10 .5 megalletron balls and executes a circular orbit having a radius equal that of jupiter's orbit.
00:34
So that is equal to distance is 7 .78 times 10 to 11 meters.
00:57
And for this, we need to obtain what is the magnetic field in that region of space.
01:05
So to calculate this, we need first to pass the energy from megal electron balls.
01:14
To joules.
01:20
So first, we know that the energy, in this case, is equal to 10 .5 mega, which means 10 to the 6 electron balls, and we just multiply this by a factor of conversion.
01:39
We know that one electron ball corresponds to one, 1 .6 times 10 to the minus 19 joules.
01:54
So if we multiply this using our calculator, we obtain a value of, the value that we obtain is 1 .68 times 10 to the minus 12 joules.
02:22
Now that we just have obtained that, and we have the radius, we know that generally the magnetic fuel is mathematically represented by the mass times the speed of the particle, or in this case the proton, divided by the charge times the radius, the radius that we have in for this problem.
02:53
So, er and m is the mass of a proton that we know the mass of a proton.
03:00
And b is the velocity of the proton, which is mathematically deduced from the formula from the kinetic energy.
03:09
We know that the kinetic energy that in this case is the only energy that this particle has is 1 over 2 times the mass, times the speed to the square.
03:19
So if we solve for the speed in order to substitute that in the other equation, we just will have that this is 2 times the energy.
03:31
Divided by the mass and we take the square root of this and this will give us the mass.
03:35
So we just simply substitute that into the equation for the magnitude of the magnetic field.
03:43
So we will have that that is the mass over the charge times the radius times the square root of two times the energy over the mass.
03:56
So we can simplify some things in here.
03:59
As you can see, we can representate m as the product between the square root of m and the square root of m.
04:05
So we can cancel one of this m with this one right here, and we will obtain that this is equal to 1 over the charge times the radius, the square root of 2 times the energy times the mass, the square root of all of that.
04:24
So we just need to simply substitute all of these values into this equation...