And this problem, the first part is asking us under is asking us to understand how moving charged also created. Okay, as we're moving feel so we want to use a particle cube moving with velocity V defying the position vector are hat. We want to show that what the magnetic field at the location is equal to. So in this case, we're gonna use the fact that the, um indefinitely indefinitely Small magnetic field D B is equal to mu, not over four pi r squared I ds times are so it's a a infinitely small region and an infinitely small magnetic field. We also know that our current is equal to, um are total number of charged particles in this area multiplied by the size of the area multiplied by charge, times velocity. And so for a, um, for a positive current charge, her first are for a positive charge. The direction of DS is going to be in the same direction as the and we also know that a D s is the volume occupied by the moving charges and it's equal toe one if we only have a singular charge. So in this case, that means that be must be equal to him, you know, divided by four pi times, Q v times are divided by R squared resorted to Cuba. Times are hat divided by r squared. And so we've shown that this part of the problem is true. The next part is that he got to find the magnitude of the magnetic field one millimeter to the side of a proton, moving at a given speed. To do this, we're once again going to use the equation that we just found. And we know in this case that, um, our that we're are sign of data so that would be here is going to be signed of 90 which is one since velocity is always perpendicular to magnetic field by the right hand rule. So we're gonna plug in are constants which really just include our, which is really just aren't radius. So that's tending the negative three meters that's gonna be plugged in for R squared as well as the charge on a proton. And when we do this out, we find a magnetic field of 3.2 times 10 to the negative 13 Tesla Now part C is asking us to find the force on the second proton at this point, if it's moving with the same speed in the opposite direction. So that magnetic force given by the equation Q v B we're gonna plug in the charge of a proton, which is 1.6 of him and negative 19 cool arms are velocity. Um, that are person is going to be moving at 2 10% of the seven meters per second and our magnetic field We do this, we get a total force of 1.24 times 10 to the negative. 24 students last for the problem is asked needs to find the electric force on the second proton eso. The electric force we know is just equal to um que e. So that's our constant K times queues have a first charge him. The second charge, divided by r squared. Now I tried to do the same. In this case is both of the murders the person of the charge of Proton. This is really just the same thing as q P River Port on squared que issued a constant and our is a distributor charges which is still, um one millimeters so we're gonna have 10 to the negative three meters squared and denominator and that's going to give us a final result of 2.30 times 10 to the negative 22 Newtons.