Question 31 has this shown with this figure here, Right? Proton is fired at the velocity given through a parallel plate capacitor. And we're told the electric field is pointing downward with this magnitude. So we're gonna find out a couple of things. For part a gonna find what magnetic field for both strength and direction must be applied to allow the proton to pass through without changing its speed or direction. So what many feel must exist such that this proton disco street Sorry, different color here. Just go straight through and doesn't seemingly is unaffected by the electric field. So because the, um, forced the election field will be pulling it, Pushing it downward means we know that the force on the magnetic field must be acting upward. This we know that be in order for that decree because in the cross product of if you think the force due to be and getting my q v cross be in order for the forced to be acting any upward direction and you know the particles traveling to the right, that means the magnetic field be must be into the page. First balls will determine the direction automatically B is into the page. Great. So in order for this to occur, you know that the force due to the electric field must equal in magnitude the force in the magnetic field. Those two balance. We have the force due to electric eels. Just charge times E and for magnetic field already kind of spoiled it. It's Q B Cross be are charges cancel out so even to solve for the magnetic field to being electric field divided by the velocity of our charge. And so we go in this plug those values in, and you can find it. Can't, um, that the harming I field turns out just by looking at night is now It's going to be 0.1 Tesla. And again, that's into the page based on the right hand roll of needing the magnetic force to be pushing the charge upward. Great perp even find the electric and magnetic field in the protons reference frames. So the protons reference frames it's the one we think of. It's the prime. It's the one moving along with the changes in the scenario. It's the electric field from an outside point of view, plus V across B and So if we have our electric field from our um, point of view for the protons point of view, the luxury field is, of course, putting in the negative direction are greater what downward, which we can call to get a J but didn't label it that way. So it's negative, plus the across B at the positive term. So, um, these 1.0 times 10 to the six we know they're particularly each other, so need aware about the cross product multiplied by be 0.1. Well, if I have a negative plus this positive of the same magnitude than in the protons from the reference there's no electric field, which makes sense because it feels an electric field, so it does not acted on it by the peril plate capacitor and similar, it was his b still not see and for the magnetic field, Big prime that's one over C squared, be across E and so they can evaluate this. Oh, is minus Sorrento looking to use their My apologies. This is be prime minus when something happens to be, say, a couple of mistakes. My apologies for that. This is being a standard cream just into the page and then remind us we have, um, 0.10 testily here, minus one over C squared and be cross. He's a one time center, six crack movie that cross mark because me very taking that to account two stories multiply times the ultra killed of 1.0, I was 10 to the five Oprah meter. So you know this is this term is times 10 to 11 but divided by term times tend the big of 16 cause it's C squared, so it's a very, very minimal effect. That term is essentially zero, and it will have a very, very minimal effect on the negative field. So this in terms of just about zero, So you can just say that this the person feel Proton experience is a 0.10 Tesla magnetic field. Rid New York. I believe that if it be so, we can look at part C now and you want to know. How does an experimenter in the protons reference frame explain that it does experience no force? Well, it's not. Accelerating is traveling at a constant speed. As we stated in the question, and be not Harvey's constant. I shouldn't care that. So, since it's not changing direction and not changing its velocity or orientation, all we know that the force acting on this part of all is, of course, sometimes a but because it experiences experiencing no acceleration and traveling at the same velocity, therefore must experience no forces. Well, that's that's what I come from. And that's why it seems kind of odd to say there's near no electric field and experience is no one should feel a little. Of course it does. Just in its reference, training is not so. Yeah, so there you go.