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AUestiohnrotom ah M1u5s 1027 kq and charge FOF19 sont with volocity ?2 10'Ms diroction reqion where Inere plecirc leld Magntude 780 Vim E Ciccon Aeeumo Ihattho...

Question

AUestiohnrotom ah M1u5s 1027 kq and charge FOF19 sont with volocity ?2 10'Ms diroction reqion where Inere plecirc leld Magntude 780 Vim E Ciccon Aeeumo Ihattho magnotic fiold has corportent &nd neglect gravitahona #Irecis Wunat IusI tne magniludo of tho uniform magnetic fioki ng {Cnion Inine proion throxtgh urdelleclexd? (In T) Wna nust Do Ino dJiroction o1Ino unitorm Mannotic Iiok Tnn (cnion nforcn naseehrornn undallecie?unilon10td

aUestioh nrotom ah M1u5s 1027 kq and charge FOF19 sont with volocity ?2 10'Ms diroction reqion where Inere plecirc leld Magntude 780 Vim E Ciccon Aeeumo Ihattho magnotic fiold has corportent &nd neglect gravitahona #Irecis Wunat IusI tne magniludo of tho uniform magnetic fioki ng {Cnion Inine proion throxtgh urdelleclexd? (In T) Wna nust Do Ino dJiroction o1Ino unitorm Mannotic Iiok Tnn (cnion nforcn naseehrornn undallecie? unilon 10 td



Answers

At a particular instant, charge $q_{1}=+4.80 \times 10^{-6} \mathrm{C}$ is at the point $(0,0.250 \mathrm{m}, 0)$ and has velocity $\vec{\boldsymbol{v}}_{1}=\left(9.20 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\imath}$ Charge $q_{2}=-2.90 \times 10^{-6} \mathrm{C}$ is at the point $(0.150 \mathrm{m}, 0,0)$ and has velocity $\vec{v}_{2}=\left(-5.30 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}$ . At this instant, what are the magnitude and direction of the magnetic force that $q_{1}$ exerts on $q_{2} ?$

Okay, We're looking at question twenty eight fifty six, where we have to point charges he won in Q two that are moving at different philosophies. So Kyu won is located at point zero point two five meters zeros. That's just right up on the Y axis here and has velocity in nine point two times, ten to the five meters per second. In the I had direction, any I had direction is B positive ex direction. So that's going to be pointing straight to the right like this Now. Charge, too, has ah charge of minus point to minus two point nine times ten to the minus six cool ums and this Quandt, and it's moving with velocity minus five point three times ten to the fifth meters per second. So that's in the J have direction. So Jay had direction is thie y direction. So since it's negative, that's going to be buying the minus y direction like this, and it's a point point one five zero zero. So that means it lies at point one five meters on the X axis, and the other two coordinates are zero. So we're asked at this point what are the magnitude and direction of the magnetic force that Kyu won exerts on cue to. So to solve this, we're gonna need our equation. That force is equal to Q V Cross B and when we think about this, we want the force that's exerted on cue too. From the magnetic field from Q ah one. So this magnetic field is going to be the magnetic field coming from Kyu won. So here we want cute too, one of the to to me. From what we're considering here now, this magnetic field, this B will be be coming from Kyu won. So here is the expression for the magnetic field. Ah, from a moving point charge. So here again, we're talking about Kyu won. So we're going to need to put in Kyu won here and again with anyone here because we're talking about this point charge creating a magnetic field that Q two is going to experience. So at this point, you to is a distance are from Kyu won. So I have ah list of that over here. So our first step is going to be to figure out what our is. So we have a nice little triangle here with Thie X coordinate is point one five. The Y is point two five so we can use any of the factory and the're, um where we know that our squared is just going to be equal to point two five squared plus point one five squared. And when we play that in and solved, we can get that our is going to be point to nine meters. So we have are are now. Ah, now next, What we need to do is think about this cross product. So we have our are here for r cubed. But we need to know this. The one cross are so we know the direction of the one and you know, the direction of our But we know the magnitude of our but the cross product. These two vectors aren't gonna be perpendicular. So the cross product is going to be equal to the one times are times the sign of this angle here. So if we label this angle, let's call it Fi. So this is the single fire right here. Now our expression for the magnetic field and we re written here in terms of the one times are times the sign of this angle. I am So now we on DH before we actually get into this, let's just see quickly that they're the factor of our appear and effective are down here. So I'm just going to quickly cancel out one of these factors and turned this into our squared. Just what we have it here. So now we need to know what sign of Fei is. We'LL sign If I If you look at this this angle, um we could make another triangle by destroying a line right over here if we wanted. So this makes a triangle. Um, this triangle that we're going to make the opposite angle. So remember that the sine of the angle is going to be the opposite side, divided by the high pop news. So the opposite side here is going to be our why component, which is point two five and the ex component is going to be, I mean by the high pot noose is gonna be our values. Our which we just found up here. So the hypothesis will be point two nine. So we know we don't even necessarily need to calculate this out. We'LL just be ableto plug this in right away Here. We know that this will be our value of sign of fire. So finally, when we ah come here to play this all in we're gonna have the force is equal Tio que tu re to cross be So I'm just gonna plug this whole thing in right here for B. Now again, we have a cross product that we're going to need to deal with. So first, let's consider let's think about the direction of the magnetic field that point at this point where que Tu is so again to find the magnetic field, we had to take the cross product of the one and our So if you use the right hand rule taking the cross product of the one and our which is going down Ah, that gives us a field that's pointing into the plane. So that's the negative Z direction or the minus k hat directions. So I'm just going toe put that here so that we you keep track of that. This is the minus. K had direction. So now when we come over here and like it vey too, we too is going in the minus J direction, the minus y direction. When we take the cross product, we want to do VI to cross minus K, and that's going to give us the value of thie. Um, the direction is going to be in the plus extraction or the plus. I had direction. So now we know our directions here, which is good. And remember, keep in mind. Ah, since the minus K hat direction and the minus j had direction are perpendicular the cross product, you don't have to deal with the sine of any angle because the angle between those will be ninety degrees. So ah will be the sign of ninety degrees is one. So we don't have to worry about taking any additional. Ah, science of any angles. So when we finally take a look at this again, we can just get rid of this cross product because we took care of that. Um, calculating the direction on the sign between the two angles is ninety. We now have this whole expression that we can just plug right in all this, re arrange it a tiny bit to make it a little bit easier, right? Essentially, at this point, all we have to do is just plug in numbers. We have you to be heavy too. We have kyu won. We have V one and we have sign of fatal. We found right up here twenty, twenty five out of five point two nine are we already found? And we found the direction. So finally, when we plug these in, one more note about thie signs the positive and negative signs We've already taken care of the directions of the one and be too because we took those into consideration when we're doing the cross product. But we have not yet taken into consideration the direction of or the sign of the two charges. So when we plug in, we want to be too. We want to use just the positive values. So we want to you just positive five point three and positive nine point to. But when we plug in Q one Q two. We do want to save these, uh, this negative sign here in Kew, too, because we haven't yet taken into account the fact that cute who is negative. So when we played us all in, we should end up with a minus six point nine six times ten to the minus six. And again, this is me. I had correction. So what? This negative sign does It essentially says that since you two is negative instead of having a force in the plus I direction, it's actually in the minus my direction minus I had direction. So Ah, we have a force with magnitude six point nine six times ten to the minus six. This isn't Nunes. Let's put that here. Um indicate that here that isn't written and again it will be in the minus. I have direction for the negative X direction.

In this problem. We have a particle entering a region of space with a uniform magnetic field and its traveling perpendicular field. When that happens, of course, the particle will travel on a circular path. In this case, it only travels. 1/4 of a circle had eased than turning around and heading due south. So let's just get a quick visualization of what that looks like. So we would have a region of space here with a magnetic field, in this case pointing outward. And then we'd have a particle heading with velocity V SE to the right. In this case, this would be due east and then down here would be due south. Now, when the particle enters this region of space with this uniform magnetic field, what happens is it follows a circular path in this case will only go 1/4 of the path and then head to color of black right for a particle. And then it's gonna head due South this way Brand. So we're asked several questions about the situation and were given several pieces of information. We know the massive heart particle. We know the speed of our particle. We know how long it spends in this region. We know the delta t here, and we also know the magnitude of the magnetic field to Richard is traveling now. This is one of those problems where it's actually easier to solve for part B first and then solve for part A. So that's what we're going to do. First part be here is asking us to find out the magnitude of the charge of this particle. Now remember, we know how to deal with them. Relate the motion of art Particle two characteristics of the particle through our equation for the radius of the path. As a particle moves through a uniform, moves particularly do a uniform magnetic field. That means if we want to find the charge of that particle we just saw for that by rearranging our equation a little bit. Keeps no mistake there. Let me. That is not gonna be our radius down here. So we have this expression here Now, what are the thing? What do we not know? Well, we don't Yeah, I know the radius here. So what? We don't know the radius of this path, but we know some information about the velocity and We know some information about the time so we can figure out how to relate those things together. So let's think about the motion that this particle is undergoing in this problem. So it's going to do travel a squirter circle of a path here and will say that that distance that it travels is a distance D. So we know from physics one how all these things relate to each other. We know that the dissed the velocity of the particle is going to be the distance the particle travels divided by how long that it's a travel. But you know that this particle the distance is traveling is only 1/4 of the circumference of the circle. A ser comments of a circle is two pi r. Divide that by four and then divide that by Delta T. So we ultimately see that the velocity of our particle is given by high times are divided by two delta t. Now we don't know what our is, so we can solve for it and plug it into our equation here, So solve for this here and then once we solve for it, we can plug it into here. So solving for our real quick, we get value of two times that Delta T how long it's in that region of space multiplied by the velocity, all divided by pi. So we're gonna take this expression here, plug it into here, and then we're going to be able to get our answer. So what is the charge of this particle? What's the magnitude of the charge of the particle? It's gonna be given by M groups, some of them m times c divided by to delta t times V times be all divided by pi that comes from plugging the expression for the to death TV over pie expression in for arms. Now, when I do some simplifications, I see the velocity goes away and I get that we have m times pi divided by two delta T times A magnitude magnetic field. It's actually gonna give us the value of the charge of our particle. So we plug everything we know into here. We have the mass of our particle, which is given as 7.2 times 10 to the negative eight kilograms pies, pie, my little calculator. Carry that out to as many decimals as we can And then we have a time that it's in this region of space 2.2 times 10 to the negative three seconds. And the magnitude the magnetic field through which is traveling is 0.31 Tesla's puggle there into our calculator. And we see that the charge of our particle is 1.7 times 10 to the negative four Coombs. So that is the answer to actually part b of this question. Now that we know the value of this charge, which in order to finish this problem, we just clear this off for a quick Now we know the value of this charge. We can then just use it to solve part A pretty quickly. So Part A s. What is the magnitude of the magnetic force that is acting on this particle? What we know the charge is 1.7 tons 10 to the negative four cools. And we know that the equation for the magnitude of the magnetic force acting on a particle is the charge of that particle multiplied by its speed multiplied by the magnet to the field. It's traveling through about a sign of the angle between the magnetic field in the velocity. We know that this is gonna be a sign of 90 because the magnetic field is perpendicular and this term is going to go toe one. So all we need to do is plug in the charge of our particle 1.7 times 10 to the negative four Coombs, multiply that by our philosophy, 85 meters per second and then multiply that by the value of the magnetic field through its the charged particles travelling 0.31 test lis. So, in the end, we had of getting that The magnitude of the force acting on this charge particle in order to cause this particular type of motion is equal to 4.4 times 10 to the negative three units. And that is the answer to part a. My problem

Hi in the given problem, a charged particle Q. Is moving do you east and after entering into the magnetic field it covers a quarter circle and then comes out of the field and it starts moving straight again. Like this. No mass of the positive charge particle is given us. M is equal to 7.2 into 10 for -8 kg. It is moving towards east with a velocity Whose magnitude is 85. Meet up our second the magnetic field Is given us 0.31 s lot within which this charge particle is moving. No it is also given that that time taking to complete one quarter circle Means 1/4 of the circular part. S. T. is equal to 2.2 into 10: 4 -3 seconds. So it's time period means time taken to complete the circle. It will be four times of this time means this is capital T. Is equal to four into 2.2 into 10 days, par -3 seconds. Which Finally comes out to be 8.8 Into 10 days 4 -3 seconds. No, In the first part of the problem we have to find the force acting the magnetic force acting on this charge particle, for which we use the expression for that time period of the charged particle which rotates in a magnetic field. That time period is given as to why AM by big you? So using that expression we obtain an expression for the product of magnetic field with the charge and that is nothing but to buy am by he is a time period. No, he will use the expression for the magnetic Lawrence force. Magnetic Lawrence force experienced by the charged particles in the magnetic field. In vector form. This has given us Q. Into we cross B. Or using the rule of vector product. Let me write it like U. V. P. Into science. Whatever theta is, the angle between the velocity vector and magnetic reflector, as the magnetic field is perpendicular to the plane of paper And velocities in Dublin of paper. So this angle is 90°.. So it comes out to be QB be signed 90°,, which is one. So using that we get an expression for the force has been to be cute. Hence finally, we into forbid you We already know this is too. I am by T. So now plugging in on non values. This is it maybe readiness to buy M. V. By tea time period. So this is two into 3.14 into mass, which is 7 to into 10 for -80 kg In the velocity which is 85 m/s, Divided by the time period, which is 8.8 into 10 for -3. So finally, this force experienced by the charged particle In this magnetic field comes out to be four 3 7-10. They should bottom -3 newton, which is the answer for the first part of this problem. No. In the second part of the problem, we have to find the magnitude of charge for which we simply use expression for the magnetic Lawrence walls. A physical to Q V. V. So the expression for the charge will be have by we seem to be For the force. This is 4.37 Into 10 days, 4 -3 Divided by velocity which is 85 NATO per second into magnetic field which was given as 0131 Tesla. So finally the value of the charges told over carried by this particle comes out to be 1.66 into intended for minus four cola, which is the answer for the second part of this given problem. Thank you.

Okay for this problem. The equation that we're gonna want to use is magnetic Field P is equal to force, divided by charge times the velocity sign of the angle between the velocity and the force. Everything that were given here is the charge. The charge is bias eight 0.3 Micro Coghlan's so one of note Micro Kuhen is the same thing as writing. You can take this micro here and just substituted four times 10 to the minus six. That's what a micro means. A very small number and that's that's cool instead of micro cool. So we'll want this in cool. So we'll want to include this times 10 to the minus six were also given the velocity velocity is equal to 7 24 times 10 to the sixth positive six in our units of velocity are in years per second were also given the angle between them. Fada is equal to 52 degrees. The last piece of information we have is what the force is by force is equal to 5.4. I mean times 10 to the minus three on force as usual is in units of Newton's. Okay, let's look back in our equation, we have force. We have charged velocity and have her angle. That's everything that we're going to need. Okay. So we can pluck all of that in straight into the formula. Take sign of the 50 to multiply that by your velocity 7.4 times cute. And then divide force by that number. And what you end up with is be is equal to 1.1 times 10 to the minus four. All right, we need our units. Now. Let's look what we have. So we have force in the top Newtons divided by charge. Cool multiplied by velocity meters per second. And we know that a Tesla is he. Newton's second divided by a cool meter. Okay, so let's see if we can get this to look like that, we need to move the second. And to do that, we can make it a fraction. That's over one. No, we have two layers of fraction here. So this is the big determined. This is the big fraction scientists with little one. So first up, we need to move it up here. That would invert it so it would look like one one over s Then we inverted again to get it up top here, which would make it look just like this again. So what will end up with, I think a little aside here is that news. What we have wept in the bottom is the cool times. The meter we took seconds out flipped it twice, so that's outside the fraction now. And it's in the exact form that we're looking for. Noontime second, divided by Coolum Kinds Meter, which is a test. And that's a unit. And there you have it, and we can box it in. And that's the answer. There you go.


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