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Questiea Co mplct ?7 Stal 5[UacVafee? nak Ruligh Eets mefootieQu5tmn 196f3HbQucstion=MacoUnetCozsider parlicle duat has d charge = mass Iatio of Im/ql-2A IO Skg C ...

Question

Questiea Co mplct ?7 Stal 5[UacVafee? nak Ruligh Eets mefootieQu5tmn 196f3HbQucstion=MacoUnetCozsider parlicle duat has d charge = mass Iatio of Im/ql-2A IO Skg C Tlepurtide entets = urifon magnetic field of B- 0.9 and makes & quarter-circular Fath belore it exits the Eeld as shown in tkefigure; Find ike time (in ns) during which the particle stays in the field037we3Matelo mattes Quribon mI atets IGpcotOlIT

Questiea Co mplct ?7 Stal 5 [Uac Vafee? nak Ruligh Eets mefootie Qu5tmn 196f3Hb Qucstion= Maco Unet Cozsider parlicle duat has d charge = mass Iatio of Im/ql-2A IO Skg C Tlepurtide entets = urifon magnetic field of B- 0.9 and makes & quarter-circular Fath belore it exits the Eeld as shown in tkefigure; Find ike time (in ns) during which the particle stays in the field 037we3 Matelo mattes Quribon mI atets IGpcot OlIT



Answers

Crussed $\vec{E}$ and $\vec{B}$ Fields. A particle with initial velocity $\overrightarrow{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{j}}$ enters a region of uniform electric and magnetic fields. The magnetic field in the region is $\overrightarrow{\boldsymbol{B}}=$ $-(1.35 \mathrm{T}) \hat{\boldsymbol{k}} .$ Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge $(\mathrm{a})+0.640 \mathrm{nC}$ and $(\mathrm{b})-0.320 \mathrm{nC}$ . You can ignore the weight of the particle.

For no deflection. The magnetic and electric forces must be equal in magnitude an opposite direction. That man's the net force should be Cedo. If we want no deflection, this man's the direction shouldn't be opposite on the magnetar should be seen now. F m is he called Cute Dying's. So the cross product off the grass mean So this is a joke. You will be deigns signed fi as the point be points in the same direction and velocity points in my direction. So they're open Nicola and scientific becomes one. So this is basically the magnitude askew Times late. I'm speaking now. The direction will be given by day cap grass. As you can see, there's a negative sign here, So negative off. Okay, cap. And this is equal dough. So this is basically given by the cross product by the right and rule. So this is negative off cap or negative X axis. Similar. That direction off is equal to the magnitude ofthe magic. Turn direction off the electricals skilled times, the electric field. So the magnitude will then be just equal to cure times the magnitude ofthe electric field on DH direction off. This should be along that electric field, like not exactly along, depending on the charge is positive or negative if tells us positive, this is a long electric field. If Giles is negative, this really opposite electric field on DH for our case, we want a distinct the direction off. This force should be along positive. I that is opposite to the direction off FM. Let's go ahead and find the magnitude ofthe the electric field so we plug these two values in this equation on DH. We have already taken the negative sign into a gown, so basically we want the magnitudes off these two to the equal now, so Q e Musical Tokyu Times The Times and this gives they electric field will be the dimes B on. We calculated that using the values who were given so notice that we have already taken this negative sign into account over here. So now you just need the absolute value off the magnetic field over here. So we know that and we find the magnitude ofthe electric field to the 7898 Newton. Welcome. Now, for part A, the judge is positive. This means Dad for the electric Force. Toby along, Plus I gap, He should be positive as well. And it should be along. Plus, I get so hee Vector should be along. Plus, I get now for five b You is negative. You. So for the force Toby along. Plus I the positive X axis The electric field should be opposite Teo. Come. Ah X axis. So this is negative, Mike now? Yeah, I think that's sick. The same configuration off electric and magnetic field walks as a velocity selected for both positively or negatively charged particles, so yeah, give that.

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.

And this problem. The concept covers the concept of the spectrum meter and the motion of a charge in a magnetic field region. So from those computer the velocity selector that is we equals the strength of the electric field upon the magnetic field strength inside the spectrometer. Yeah, so this is a question but and the radius of the circular path outside the spectrum meter equals mv upon the charge Q. Into the magnetic field outside. That's what this is the question too. So from one and two we can write the radius equals the mass into the electric field upon the judge to into be in and to be out are the mass of the particle equals the radius r. Into the charge to and to be in be out Apple. T now substitute the value So the mass of the particle equals the radius is four centimeter into The charge use 1.6 into Tenders -19 Cooler. And the magnetic field inside is uh the magnetic field Inside a 0.0053 Tesla. and the magnetic field outside is z required 00-4 to Tesla. Upon The electric field strength that is 92 and as full bolt or meter. But the mass of the particle is 9.12 and two tenders -31 kg


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