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MULLILE CHQICEQUESTIONS shown: What is the direetion of uniform magnetic - field positive charge entets magnetic force? of page XxXXX B) into the page XXI Downward...

Question

MULLILE CHQICEQUESTIONS shown: What is the direetion of uniform magnetic - field positive charge entets magnetic force? of page XxXXX B) into the page XXI Downward 90 Upward XXXXXX E) to the lelt path will the atoms follow"? beam of atoms enters magnetic field region: What 8) 9) EIIIIIIX}magnetic field with the same specd and follow Two particles of the E mass enter the paths shown Which particle has the bigger charge?C) both charges are equal impossible tell from the picturerectangular cu

MULLILE CHQICEQUESTIONS shown: What is the direetion of uniform magnetic - field positive charge entets magnetic force? of page XxXXX B) into the page XXI Downward 90 Upward XXXXXX E) to the lelt path will the atoms follow"? beam of atoms enters magnetic field region: What 8) 9) EIIIIIIX} magnetic field with the same specd and follow Two particles of the E mass enter the paths shown Which particle has the bigger charge? C) both charges are equal impossible tell from the picture rectangular current loop with clockwise current What the direction of the net force on the loop? uniform magnetic field_ zero



Answers

The magnetic field $\overrightarrow{\boldsymbol{B}},$ at all points within a circular region of radius $R$, is uniform in space and directed into the plane of the page as shown in Fig. $\mathbf{P} 2 \mathbf{9 . 6 1}$. (The region could be a cross section inside the windings of a long, straight solenoid.) If the magnetic field is increasing at a rate $d B / d t,$ what are the magnitude and direction of the force on a stationary positive point charge $q$ located at points $a, b,$ and $c ?$ (Point $a$ is a distance $r$ above the center of the region, point $b$ is a distance $r$ to the right of the center, and point $c$ is at the center of the region.

Uh huh. In the human problem there is a magnetic field directed into the plane of paper. Like this. A positive charge Q. Is moving towards right with the velocity V. Bar the magnetic field is be back. So in the first part of the problem we have to find the direction of the force experienced by the charged particle. The magnetic Lawrence force. So using the expression four magnetic laurels force experienced by this charge particle. Yes. In better phone is equal to Q. Into We cross beat. Hence using right and screen rule over is we cross B. The direction of force experienced by the charged particle will be upward in the plane of paper mm. Hence we can stay here. Our option four is correct. Not in the 2nd part of the problem. The Maget Europe charges given a 0.25 Coolum. The speed with which it is moving Into the magnetic field is 2.0 Into 10 days, part two m/s. The force experienced by eight is 20 new plan. So using the same expression for the magnetic Lawrence force again, F equals two Q B. B. Scientific to in scaler forms. But data here is 90 degree as a child particle is entering into the magnetic field perpendicular early. So it consulting Q. V. P. Sign 90. So plugging in all other known values here. 0.25 colon for the charge 210 into 10 days. Part two. Or it can also build and has 200. Need a per second, multiplied by the magnetic field which is not known to us. and for sign 90 this is one and the force is given as 20 newton. So this magnetic field will be given by 20 divided by 0.25. multiplied by 200. So finally it comes out to be 0.4 Tesla, which is an answer for the second part of this problem. Thank you.

Hi in the given problem there is a magnetic field in the plane of paper and acting in downward direction like this. In vector form, we can represent it by the bar. Then a positive charge Q. Is moving towards right with the velocity V. Bar and enters into this magnetic field. Then we have to find the direction of magnetic Lawrence force acting on it. So using the expression or magnetic Lawrence force acting on this charge particle which says F. Bar. In vector form is equal to Q. Into we cross be then the direction of this week. Crosby will be the direction of force. So using right and school the direction of we gross B will be into the plane of paper, then started into the plane of paper. So option one is correct here. No. In second part of the same problem. The magnitude of this charge It's given us Q. is equal to 0.25 Coolum. The speed with which it is moving towards right, He's given us 2.0 into 10 part two Nita per second. Which can also be given as 200 meter per second. The force magnetic force calculated to be 20 new gun. So we have to find the magnitude of magnetic read. Hence using again the same expression for the magnetic Lawrence force F equals two Q. V B. Santa. For the cross product of Crosby VND. Now the value of this theater is 90° here. So it is Q B B into sign 19 degree. Now Plugging in all other known values. This is 0.25 colon for the charge for velocity. This is 200. Meet up for a second and magnetic release. This is missing. We have to find And sign 90 is one and the force is 20 Newton. So the magnetic will will be given by 20 newton Divided by 0.25 Ghulam multiplied by 200 m/s. So finally, the magnitude of this magnetic field comes out to be 0.4 tesla, which is an answer for the second part of this problem. Thank you.

This is Problem 31 chapter 19. And in this problem, we're looking at a 15 7 meter length of conductor. But you're gonna draw first. And this 15 centimetre length of conductor is moving vertically up and down along too thin vertical conductors that I'm drying along the sides. And in this problem, we're given a figure that shows that this length of conducting wire is 15 centimetres across were also given that there is a five amp your current running up this wire cross our horizontal wear and down this wire here, some other things that were given in this problem are the mess of the horizontal conducting wire here. So the mass of that wire is given to us as 15 grams and just help us out later on in this problem, I'm gonna convert that over into S I unit. So 15 grams is equal to 0.15 kilograms. I found that by just dividing 15 grams by 1000. Another thing that we're given in this problem is the length of the wire. And I'm gonna actually write that down here flowing through the wire or l is equal to 15 centimeters in centimeters or not s i units. And so I'm gonna actually convert that as well. 15 centimeters is 0.15 meters, and I found that by dividing it by 100 And just so I can keep all of my given things straight, I'm gonna write down beat current just I The current here is 5.0 amperes, which isn't us I unit. So we can keep that as is. And then a few other details they give us is that there is a magnetic field in its uniform and as acting perpendicular to the page, so our magnetic field or the direction over magnetic field be is out of the page. So out of the pages also meaning perpendicular to the page. So the magnetic field, if you can imagine, is coming out of your screen right now. One other thing we're given on this problem is that our thin are horizontal. Conducting wire is able to move free vertically, and when it is in the presence of gravity, it moves upward at a constant velocity. I'm gonna write that as constant velocity just as a reminder. All right, so the first part of this problem wants us to find the forces that are acting on the horizontal wire. We'll look back at this wire and we look at the problem. There are really only two forces that are present in this situation. One of them is told to us, and it's told to us in the part where it says that this wire will move upward at a constant velocity in the presence of gravity. Gravity is our first force that were looking up. I'm gonna call gravity F G. In this problem of sub G. The other force that's present here is a force that's created due to the current moving throughout these wires and that force is the magnetic force. And I'm gonna be calling the magnetic force stuff sub mag for this problem so that when we're doing some of the calculations later on, we have some easy shorthand to use. The second part of part is asking us under what conditions that wire is able to move upward at a constant velocity. And, um so which conditions? And for that part of the problem, we really should think about what it means to have a constant velocity. You having constant velocity. It means that you are not accelerating, so your acceleration is equal to zero. So if you have a constant velocity A. Your acceleration is equal to zero. Which means that you're not changing any velocities, and to figure out when you can have a acceleration equal to zero, we're going to use Newton's second law. Newton's second law is written as the sum of the forces equal to mass times acceleration, and in this problem, we know that is going to be equals zero, which means that are some of the forces should also be equal to zero. For that to be true, we would need are two forces that we just talked about gravity and the magnetic force to add up to zero, which happens if they're equal in size and opposite direction. So force of gravity. If some G is acting on the wear in a downward direction and F mag, the magnetic force is acting on the wire in an upward direction, and so these air going to be equal in size and opposite direction. And under those conditions we have a zero acceleration, which means we have a constant velocity, and so to answer part A we have to force is gravity and the magnetic force. And if those two are equal in size and opposite direction, then we can have the horizontal wire move upward at a constant velocity. Herbie of this problem is asking for the magnitude and direction of the minimum magnetic field that you would need in order to have the wire move at a constant speed. So if a part me we're looking for a magnitude and direction of our minimum magnetic field, I'm going to write the magnetic field as B to stand for magnetic field. So for this problem or for this part of the problem, we'll need to use an equation to give us the magnitude of our magnetic fields using the force and some of our other given quantities. That's a question is the force of the magnetic field is equal to the magnitude of the magnetic field times The current times, the length of the wire times signed Fada worth data is again that angle between the current or the direction of the current and the direction of the magnetic field. So if we go back to our givens, we can see that we have all of these quantities that we need in order to solve this problem for beak. I'm going to rearrange this equation first so that we can solve for the thing we want, which is be which will give us the magnitude of the magnetic field. So it's really just a problem. I'm going to divide both sides by current times length time signed data and rewrite this so that it's easy to look at. We have b is equal to, uh, divided by I times l. A. Times science data. At this point, we can plug in the quantities that we already know and the one that we solved for in the previous part, the force f is going to be the magnetic force which we know is equal to the force due to gravity. And so to find the force due to gravity, we're going to need toe know the mass and G, which is the acceleration due to gravity. The mass of this object is a given 0.15 killer Gramps, I'm gonna plug that in 0.15 kilograms And G is the acceleration due to gravity which is given for all problems on Earth, which is 9.8 meters per second squared. When you multiply those two numbers together, you end up with 0.147 Newtons. And since F G is equal to the magnetic force, we can say that, uh, mag is equal to 0.1 force of it units as well. And so I'm going to plug that in for our f right here. I stands for current and our current in this problem going back to their given is five amperes the length of our wire. It's also a given 0.15 meters. Inside data is the same thing as sign 90 degrees because we know that the direction of our magnetic field is out of the page. And the direction of our current is the direction I'm going to highlight here in green, this direction going up this wire across and down, which is in the page. And since this is perpendicular to one another, it's data is equal to 90 degrees and sign of data, then assigned 90 degrees just equal toe one. If you plug that into your calculator, this equation right here, you'll end up with 0.196 And the units for this or the units for the magnetic field, which is Tesla or tea, and this gives us the magnitude for the direction were actually given that in the beginning of the problem in the problem, it tells us the direction of the magnetic field is out of the page, and so that carries over to this problem as well. The direction of the magnetic field is out of the page, so this magnetic fields that we solved for is a minimum magnetic field, and I know it's a minimum of mine. I haven't field because I found it using the force of gravity needed toe balance out this force diagram. So when I say that, I mean that it's, um, equal in size and opposite direction to the force of gravity in the third part part See of this problem. They're asking us what would happen if that magnetic field that we just found right here what would happen if that exceeded that value. So if B were larger than 10.196 tests and in that situation, if we had a larger magnetic field than we would have a larger force on the wire. I'm gonna write that out here. So we're looking for what happens if be the magnetic fields exceeds 0.196 Tesla. If that were the case, you look back at this equation that we were given on the forces of is equal to be I l sine theta. And if this quantity increases of be exceeds one or 0.196 Testa, then f the force, the magnetic force is also going to increase. So if B goes up, F goes up. And let's think about what that would look like for our wire. So I'm gonna drop another quick force diagram if f, um upward is larger with f. The force of gravity are of the magnetic field, is larger than the forced you to gravity than the sum of our forces is going to be in the upward direction. And if that's the case and our acceleration is also in the upper direction because the sum of the forces is equal to mass times acceleration. And I know that because of Newton's second law, and so if be exceed 0.196 Tesla, then the wire will accelerate upwards. All right,

You know that the magnetic force is given by I times a lacrosse being. But that's the case where the magnetic really is. Uniforms far. A case of magnetic field there is with the land. The change in force can entertain us. I times the magnetic fields. And this is the question that we are going to use on. We're going to apply this equation to eat side off the loop now for each side off the loop. The end is parallel to that side. Off the look on, if that isn't the direction off, right, since the loop is in the excellent plan, So Z zeros at the look on dh B y is zero because, uh, be light is basically we just plug is equal to zero in this equation. Then the director will be quarto. You see the light over end Thanks, Ski cap. Because this just God's Edo here. Now you're going to apply this on DH into this equation on DH. We're going tto do the first part now. So for the first part, the magnetic field lines in the plane Ah, to be sketched, son, it's true that so let's see. This is the X and this is the lie exits. As magnetic field is along, the positive key access on it varies with white. So there is no need for the X axis to be here. So you're gonna use why end, See, Axis? Because those are the only components contributing to the magnetic field Here Z is the direction on why he is a distance with which the magnitude ofthe magnetically changes on. You. See, that, uh, the magnet treated b is that is linear. Live it white. That means this is going to be like this. I'd for why? For the audition. Now, if the light changes, this is how it's going to look. Yeah, so that this part e now 45 we have to find the force. So let's find it for side one. Let's mark the site first. So this is side one. Elder's Phil. Let's call this toe beside mine this toe beside toe. This will be side three on DH. This to beside for and for side one. It's called a force to be F one. So this is given by this equation over here so messily integral over this equation. I Times deal Times B signed Fly and the integral with that. But sci fi is basically one because FIEs 90 degree as the magnetic field is hoping to go large. No, the segment here, basically magnetic field is perpendicular to the plane. Off the whole, it's a stone. So for every so far, every, um, case this was is going to Brazil. Sorry, this angle is going to be 90 victory. This gives So Isaac Constance So it comes out on. We know there's a constant Eliza. Lt's a constant. So they come out a cell on we have. Why on the l is basically the change in lentil on this directions or deal is going to be called to the why? Because this segment is along the lie Exes on the integration goes from single toe on my right and rule. So let's do that letter was Let's find the force that they're fantastical dough, half dimes, the knot for l times I on DH. We apply the right and rule, so that gives the force two words. Take AJ two lights, the positive exact sis, This is the force for the first segment. And for those who don't know what this integration is. So I'm just going to write it over here. So let's go on this way to the party and where Anderson in the jet other than negative one. So if this goes from a Toby to the integration of this is basically right at the part and plus one over N plus one on your slide animals from it. This Kips bay to the parent less one minus eight. A guard and place one forward and placement. That's what I did over you. And if n is minus one, the integration is just going, Toby, the lager them off. Why? And then you have to apply the limits too. So now we do the same for you. F door. So after is equal to integration over I b l b sci fi, this one on the limit side and my sins the limits are staying with told me to explicitly can't colored disentangle. This is Okay, so this this man old Sorry, it's Ah. Okay. So the way we have to see what the The surveillance Basically, Xia date, Because deal is a battle to the x axis. So Oh, our limits are going to be the limits off eggs and not the limits of Wei. That's going to be CTO do l Andi. I is a constant. We notice the constants. Also, l is that the limits off the end on deal is basically the eggs as myself. There on, we have a live from the magnetic field and we find this value. Toby, I thank thee not over l So the value ofthe y He's basically a constant here and is equal to Elsa. That comes out and we just have to integrate with the d X not. And this gives the value, Toby. My time's not times l on and off. The direction is given by right and rule being applied toe the direction of current and magnetic field on DH. You see that this direction is going to the towards now good day like the addiction. So I'm forces gonna hit this. Now we do the same thing for three of three. So far three we have I d being I'm sci fi is one on. So the constants come out off the integral on DH. We have revived from the magnetic field on. We have to find see what be Elise. So for the third segment here. Deal is basically negative. Why you again? Saudi deal is basically Dubai again. So you think that on DH is to be Quinto? I be not. And over too. This is basically same magnitude to what we found here on. Let's find the direction now. So they applied. I tend rule Andrei, get in direction. Toby along. Okay, so one thing to note this deal is actually negative way negative delight Because this is going towards the negative I exist, or so b l there's going to be negative off the light. So this is basically mimic on the direction If you apply if you apply the right and girls to see that direction is so I is, uh, okay, Cap Jake on DH, I'll be is cancer. Jacob Cross ski cap is basically I cap. We can also not include the negative sign on DH. Do the cross for that as negative. J cap cross pendant. It was stiff cake and you get killed in the same answer. So for the last part, we have I forgot to write. The elements of this is just a little over seal on DH. Let's make this place on the arrival of distress or so. But I think this is simpler, Dan. Whatever strange value you had to stake the limits As the current flows to zero and take the we control the white Now we find a Ford. So that is I times, Theo times be again on and on the constants come out ofthe day Cushion We have a life from the magnetic field on the l is the ex again And we have the limits from L. A to Theo So the at the system Now we get the value to be I be not Oh well on In this case you see that Why is see you So for this segment of food segment, that means that interval a zero and this gives the force to proceed next, the neck force is going to be Victor sunk off all the forces. So that gives if one the next F net to be equal do F one vector plus two. It's stored and the next page so f net and the corridor f one miss after less f three less before this is basically serial, as we calculated on F one and every have the same magnitude You on his three year, but they're there opposite in direction. So they add up to zero and we're just left with after. That is a negative. All high times they're not. Thanks. And this is the net force.


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