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An ionized deuteron (a bound proton-neutron system with a net +e charge) passes through a velocity selector whose perpendicular magnetic and electric fields have ma...

Question

An ionized deuteron (a bound proton-neutron system with a net +e charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of 65 mT and 3.8 kV/m, respectively: Find the speed of the ion. Present the answer in km/s

An ionized deuteron (a bound proton-neutron system with a net +e charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of 65 mT and 3.8 kV/m, respectively: Find the speed of the ion. Present the answer in km/s



Answers

An ionized deuteron (a bound proton-neutron system with a net $+e$ charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of $40 \mathrm{mT}$ and $8.0 \mathrm{kV} / \mathrm{m}$, respectively. Find the speed of the ion.

Solving party of this problem. So here I will use the formula to multiplication, be multiplication B equal to Q. E. Simplifying it further, I can write the value of P. B. Is equal to beat us by the solving it further. I can write the value of BDS is equal to be multiplication. Be multiplication D. Which is equal to eight. Multiplication 10 to the powerful multiplication, 1.5, Multiplication .015. On simplification. I get the value peter surged 1.8. Multiplication 10 to the power three B. This is our answer for party. Now for part B. As here a W charge particle is traveling at the same speed as we can see that this election of a specific velocity does not depend on the value of the charge as it is independent of the charge. As 88 in the dependent of the charge. Adjective, independent of the geology. So the value of P will be same as PDS which is equal to 1.8. Multiplication, tend to depart. Three P.

So in this question, we know that we have both electric field and magnetic field present. Um, the magnetic field has a magnitude of 0.635 Tesla, and the electric peeled, has a magnitude of 2.68 times 10 to the sixth Wilkes per meter. And the problem also tells us their direction. In fact, they their direction is such that they crossed each other at a 90 degree angle. So for the magnetic field, the direction is vertically downward. And for the electric field, it is horizontally east, horizontally due east. And then the problem tells us that we have an electron and it travels in the direction of horizontally north. So this is painting a three D picture in my mind because think about a compass thought is lying flat and how not only do you have the four directions on that plane on that flat plain, you also have the direction of going up and down vertically crossing that plane. Okay, so keep that in mind as we go further in this question, and um, they tell us that this electron will go through both field with zero net force, and it will continue in a straight line, and they're asking, Okay, given all of these conditions, what must be its velocity? So before I do any more mass with this problem, I want to draw a diagram. So we have a better idea of how every every factor, a place in here. So first, I will establish my directions. So this is the compass I'm talking about. You have north so west east. And I want you to think of this as a plane on its own. And this plane is horizontal. So imagine you are kind of laying this flat. Then you should also have the direction that is, um, into the page and out of the page. If you're following the way I'm doing it into the page will represent downward critically downward. Okay. And out of the page will then represent vertically upward. So what is happening right now is that we have, um, electric field that is horizontally due east. So the straight horizontal lines that I'm drawing in the east direction are the electric field lines, and at the same time, you have, um, the magnetic field vertically downward. So based on our legend, downward means you're coming out of the page or start into the page. Okay, so here you have a crossed magnetic and electric fields, and all the while you have a an electron moving vertically. Do you start moving horizontally north? So that means following this legend horizontally north means this direction. Okay, so here you have an electron, and that's your velocity right here. Maybe I should specify what each of these means. All right, so And this is your electron right here. Okay? And it's asking, how can it How can all the forces on it be balanced such that it will move in a straight line continually without changing directions or speed. So because this is an electron there, let's review a couple of things. Okay? So for electrons, it experiences A and, um, electric field electric force. So that is my electron. The electric force should point in the opposite direction. Asked the electric field so it will experience an electric force that is pointing to the left. At the same time, you have to use the right hand rule to figure out the magnetic force, got the electron experiences within this field that is going into the page. So, as always, you should have your index finger in the direction of velocity. You're, um, middle finger in the direction of the magnetic field. And your stomach should point in the direction of the magnetic force. So when you use right hand rule right now, you will notice that. Hey, my film actually points to the left. So doesn't that help the electric force and wouldn't up, Then accelerate the velocity, accelerate the electron, You know, with a radio acceleration to the left. Um, but remember that we're talking about an electron, so whatever your thumb, whichever direction your thumb points in as a result of right and rule, the direction for an electron should be the opposite. That so is that a pointing to the left the, um, magnetic force from feel a magnetic field. It's actually pointing to the right. Okay, so, um, basically, remember that electric field depends on sorry. Electric forest depends on the electric field. Right? Where asked, the magnetic force depends on not only the magnetic field, but also the speed of the electron. So, essentially, this question is saying how fast must be electron be moving through this crossed field, so that, um it can maintain the speed it has. So let's set these two equal to each other, right, because that's what it means to have a balanced force. Um, and the left side is Q B B, The rights that is Q E. And of course, we can cancel out que and mathematically. This question is quite simple, because velocity is just equal to eat about a B, and we are given the magnitudes of both fields. So go ahead and substitute. Um, velocity then becomes 4.22 times 10 to the sixth meters per second.

Okay, folks, in this video, we're gonna be talking about this problem. A size muse. E particle Thistle's a unit of charge. It's micro columns five micro column. Particle moves through a region containing the uniform Byfield Negative 20 military surplus and the uniforms e felled 300 volts per meters at a certain instant. The velocity of the particle is this. And at that instant and in getting veteran notation what use the net PNM force on the particle. Well, this is a relatively straightforward problem to solve. Basically, where we're gonna be doing is we're gonna be doing the Lawrence Force, which is given by cute. I mean, sorry, Force Lawrence Force is gonna be given by q E plus V cross beat. So when when you have the particle that has a non zero charge Q. And it's moving in a surrounding the veil and a surrounding be filled than the amount of force. The amount of force acting on the particle is given by this formula, and that is what we're doing for this problem. So So let's plug numbers in here. We're the reason I say this is a relatively straightforward problem to solve. Is because we're already given a lot of the information we know. Q. Because Q is five mu columns, which is five times 10 to the negative six columns. And we know E because E is is in the wind direction. There's 300 votes per meter and we also know B. That means we know everything we know. B is where is be. Okay. B is here. Negative. 20 m. T. So, Anderson the ex directions or negative? 20 times 10 to the negative three and then 00 which is gonna be negative two times. 10 to the negative. Two jewelers, you know. Okay. And what about the velocity vector? Well, we also know the velocity vector, and that is 17 minus 11 and then seven. The whole thing times tend to the three because this is kilometers, not readers. So we're given these three vectors all we're looking for. It's the cross product of these two vectors, plus this vector, and that is the force. So this is a very, very simple problem. Too soft. Let's start plugging in numbers. I'm so cute. Five times 10 to the negative six. I'm he is 0 300 0 plus 17 minus lemon and then seven times Who let Mia right that somewhere else. Cross product with B Negative. Two times 10 to the negative, too. And then zero and zero, um, multiplied by 10 to the cute. Okay, this is the answer. All you need to do is do the cross product, and then you add it to here, and then you multiply it by this number and you're done. Let's set up the product. The cross product without actually doing it. Um, the cross product between B and B is I j K. And then V X is 17 10 cubed, minus 11. Time attendant cute seven times 10 today, cubed and then be is negative two times 10 to the negative, too. 00 This is the cross product. You're gonna evaluate this determinant. And if and then after that, added to the field and then you multiply by this number, that's all we're doing for this problem. And when you do that, you're going to find that the answer. Let me, uh, let me pull out the answer for you. Where is that? Um, that answer is Lydia is gonna be a very simple expression. That's 0.8 million Newtons in the deejay hat direction and then minus 1.1 million Newtons in the Z hat direction. This is the force that will you get this force is like I said, You do this cross product, you added to the E field and you multiply it by the Q. And you know, I don't I think this problem is mainly computational. There is not a big deal. Okay. Um all right, so that's the answer. I was seeing the next video.

So there are a couple approaches for this problem. Um, let's start with just writing down everything that we're given. We have a mass i m It is equal to 6.6 times 10 to the minus 27 kilograms. 27. And it's a good idea to keep track of S I units, make sure we don't need to convert anything. This is in kilograms, which is in units that are as I so that's good. There we're given the velocity of this traveling mass is 4.4 4.4 times 10 to the fifth and velocity is given in meters per second. Also, outside units don't need to convert anything. There were given a magnetic field that this particle is in be that is equal 2.75 Tesla. That's also it's a unit for magnetic field. So we don't need to change anything. And lastly were given a radius that's this particle. This mass is traveling in a circle, so it were given a radius. Little are This is 0.0 12 This is in meters also s I units. That's good. We don't need to change anything. All right. Off the bat. So now another point to note before we we start doing any any of the math for this problem is that everything that we're given has two significant figures. So that's what we want to keep all over answers to. We want to have two significant figures. So now, as I said at the beginning, there are two approaches to this problem. One of them is a little bit more complicated. Um, let's read out the equations that we have. So we have that force. The centripetal force F sub sea is equal to this is mass times velocity squared, divided by the radius. How centripetal force is the force that keeps it traveling in a circle. If there was no centripetal force, it would just fly off straight instead of continuing to travel in a circular motion. That's what this forces and we know that this is caused by this magnetic field. So this eps of see this f sub sea is the same f used in the magnetic, the force, the magnetic force equation. I'm just gonna rated f subsea good rytas as FCB. But since they are the same thing, I'm just gonna use the same notation here. So we have this other equation. Have subsea is equal to be, which is the magnetic field times to charge giant times the speed that it's traveling at the velocity. You might have seen this in another form, whereas it solved for B instead. It's just simple algebra moving this equation around. If you wanted to write it in that form, you keep the be on this side and move. Divide both sides by Q V. That would give you in the form B is equal toe FC divided by Q B. So these are the two equations that we have. The harder method would be to look at it, and we see that we have everything we need to calculate this centripetal force. Here. We can go through and plug in all the numbers and find this force. And then the physics of this problem is to equate this F with this F and know that they are the same force. It's the force that's causing it to stay in circular motion. The easier way to do this is we can write that if these EFS are equal to each other, we can simply just right. The two equations equal to each other and do algebra and make it a little bit easier. So these two Esther equal, which means he's the right hand side of the equations, are equal to each other. We have M V squared, divided by R is equal to the other right hand side. B Q V. Magnetic field times charge. Times velocity. I know algebra can be a little scary sometimes, but this saves us from having to plug in all of these numbers twice. And since some of these air really bigger, really small numbers, it can get really come for someone it saves on possibly making an in arithmetic it mistake. So now we could make this a lot nicer by eliminating some things here. So we see that this V will cancel with this and then all we're left with now will just divide both sides here. We want Oh, what? We're trying to find his cue. So we'll divide this be out. So this equation turns into I'm gonna flip it around, Alan. Right, Q. Because that's what we just saw. Four on the left hand side, which doesn't change anything in the equation. I'm just flipping it around que We've divided by B is equal to n divided by are be And another point. This was a velocity squared this since we divided by this velocity, it only cancelled out the square. So we still have a velocity in the numerator here. Now, this is much simpler to plug in to then doing this and then doing this. Now we just have one equation so we can plug in these numbers and it's all for the charge. So what? We get you plug in the mass the 6.6 times 10 to the minus 27th for em of the velocity. We have the radius and we have the magnetic field that gives you he charged. And we know these are all s I units. So this will come out to be in cool ums. Um, we get a factor of 3.2 times 10 to the minus 19. This isn't cool ums. We know that an electron is 1.6 times 10 to the minus team might not minus 19th Rather, um and this is exactly double that. So it's exactly two times the charge of electron or um, A plus in front of that just, basically, is like trying protons share the same charge. That's just basically talking about a profound. So it's It's twice the charge of a proton, so we have for a final answer. Que is equal to will. Write a plus just to indicate that it's a proton. Because electrons have a negative charge, this is equal to twice the charge, often electron, and there you have it.


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