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(T) A charge pC has an instantaneous velocity 2 X 106i 3 X 106j IO6k m/s in a uniform field B 10-2i + 5 X 10-2j 3 X 10-%k T. What is the force on the charge? 8. (II...

Question

(T) A charge pC has an instantaneous velocity 2 X 106i 3 X 106j IO6k m/s in a uniform field B 10-2i + 5 X 10-2j 3 X 10-%k T. What is the force on the charge? 8. (II) When charge 4 -2 uC has an instantaneous veloc- ity (~i + 3j) X 106 m/s, it experiences a force F 3i j + 2k N. Determine the magnetic field, given that B, 0_ 9_ (I) An electron experiences a force F (-2i + 6j) x 10-13 N in magnetic field B ~1.2k T_ What is the velocity of the electron given that Uz 02 10. (I) An electron moving at 1

(T) A charge pC has an instantaneous velocity 2 X 106i 3 X 106j IO6k m/s in a uniform field B 10-2i + 5 X 10-2j 3 X 10-%k T. What is the force on the charge? 8. (II) When charge 4 -2 uC has an instantaneous veloc- ity (~i + 3j) X 106 m/s, it experiences a force F 3i j + 2k N. Determine the magnetic field, given that B, 0_ 9_ (I) An electron experiences a force F (-2i + 6j) x 10-13 N in magnetic field B ~1.2k T_ What is the velocity of the electron given that Uz 02 10. (I) An electron moving at 106i m/s in magnetic field expe- plane at 308 to riences force of 4 10-I4j N_ (a) What can you deduce +z direction_ about B with the given information? (b) Suppose the given +y axis_ the force were the maximum possible in the field_ What can you magnitude as deduce in this case? c field? M1_ (II) When a proton has a velocity v = (2i + 3j) X 10 m/s, it 2 X 106 m/s at experiences force F -1.28 10-13 k N. When its in Fig. 29.41. velocity is along the +z axis, it experiences force along T. (a) If B is the +X axis. What is the magnetic field?



Answers

What objects experience a force in an electric field? Chapter 23 gives the answer: any electric charge, stationary or moving, other than the charge that created the field. What creates an electric field? Any electric charge, stationary or moving, also as discussed in Chapter 23 . What objects experience a force in a magnetic field? An electric current or a moving electric charge other than the current or charge that created the field, as discovered in Chapter 29 . What creates a magnetic field? An electric current, as you found in Section 30.11, or a moving electric charge, as in this problem. (a) To display how a moving charge creates a magnetic field, consider a charge $q$ moving with velocity $\mathbf{v}$. Define the unit vector $\hat{\mathbf{r}}=\mathbf{r} / r$ to point from the charge to some location. Show that the magnetic field at that location is $$ \mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{q \mathbf{v} \times \hat{\mathbf{r}}}{r^{2}} $$ (b) Find the magnitude of the magnetic field $1.00 \mathrm{~mm}$ to the side of a proton moving at $2.00 \times 10^{7} \mathrm{~m} / \mathrm{s}$. (c) Find the magnetic force on a second proton at this point, moving with the same speed in the opposite direction. (d) Find the electric force on the second proton.

Hi. In the first part off the given problem here we consider charge in motion. Suppose the part off this George along which it is moving with a velocity V is given by this line segment the observation point here at which we have to find an expression for magnetic field at any instant. Their position off observation point is this given by are but the angle between the velocity vector and the position vector let it be t talk. So as for biotin severed slaw, the magnetic field do do a current element at an observation point is given us you not by 45 into i d l or we can say I'll the product off the cross product off the vector product off current element with the position vector the unit director off position vector then and divided by risk wired the magnet euro the distance off opposition point from the current element. So here we can replace this I with the formula for the mathematical formula for the current And as we know, the current is given as the time rate off flow off charge. So this is Cuba. I t and cross our gap divided by R Square. So if we rearrange the terms like this is Kyul El Bar divided by t across our camp divided by R Square. So this is the displacement off the charge per unit time which can be given us the velocity off discharge particle. Hence, the magnetic will produced by this moving charge will be given us new, not by four pi in tow. Q. We cross are kept divided by the square, and this is the answer for the first part off the problem. Now, in the second part of the problem, we have to find this magnetic field the magnitude of magnetic field at a distance of 1.0 millimeter to the side off proton if the charged particle is proton and it is moving with a speed off 2.0 into 10 days per seven meter per second. So in the second part of the problem, this is we having a magnet. You're off 2.0 in tow, 10 inch per 7 m per second. The distance off the observation point is 1.0 millimeter, So using the same expression, be bar is equal toe. Do not buy for pie que the bar across are kept divided by R Square. And as the charged particle is moving just for particularly, it is perpendicular to the observation point. So the displacement, the angle between this position rector and the velocity vector will be 90 degrees. So we can say this B is equal to mu, not by 45 into Q V. Sign 90 degree divided by R Square. So here it will be if you plug in all the known values here. This is B is equal to and dish par minus seven multiplied by charge, which is having a magnet. Europe 1.6 into 10 dish par minus 19 for the speed. This is 2.0 in tow. Tentage part 7 m per second for sign 90. This is one. And finally for our this is then raised to the power as the distances one millimeter. So this will be 10 dish par minus 3 m. And if the square is standish par minus six neither. So finally this magnetic field here comes out Toby equal to 3.2 in tow. 10 days to the power minus 13 Tesla. So this is the answer for the second part off the problem Not in the third part of the problem. We have to find the force on a second proton moving from this point. So this will be given as f B is equal. Toa que we be the magnetic Lawrence forced so as another charge is also a proton. So this is 1.6 in tow, 10 days par minus 19 Collomb again and it is passing through the same speed passing with the same speed two into 10 days for seven multiplied by the magnetic field which was 3.2 in tow. 10 dish par minus 13. So finally this force here comes out Toby one point 0 to 4 into 10 days apart. Miners 24 Newton The force experienced by another proton passing through the observation point. Finally, in the last part of the problem, we have to find electrostatic force off attraction off repulsion between these two protons at the same distance. So here it is. F e is going okay. Que square by our square. So it becomes nine in 2 10 days apart. Nine multiplied by 1.6 into 10 dish par minus 19. So the whole square divided by one into 10. Dish par minus three, the whole square. So finally this electrostatic force of revulsion comes out to be. Do point three in tow, 10 dish par minus 22 Newton. And this is the answer for the last part off the problem. Thank you.

First, we're going to find that ideas are just given by everything over your hand. This a question Guns from the Newton's law, which is, like forces a greater mass times acceleration on DH. The case where the Spartans are travelling along a circle allows them circle do get imaginative field, since equal toe to force is given by Q times tanks being because sign Fei is one here. This is going to the centripetal force, so master insistent it of acceleration. With this, we scare over I and this gives after the credible and three bullets cubing they questioned that you're using here. So Eustis accretion and regard this value off our radios to be 5.1 for Peter now for the second part, if you see figure 27.81 from the spoke, uh, distance along the guard are small is given by capital at times, So this is the smallest physically the Ark. Life is given by the radius times the angle that the are accepting substance at the center of the circle, so this gifts tinta to equal door cedar point 25 Move 5.14 meter. Aunt Tita is nearly called Assigned Tater for small angle measurement, which is It's over here on this gifts. Sign Tita to be equal to 0.5 meters over 51 one for me. And get the scripts to tactically 2.79 Decree our ingredients. This will be okay. Tita is basically not small. Really small here. It's okay. Not today. Small. So heads going back to what we did? Yes, on ingredients. This's equal to zero point. See, You're for eggs. Six radiance. I'm Since the velocity is constant here, the time will be given by ah distance over the velocity on DH. Get this distant. We get the time to be 1.72 thanks. Tank to defy minus six tickets. But the third part, we have to find the X one on that is given by So don't explain is the horizontal deflection at the point off exit from the from the field on DH. So the fine died. We have to multiply that damn teeth out. All right. And this is and we used the data from here. We get this new often paddocks one to be equal to six points to donate. Langston did the body minus three meter. For the last part, we have to find out that exit is still excellent. Plus Delta X two and their legs to is the horizontal displacement of the particle from married exits. The field region toe where it hits the water on their likes to should be given by Capital D dying stamp, Tita And we can't dispel youto be syrup ones. Either two or four meter Andi, we use the value that we found here and here over this. On this, a question and we get an X to the cold Cedar point, please. I think you know this over here is being smart. Lee is much less than capitalized. So the horizontal reflection off the particle is much smaller than the distance it travels in the lie detection.

To see something put. Mission is given in this problem. Your source off Electric Village Judge I did it. Rest own in motion. It doesn't matter. But sort off my great village moving judge are cutting. Not in this question. In the first part, we have toe calculate magnetic field intensity due to a movie judge and chicken parts. We have a local public value up magnetic will intensity do toe movie bit validity off doing 10 to the power seven. Meet up our ticket after white but millimeter promised spot in turn or C park Be able to calculate the ports off interaction between true prevalence, Moving it upload your direction and in the deeper they have to calculate the forethought. That is, if this is magnetic force electric ports, we have to calculate on the second proton due to past due to second on the second platoon that these are the the best part of this question which I put when we start solving with obstinate support charge you it moving along this but bit by little TV on. We are finding the magnetic field intensity at three point having w Shin bet that hot as you know that better. The very current element and I couldn't is flowing through it than magnetic field intensity at any point can be given by I be it because our unit vector divided by r squared Why should I loved this concept? Give will apply here. Let us see you the product off current and it's element be it's can be originate as you know Currently defined it Charge flowing body written right and be a support duty The velocity covered by the charge But in the time so here be your support beauty candidate in it velocity of the judge Here I am writing separately Felicity can be defined a the spits covered by the church But he did time So this concept we are using substitute this value Any question But the magnetic reading density at any point due to moving judge, we can write you not upon poor fight I can get in it You Are you in ideas in Britain as you we crossed be 30. Would you submit that are divided bite? How do you square? So this is about as there are a bust. So if any charges moving, then it will creates a magnetic field around and which could be measured by this bar. So this is very important relations. No integral part in group would want odd moving Cuddy in second part. If Proton is moving, been desperate up. Can do the power. 6 30 With that speed up, two into 10 to the power of it. Meet up our ticket. Then we have to calculate the magnetic. Me? Let me point. Which is that a separation off? Which is that the separation off? One millimeter That is 10 to the power off minus three. Bitter. We used the same relation. Magnetic field can be given. What you not. Upon bullfight, you re cross me, uh, unit acted upon. Are you square in our substitute to calculate the magnetic Quit. We're calculating here. Only the magnitude. So Ford by in tow tend to de power off minus seven upon port by charge on the protons. But 0.6 in 10 departments 90. The spirit is given two into 10 to the power seven on. Dispenses tend to depart my industry only square on bill. Between these two, it's 19. But this is a sign off. 90 on solving you will get Magarri feeling pencil. It will be 3.2 into Tend to deposit off 30 Islam. So this is the answer up. Second part? No. In the third particles given, unforgettably moving in this direction. Big publicity. Two into 10 to develop. Save it, meet a per second. And that is moving in a positive direction with that same validity at the separation off one millimeter. So you have to cooperate the force off interaction with me them before disability. You Sorry. Just a moment, please. Who is going to find it? She window the cross Me, They will find the magnitude charge on the protons spirits doing 10 to the power seven on magnetic will be every calculated on angle will be 90 degree. So this course you will get but white You don't go into 10 to the power up minus country for nuclear and they will depend each other. So they're moving in a positive direction. So they are repelling. So there will be weapons imports between that in their deeper we have to calculate forced to do electric field would be there. There will be electricity, so they will be there will be a port with them. Okay, Cuban butto by artists quit nine in 10 to the other night, 1.6 into 10 to the power off minus 90 1.6 in 10 to the power off 90. His but millimeters that it's tend to the poor by the three Holy square. So the port on. But we're done on other booby 2.3 and 10 to the power minus T. And they are repelling that This is the repelling puts here. I have to make a correction. I have written drunk because they are moving in a positive direction. The force will be attractive. I'm really very sorry. I am doing the correction here. Magnetic ports will be attractive, but electric port will be repulsive.

And this problem, the first part is asking us under is asking us to understand how moving charged also created. Okay, as we're moving feel so we want to use a particle cube moving with velocity V defying the position vector are hat. We want to show that what the magnetic field at the location is equal to. So in this case, we're gonna use the fact that the, um indefinitely indefinitely Small magnetic field D B is equal to mu, not over four pi r squared I ds times are so it's a a infinitely small region and an infinitely small magnetic field. We also know that our current is equal to, um are total number of charged particles in this area multiplied by the size of the area multiplied by charge, times velocity. And so for a, um, for a positive current charge, her first are for a positive charge. The direction of DS is going to be in the same direction as the and we also know that a D s is the volume occupied by the moving charges and it's equal toe one if we only have a singular charge. So in this case, that means that be must be equal to him, you know, divided by four pi times, Q v times are divided by R squared resorted to Cuba. Times are hat divided by r squared. And so we've shown that this part of the problem is true. The next part is that he got to find the magnitude of the magnetic field one millimeter to the side of a proton, moving at a given speed. To do this, we're once again going to use the equation that we just found. And we know in this case that, um, our that we're are sign of data so that would be here is going to be signed of 90 which is one since velocity is always perpendicular to magnetic field by the right hand rule. So we're gonna plug in are constants which really just include our, which is really just aren't radius. So that's tending the negative three meters that's gonna be plugged in for R squared as well as the charge on a proton. And when we do this out, we find a magnetic field of 3.2 times 10 to the negative 13 Tesla Now part C is asking us to find the force on the second proton at this point, if it's moving with the same speed in the opposite direction. So that magnetic force given by the equation Q v B we're gonna plug in the charge of a proton, which is 1.6 of him and negative 19 cool arms are velocity. Um, that are person is going to be moving at 2 10% of the seven meters per second and our magnetic field We do this, we get a total force of 1.24 times 10 to the negative. 24 students last for the problem is asked needs to find the electric force on the second proton eso. The electric force we know is just equal to um que e. So that's our constant K times queues have a first charge him. The second charge, divided by r squared. Now I tried to do the same. In this case is both of the murders the person of the charge of Proton. This is really just the same thing as q P River Port on squared que issued a constant and our is a distributor charges which is still, um one millimeters so we're gonna have 10 to the negative three meters squared and denominator and that's going to give us a final result of 2.30 times 10 to the negative 22 Newtons.


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