5

A proton (charge 1.6 X 10-19 € and mass 1.67 X 10-27 kg) moves with a velocity v through a region where both a uniform magnetic field (B = ~j + k T) and a uni...

Question

A proton (charge 1.6 X 10-19 € and mass 1.67 X 10-27 kg) moves with a velocity v through a region where both a uniform magnetic field (B = ~j + k T) and a uniform electric field (E ~2i +j + 3k NJC) exist: Calculate: (a) the velocity of the proton so that the proton moves in a circular trajectory of constant speed: () the radius of the circular path

A proton (charge 1.6 X 10-19 € and mass 1.67 X 10-27 kg) moves with a velocity v through a region where both a uniform magnetic field (B = ~j + k T) and a uniform electric field (E ~2i +j + 3k NJC) exist: Calculate: (a) the velocity of the proton so that the proton moves in a circular trajectory of constant speed: () the radius of the circular path



Answers

A proton (charge $+e,$ mass $m_{p} ),$ a deuteron (charge $+e$, mass $2 m_{p} ),$ and an alpha particle (charge $+2 e,$ mass 4$m_{p} )$ are accelerated from rest through a common potential difference $\Delta V$ . Each of the particles enters a uniform magnetic field $\overrightarrow{\mathbf{B}},$ with its velocity in a direction perpendicular to $\overrightarrow{\mathbf{B}}$. The proton moves in a circular path of radius $r_{p}$ . In terms of $r_{p},$ determine (a) the radius $r_{d}$ of the circular orbit for the deuteron and (b) the radius $r_{\alpha}$ for the alpha particle.

For this question. Rats to find the radius of circular orbit for do Theron in part A in terms of the radius of orbit of proton in a magnetic field where the velocity is perpendicular to the magnetic field. Okay, so to do this first, let's find out what the radius of the proton or the radius of a charged particle in general and a magnetic field is going to be. So the law of conservation of energy states that the absence of non conservative forces, the total energy of the system, remains constant. So therefore, the change in kinetic energy Delta K plus the change of potential bell to you is equal to zero. Or, in other words, the potential energy Delta U was equal to negative Delta K work Delta K is, um okay, final minus K initial finals. The final kinetic energy and K initial is the initial kinetic energy. Okay, but since the particle comes at rest, its final kinetic energy is zero, so we can let that go to zero. So therefore, the change of potential energy dull to you is equal to minus negative K I or just K II, which is the initial kinetic energy. So the energy possessed by a charged particle moving by virtue of motion. The kinetic energy here is equal to 1/2 in the squared. So dull to you, then is equal that, since it's equal to case of eye, is equal to also equal to 1/2 and the square okay. But the potential energy acquired by a charged particle that was accelerated through a potential difference is equal to Q times. The change in, uh, voltage potential Delta V. So we have dealt to you. There's also equal to the charge of the particle Q times The change in potential voltage Delta V We can set these equal to each other and sulfur V. Let me find that V here is equal to the square root two Claims Cube. I'm still TV, and this is velocity. So let's maybe indicated difference between velocity and voltage here, so we'll use little V for velocity and big B for voltage. That way you can see the distinction between them, so velocity is equal to two q Q times to change in the potential voltage Delta V, divided by the Mass m. Okay, but we're asked to find the radius of the orbit. So the magnetic field in this temple centripetal force balance in this situation. So therefore we have F sub C is equal to half seven, which f sub C is equal to in times V squared over our and F's and M is equal to the charge times. Um, the velocity V yeah, this is also velocity. So let's keep using little B times magnetic field times the side of the England between. But we're told the angle is data or is 90 degrees and the sign of 90 is one, so we can ignore that. We can rearrange this to Fox Sulphur, are we find that are here is equal to mm. Divided by Q B. Um, we can cancel out to use some of these voltages velocities here. Okay, And then for the velocity will plug in the term we found above. So this is multiplied by the square root of two times Q times. Delta V over in. Okay, and we're gonna use some simple algebraic properties here to simplify this a little bit more. So, for instance, route Q over Q is equal to one over route Q and em over route M is equal to room em. So using those facts we're gonna simplify are a little bit more We find that are here is equal to two times em times Delta V divided by q b squared all under the square root. So what we found at the bottom of this page is a public to that just with a little bit of algebraic rules there. So now we can substitute in our value for let's say, this is starting part ay, here we can substitute in our value for our sippy since you just replace everything in that with proton related. So this is gonna be two times the mass of the proton m stumpy times Delta v divided by, uh, the charge of the proton, which is the same as the charge of the electrons. So we could just write e times b squared. Okay, well, now, let's compare that to what would be the radius of orbit for the deuterium. Our sub d is going to be equal to the square root of two times the mass of its teary, um m sub d. The potential difference is the same Delta V divided by the charge of deuterium called excuse of D A times B squared. Okay, well, you can look these values up. Um, but I'll just tell them to. You hear the mass of deuterium is two times the mass of the proton. So this is two times s a p. I'm still TV divided by the charge of deuterium is the same as the charms of the proton. It's e times b squared. So if you compare that to arson p above you find that this is route two times more than next. You can pull that route to that gets multiplied by the M C. P out front so would take this route to we can pull it out front. And this is just gonna be rude to times arson p makin box. It is their solution for a for part B rasa do the exact same thing except for this time with an Alfa particle. So we have our save. Alfa is equal to the square root. It's right there of two times in some Alfa. Sometimes the change of potential divided by the charge of Alfa Times. Be square. Okay, so now let's put these in terms of the mass of the proton in the charge of the proton. So this is too times the mass of the alpha particle is four times the mass of the proton. Okay, time still to be divided by the charge of the Alfa particle is two times the charge of the proton. So this would be too time z times b squared. Okay, so we can pull that the extra four and the extra two that we get out. So we have root for over root to the root for over route to is just route to So this is also equal thio route too times the radius of the proton making box that in is their solution to part B.

Hello, everyone In this problem, it is given. There is a Proton having that judge. Let's see. And much empty. Yeah, you thought same charge as that Operadora. But mustard two times that of Porto And at a particle started last week, Mass is four times that of proto de each Particles are excellent. It all These are accelerated by, say, importation different off dead people and productive in the same magnetic pre law, we have to compare the radius off road you, Katya and Bob article first in general, people discuss in any charged particles is to be accelerated at getting people to eat elephants then and our chicken by it his photo hop and be split. So velocity off the charged particle is you know No, you build a B. Why am if any charge? Particulate productive, but particularly to magnetic mean? Then it Describe the circular part off radius and be a point. Thank you. Secure the velocity. Here go. No que delta v upon em and be ready by you. So really, it's after player. But you will get school and they're Dubbie to wear goodbye. We took you So this relation people use to find the radios off road. You'd run and grab our tickets to Marsa Proto upon really angry for everybody. Because really, it's often sorry you draw is two in tow. Don't be apart reading. Hi, It is off. Robotic it in tow four times off parts of Britain. They ve apart being took we if we compare in question Want to every comparing what? Who Every we will get radius off you trot to be rude Full time top Really? It up wrote one and really is off Help articulate Also your time's up. Really? It's off photo. So this is the answer for this problem. What?

In this problem on the topic of magnetic fields were told that a proton with the charge E. And mass M. P. A. Due to Iran with charge E. And mass to Mp. And an alpha particle with the charge to E. And mass four times that of the proton are all accelerated through a common potential difference, which is doubt avi the particles enter a uniform magnetic field B. The velocity that is perpendicular to the direction of the magnetic field. The proton has a circular path of radius R. P. And we want to find the radio of the circular orbits for the veteran and the other particle in terms of the radius of the path for the proton. Now we know that the that Q times delta V is equal to a half M. V squared. And from here we can find the speed of the particle V to be the square root of two times its charge. Q times the potential difference delta V divided by the particles mess em And we also know that the magnetic force Q V B provides a centripetal motion. So it's equal to the centripetal force M V squared over R. Which means that the radius of the circular motion R. Is equal to M V over Q B. Which we can write as M over Q B times the square root of two Q times delta V divided by M. Which we can simplify as the square root of to M times delta V divided by Q B squared. And so for the proton we have R P squared is equal to two M P delta V divided by the charge of the proton, E times the magnitude of the magnetic field B squared. And then for the day Tehran we have our D squared is equal to two mg times delta V divided by he charged the due to Rome times B squared. And according to the given values, this is too into two times the mass of the proton. And the potential difference is the same dot A V divided by the charge of the doctrine, which is E times B squared. And so this is equal to to into two mp delta V of E B squared, which we can see is two times the radius squared for the proton too, R P squared. And then lastly, the radius for the alpha particle squared is equal to two times the mass of the alpha particle. And the potential difference delta V divided by the charge on the alpha particle times B squared. And this is to into four times the mass of the proton, times the potential difference data V divided by two times the charge of the electron times B squared. And so this simplifies into two. Did too, M P delta V divided by E B squared, which again is to times are P squared. And so therefore the radius of the circular orbit for the alpha particle is equal to the radius of the circular orbit for the do Tehran, and this is equal to you, scared out of two times the radius of the circular orbit for the proton.

The magnetic forces given by Q V B who it should act as the centripetal force and three square over our So from here, we can find that V is given by you be are over em. So in this particle er case, the velocity of the protons will be charged for Proton magnetic field radius and the mass of a proton solving for it. It comes out to be Siriporn 072 time stand to the eight meters per second, so that is 7.2 times turned to their six meter per second.


Similar Solved Questions

5 answers
ExBlms Khe mtegrand as & sum of partial iractions and evaluate the integral Ixb [7 110xT25 dxIx2 IDIAL 125 m2| Hiox + 75 In |x +5/ + x+5 +C 175 M125 KDHBi75 In |xm 25|+ T C Ix +5 Kxi+ 5)24 4 1125 Dlci Mox] 175 In |x+5|1 #C 2 Kx+ 5424 @DL 7 25 MtOx + 15 In |x +5/ X+5 + C0
ExBlms Khe mtegrand as & sum of partial iractions and evaluate the integral Ixb [7 110xT25 dx Ix2 IDIAL 125 m2| Hiox + 75 In |x +5/ + x+5 +C 175 M125 KDHBi75 In |xm 25|+ T C Ix +5 Kxi+ 5)24 4 1125 Dlci Mox] 175 In |x+5|1 #C 2 Kx+ 5424 @DL 7 25 MtOx + 15 In |x +5/ X+5 + C 0...
5 answers
Find a29 the twenty-ninth term of the geometric sequence whose first term is a1 6,000 and whose common ratio is r = 2a29(Type decimal rounded to nine decimal places )
Find a29 the twenty-ninth term of the geometric sequence whose first term is a1 6,000 and whose common ratio is r = 2 a29 (Type decimal rounded to nine decimal places )...
5 answers
0 44Cecmneat 5 thc >equry d thr coudl o7 tor prduting % Tahoh prleend lon coeficn-t d destnrtan Ininto #hat +e LC 4 pl 0 uerzIraton nJcntetheIr Icy cllct d ducrnat n AchoRtnt [email protected]=Cmtty Onhrede lio 40-Hienc Dido C D5| Tha hijh pyaua rd law 0-E0e9E deemmnaten |edicita [Rat pe urd tgh cerictnt $ duet neton ndearathat bd pdurdlr cealljent d eeetinjtkc Et elesF = bm Draalurdhoh coonaerr ddetdrnratin Irdcenetn
0 44 Cecmneat 5 thc >equry d thr coudl o7 tor prduting % Tahoh prleend lon coeficn-t d destnrtan Ininto #hat +e LC 4 pl 0 uerzIraton nJcntetheIr Icy cllct d ducrnat n AchoRtnt ddeceetn FVLiti @Faketm= Cmtty Onhrede lio 40-Hienc Dido C D5| Tha hijh pyaua rd law 0-E0e9E deemmnaten |edicita [Rat pe...
5 answers
Section 4.9 Math 1550, section 33, Fall 20190/7.69 points Previous Answers SCalcET8 Find f.f '(x) = 2/V1-x2, (2) = 8f(x)Enhanced FeedbackPlease try again, keeping in mind that an antiderivative arbitrary constant C:Need Help?Retd ItJkid Qer~17.69 points SCalcET8 4.9.037.Find ftanttul1 <t< %,
Section 4.9 Math 1550, section 33, Fall 2019 0/7.69 points Previous Answers SCalcET8 Find f. f '(x) = 2/V1-x2, (2) = 8 f(x) Enhanced Feedback Please try again, keeping in mind that an antiderivative arbitrary constant C: Need Help? Retd It Jkid Qer ~17.69 points SCalcET8 4.9.037. Find f tanttu...
5 answers
10) A study of the system 4NHxg) 502(g) ANO(g) 6HzO(g) , A system was prepared with was carried out '[NHs] [O:] = 5.60 M as the only At equilibrium; INO] is 0.36 M comporents Initially. Calculate the oquilibrium coricentralion of 0z,
10) A study of the system 4NHxg) 502(g) ANO(g) 6HzO(g) , A system was prepared with was carried out '[NHs] [O:] = 5.60 M as the only At equilibrium; INO] is 0.36 M comporents Initially. Calculate the oquilibrium coricentralion of 0z,...
1 answers
In Exercises $17-58,$ solve the given problems. Part of a railroad track follows a circular arc with a central angle of $28.0^{\circ} .$ If the radius of the arc of the inner rail is $93.67 \mathrm{ft}$ and the rails are $4.71 \mathrm{ft}$ apart, how much longer is the outer rail than the inner rail?
In Exercises $17-58,$ solve the given problems. Part of a railroad track follows a circular arc with a central angle of $28.0^{\circ} .$ If the radius of the arc of the inner rail is $93.67 \mathrm{ft}$ and the rails are $4.71 \mathrm{ft}$ apart, how much longer is the outer rail than the inner rail...
5 answers
QUESTION 15Which of the following statements about polar molecules is false? OA a molecule with polar bonds can be nonpolar. 0 B. A molecule with polar bonds can be polar C A molecule with only nonpolar bonds can be polar: D.A molecule with 1200 bond angles can be polar E A molecule with 1809 bond angles can be polar;QUESTION 16How many lone pairs of electrons are on the Xe atom in the XeFa molecule? 0A oneC threeE: zerotWofour
QUESTION 15 Which of the following statements about polar molecules is false? OA a molecule with polar bonds can be nonpolar. 0 B. A molecule with polar bonds can be polar C A molecule with only nonpolar bonds can be polar: D.A molecule with 1200 bond angles can be polar E A molecule with 1809 bond ...
5 answers
METALLIC CRYSTAL STRUCTURE. A newly discovered metal has density5.36 g/cm3, and crystallizes as a face-centered cubic with a celledge length of 565 pm. What is the molar mass of thismetal? Express your result in g/mol.
METALLIC CRYSTAL STRUCTURE. A newly discovered metal has density 5.36 g/cm3, and crystallizes as a face-centered cubic with a cell edge length of 565 pm. What is the molar mass of this metal? Express your result in g/mol....
5 answers
XeCl5 hybridization
XeCl5 hybridization...
5 answers
Che derlyatiyetne functlcn_64 arcsInNeed Help?snow Ely "k kuanSioMt AMiet[-71 Points]DETAILSLARCALC1O 5.6.059.MI:equatlon the tangent Iinegraph of the functlon at the Diven Pol7t3rcsirNeed Help?snow Ely "k kuan33.0 [-/1 Points]DETAILSLARCALC1O 5.6.071.ML:any relatlve extrem?function. (Round Your answersthree declmal places.)ix) arctan x) ~rcanirrelatlve Maximum (x, Y)
che derlyatiye tne functlcn_ 64 arcsIn Need Help? snow Ely "k kuan SioMt AMiet [-71 Points] DETAILS LARCALC1O 5.6.059.MI: equatlon the tangent Iine graph of the functlon at the Diven Pol7t 3rcsir Need Help? snow Ely "k kuan 33.0 [-/1 Points] DETAILS LARCALC1O 5.6.071.ML: any relatlve extre...
5 answers
74x2 +74x+74 If x3_x+X-1Bx+C x2+1X-1thenA =
74x2 +74x+74 If x3_x+X-1 Bx+C x2+1 X-1 then A =...
5 answers
75. Five PointsFrom the interval (0,1), five points are selected at random andindependently. What is the probability that(a) at least two of them are less than 1/3;(b) the first decimal point of exactly two of them is 3?
75. Five Points From the interval (0,1), five points are selected at random and independently. What is the probability that (a) at least two of them are less than 1/3; (b) the first decimal point of exactly two of them is 3?...
5 answers
Warren wants to enclose rectangular field with five internal partitions as shown_ If there Is 13600 feet available for fencing, what dimensions will produce the greatest area? Note that the Internal partitions also require fencing-lengthThe width should befeet and the length should be(eet: (Round your answers to two decimal places)
Warren wants to enclose rectangular field with five internal partitions as shown_ If there Is 13600 feet available for fencing, what dimensions will produce the greatest area? Note that the Internal partitions also require fencing- length The width should be feet and the length should be (eet: (Roun...
5 answers
Which of the following Is not a valid solution for the equation sin 0 cos 0 + San 0 = 0 M 0 < 0 < 270 ?
Which of the following Is not a valid solution for the equation sin 0 cos 0 + San 0 = 0 M 0 < 0 < 270 ?...

-- 0.020848--