# Problem Set 6: Solutions

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1 UNIVESITY OF ALABAMA Depatment of Physics and Astonomy PH 16-4 / LeClai Fall 28 Poblem Set 6: Solutions 1. Seway Potons having a kinetic enegy of 5. MeV ae moving in the positive x diection and ente a magnetic field B =.5 ˆk T diected out of the plane of the page and extending fom x = to x=1. m, as shown below. (a) Calculate the y component of the potons momentum as they leave the magnetic field. (b) Find the angle α between the initial velocity vecto afte the beam emeges fom the field. Note that 1 ev= J. Bout Figue 1: Poblem 1 Fist of all, we know that once the poton entes the egion of magnetic field it will follow a cicula path of adius, and once it leaves the egion it will once again move in a staight line path, tangential to the cicula path in the field egion. We will need to use some geomety to elate α to the given distance x and the adius of the cicula path. Then we will detemine the adius of the cicula path in tems of the known kinetic enegy, and we will be good to go... efe to the figue below: d 9-α β α Figue 2: Poblem 1 Solution The path of the potons in the egion of magnetic field is cicula, and descibed by a adius. The potons will move though the egion of magnetic field acoss its lateal distance d, and this will define an angle α, caving out an ac of a cicle of adius. We can define a ight tiangle by the cente of

2 the cicle defining the path in the magnetic field egion, the point at which the potons ente the field, and the point at which they leave. The angle at which the potons leave the field with espect to the hoizontal is then β =α based on the geomety of the figue above. Once we know, given d, we can find α by noting sin α=d/. The adius of the cicula path of the potons can be found by noting that the centipetal foce (keeping them in a cicula path) must be povided by the magnetic foce. Note that a poton has chage e and mass m p, and let the poton s velocity be v. Also ecall that magnetic foces do no wok, so the potons velocity will not change in magnitude afte passing though the egion of magnetic field. Since the motion of the paticle is always at a ight angle with espect to the field, we can just deal in magnitudes. F cent = m pv 2 = F B = evb sin θ Bv = evb = = m pv eb = p eb Hee p is the magnitude of the potons momentum. Now we have the adius of the path in tems of the field, the chage on a poton, and the potons momentum. We ae given the potons kinetic enegy K, which is elated to its momentum by K =p 2 /2m p. Thus, = p eb = 2mp K eb 6.46 m emembe that 1 ev= J to make the units come out popely. Given the adius, we can now find the angle α: sin α = d = α = sin 1 d 8.9 Now, since the magnetic foce does no wok, the potons momentum does not change in magnitude, so the initial and final momentum ae the same. The vetical (y) component of the potons momentum can thus be easily found: p y =p sin α p y = p sin α = 2m p K sin α kg m/s 2. Seway Conside an electon obiting a poton and maintained in a fixed cicula path of adius = m by the Coulomb foce. Teating the obiting chage as a cuent loop, calculate the esulting toque when the system is in a magnetic field of.4 T diected pependicula to the magnetic moment of the electon. Fist, need to know the cuent that coesponds to one obiting electon. Fom the cuent I, magnetic field B, and the obital adius we can find the toque. An electon in a cicula obit of adius has a peiod of T = 2π/v, whee v it he electon s velocity. If a single electon chage e obits once evey T seconds, then the cuent is by definition I = q t = e T = ev 2π

3 We can find the velocity fom the condition fo cicula motion. The only foce pesent (that we know of) is the electic foce, which must then povide the centipetal foce on the electon. The electic foce is just that of two point chages e and e sepaated by a distance. F cent F E = m ev 2 k ee 2 k e e 2 = = v = 2 m e We can now substitute this in ou expession fo cuent above: I = ev 2π = e k e e 2 2π m e = e2 ke 2π m e 3 Finally, since the magnetic field is pependicula to the electon s magnetic moment, the magnitude of the toque is given by τ = IAB whee A is the aea of the cuent loop" fomed by the obiting electon, A=π 2. Thus, τ = IAB = e2 ke 2π m e 3 π2 B = 1 ke 2 e2 B m e N m The negative sign eminds us that cuent is the diection that positive chage flows, and thus the diection of the toque is given by the ight hand ule consistent with the cuent, which is opposite the diection that the electon obits. 3. Ohanian 29.5 A wie lying along the x axis caies a cuent of 3 A in the +x diection. A poton at = 2.5 ŷ has instantaneous velocity v = 2. ˆx 3. ŷ + 4. ẑ, whee is in metes and v in metes pe second. What is the instantaneous magnetic foce on this poton? If the cuent flows along the ˆx diection, and the poton is diectly above the wie in the ŷ diection, then the magnetic field must be pointing along the ẑ diection at the poton s position. Thus, the magnetic field at the poton s position is given by B = µ oi 2π ẑ B z ẑ Finding the magnetic foce is now just a matte of calculating the coss poduct v B and multiplying by the poton s chage e. Fist, the coss poduct: ˆx ŷ ẑ v B = det v x v y v z B x B y B z = det ˆx ŷ ẑ 2 m/s 3 m/s 4 m/s B z = 3B z ˆx 2B z ŷ m/s Thus, the magnetic foce is F B = q v B = 3eB z ˆx 2eB z ŷ m/s ( ) (3 ˆx + 2 ŷ) N If you note that 1 T = 1 kg/s 2 A, you should be able to make the units come out popely. Fo

4 completion, the magnitude of the foce is then F B = ( ) N 4. Ohanian The electic field of a long, staight line of chage with λ coulombs pe mete is E = 2k eλ whee is the distance fom the wie. Suppose we move this line of chage paallel to itself at speed v. (a) The moving line of chage constitutes an electic cuent. What is the magnitude of this cuent? (b) What is the magnitude of the magnetic field poduced by this cuent? (c) Show that the magnitude of the magnetic field is popotional to the magnitude of the electic field, and find the constant of popotionality. The cuent can be found by thinking about how much chage passes though a given egion of space pe unit time. If we wee standing next to the wie, in a time t, the length of wie that passes by us would be v t. The coesponding chage is then q =λv t, and thus the cuent is I = q t = λv t = λv t Fom the cuent, we can easily find the magnetic field a distance fom the wie. B = µ oi 2π = µ oλv 2π If the wie wee sitting still (o we wee taveling paallel to it at the same velocity v), it would poduce the electic field given above. eaanging the given expession, we can elate λ and E, λ = E/2k e. Substituting this in ou expession fo the magnetic field, B = µ oλv 2π = µ oev 4πk e = µ oɛ o ve Fo the last step, we noted that ɛ o =1/4πk e. 5. Pucell 7.14 A metal cossba of mass m slides without fiction on two long paallel ails a distance b apat. A esisto is connected acoss the ails at one end; compaed with, the esistance of the ba and ails is negligible. Thee is a unifom field B pependicula to the plane of the figue. At time t=, the cossba is given a velocity v o towad the ight. What happens then? (a) Does the od eve stop moving? If so, when? (b) How fa does it go? (c) How about consevation of enegy? Hint: fist find the acceleation, and make use of an instantaneous balance of powe.

5 B in v o b Figue 3: Poblem 5 The moving od foms a closed loop with the ails, and once the od stats moving, the aea of this loop inceases with time. With a constant magnetic field, this means that the magnetic flux is inceasing with time, and theefoe thee must be an induced voltage. Let the position of the od be x, with the ˆx diection being to the ight, and the ŷ diection upwad. This means the magnetic field points in the ẑ diection, giving in a magnitude B. At time t=, we will say the od has velocity v o and position x o. Fo any time t, we will just call the velocity v and position x, since we don t know what they ae yet. The induced voltage can be found fom the magnetic flux though the loop, which is itself is easily found, since the magnetic field is constant and eveywhee pependicula to the plane of the loop. We only need the aea of the loop. If the width of the loop is b, and the position of the od is x, the aea is just bx, and that is enough to find the flux: Φ B = B d A = B d A = BA = Bxb loop loop The induced voltage is found fom the time vaiation of the flux via Faaday s law. The induced voltage - which is now applied to the esisto - will lead to a counteclockwise cuent in the loop, since it wants to stop the incease in flux by ceating a magnetic field opposing the extenal magnetic field. V = dφ B dt = Bb dx dt = Bbv = I The pesence of a cuent in the conducting od will lead to a magnetic foce. Since the field is into the page ( ẑ diection), and the cuent is flowing up though though the od (ŷ diection), the foce must be in the ŷ diection. F B = I L B = IbB ŷ ( ẑ) = IbB ˆx = B2 b 2 v = ma ecall that the diection of L is the same as the diection of the cuent. Since the magnetic foce is the only foce acting on the od (in the absence of fiction), it must also give the acceleation of the od, as indicted in the last step. Incidentally, we could have gotten hee much moe quickly with a little intuition. If we ecognize that thee must be a cuent flowing in the esisto due to the induced voltage caused by the motion of the od, then we know thee is powe dissipated in the esisto. This powe must be the same as that supplied to the od. The mechanical powe is F v, and the electical powe is I 2. Consevation of enegy equies that these two powes be equal, which along with the motional voltage leads diectly to the equation above.

6 Anyway: now we have a small equation elating v and its ate of change, dv/dt=a. We can solve it by sepaation of vaiables, which is totally cool since none of ou quantities ae zeo. Dividing by zeo is not cool. B2 b 2 v = mdv dt m dv B 2 b 2 v = dt Now we ve got something we can integate. Ou stating condition is velocity v o at time t ==, going until some late time t whee the velocity is v. v v o m dv t B 2 b 2 v = m B 2 b 2 ln v v v o = t m B 2 b 2 ln v v o = t dt t = v = v o e t/τ with τ = m B 2 b 2 The velocity is an exponentially deceasing function of time, which means it neve stops moving - the velocity appoaches, but does not each, zeo. The od will also appoach a final taget displacement in spite of this fact, which we can find eadily by integating the velocity. x = o v dt = o ( ) v o e t/τ = v o τe t/τ = v o τ = mv o B 2 b 2 Once again, if you note that 1 T = 1 kg/s 2 A and 1 V 1 A = 1 W, you should be able to make the units come out coectly. Finally, we can calculate the total electical enegy expended. The electical powe dissipated in the esisto is P = du/dt = I 2, so the tiny bit of potential enegy du expended in a time dt is du = I 2 dt. We can integate ove all times to find the total potential enegy. o U = I 2 dt = = B2 b 2 v 2 o = 1 2 mv2 ( Bbv ) 2 dt = B2 b 2 e 2t/τ dt = B2 b 2 v 2 o v 2 dt ( τ 2 e 2t/τ ) = B2 b 2 v 2 o ( ) m 2B 2 b 2 As we would expect fom consevation of enegy, all of the initial kinetic enegy of the conducting ba ends up dissipated in the esisto.

7 6. Seway A unifom magnetic field of magnitude.15 T is diected along the positive x axis. A positon (a positively-chaged electon) moving at m/s entes the field along a diection that makes an angle of 85 with the x axis. The motion of the paticle is expected to be a helix in this case. Calculate the pitch p and adius of the tajectoy. y v z 85 p B x Figue 4: Poblem 6 The fist thing to ealize is that a helix is basically a cuve descibed by cicula motion in one plane, in this case the y z plane, and linea motion along the pependicula diection, in this case the x axis. A helix of cicula adius a and pitch p can be descibed paametically by x(t) = pt 2π y(t) = a cos t z(t) = a sin t As we can see, the motion in the y z plane obeys y 2 + z 2 = a 2, descibing a cicle of adius a, and along the x axis we just have constant velocity motion. Since the x, y, and z motions ae uncoupled (e.g., the equation fo x(t) has no y s o z s in it), things ae in fact petty simple. The cicula motion comes fom the component of the velocity pependicula to the magnetic field, the component of velocity lying in the y z plane, which we will call v. The pitch is just how fa fowad along the x axis the paticle moves in one peiod of cicula motion T. Thus, if the velocity along the x axis is v x, p = v x T = (v cos 85 ) T We have aleady discoveed that the peiod and adius of cicula motion fo a paticle in a magnetic field does not depend on the paticle s velocity, it only mattes that thee is always a velocity component pependicula to the magnetic field T = 2πm qb and = mv qb Putting eveything togethe,

8 p = 2πmv cos m Bq = mv qb sin m By the way, hee is an inteesting tidbit fom MathWold: i A helix, sometimes also called a coil, is a cuve fo which the tangent makes a constant angle with a fixed line. The shotest path between two points on a cylinde (one not diectly above the othe) is a factional tun of a helix, as can be seen by cutting the cylinde along one of its sides, flattening it out, and noting that a staight line connecting the points becomes helical upon e-wapping. It is fo this eason that squiels chasing one anothe up and aound tee tunks follow helical paths. i

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