University of Albm Deprtment of Physics nd Astronomy PH 102 / LeClir Summer 2012 nstructions: Problem Set 4: Mostly Mgnetic 1. Answer ll questions below. Show your work for full credit. 2. All problems re due Tuesdy 24 July 2012 by the end of the dy (11:59pm if electroniclly submitted, by 5pm s hrd copy) 3. You my collborte, but everyone must turn in their own work. 1. A uniform mgnetic field of mgnitude 0.150 T is directed long the positive x xis. A positron ( positively-chrged electron) moving t 5.00 10 6 m/s enters the field long direction tht mkes n ngle of 85 with the x xis. The motion of the prticle is expected to be helix in this cse. Clculte the pitch p nd rdius r of the trjectory. y v z 85 r p B x The first thing to relize is tht helix is bsiclly curve described by circulr motion in one plne, in this cse the y z plne, nd liner motion long the perpendiculr direction, in this cse the x xis. A helix of circulr rdius nd pitch p cn be described prmetriclly by x(t) = pt 2π y(t) = cos t z(t) = sin t As we cn see, the motion in the y z plne obeys y 2 + z 2 = 2, describing circle of rdius, nd long the x xis we just hve constnt velocity motion. Since the x, y, nd z motions re uncoupled (e.g., the eqution for x(t) hs no y s or z s in it), things re in fct pretty simple.
The circulr motion comes from the component of the velocity perpendiculr to the mgnetic field, the component of velocity lying in the y z plne, which we will cll v. The pitch is just how fr forwrd long the x xis the prticle moves in one period of circulr motion T. Thus, if the velocity long the x xis is v x, p = v x T = (v cos 85 ) T We hve lredy discovered tht the period nd rdius of circulr motion for prticle in mgnetic field does not depend on the prticle s velocity, it only mtters tht there is lwys velocity component perpendiculr to the mgnetic field T = 2πm qb nd r = mv qb Putting everything together, p = 2πmv cos 85 1.04 10 4 m Bq r = mv qb sin 85 1.89 10 4 m By the wy, here is n interesting tidbit from MthWorld: i A helix, sometimes lso clled coil, is curve for which the tngent mkes constnt ngle with fixed line. The shortest pth between two points on cylinder (one not directly bove the other) is frctionl turn of helix, s cn be seen by cutting the cylinder long one of its sides, flttening it out, nd noting tht stright line connecting the points becomes helicl upon re-wrpping. t is for this reson tht squirrels chsing one nother up nd round tree trunks follow helicl pths. 2. Find the mgnetic field t point P for ech of current configurtions shown below. Hint for : Mgnetic due to the stright portions is zero t P. Hint for b: Two hlf-infinite wires mke one infinite stright wire. Hint c: use superposition nd symmetry! () The esiest wy to do solve this is by superposition our odd current loop is just the sme s two qurter circles plus two smll stright segments. We know tht the mgnetic field t the center of full circulr loop of rdius r crrying current is B = µ o 2r (loop rdius r) Since the mgnetic field obeys superposition, we could just s well sy tht our full circle is built i http://mthworld.wolfrm.com/helix.html
R P P b () (b) out of four equivlent qurter circles! The field from ech qurter circle, by symmetry, must be one qurter of the totl field, so the field t the center of qurter circle must simply be B = µ o 8r (qurter circle, rdius r) n other words: qurter circle gives you qurter of the field of full circle. Here we hve two qurter circle current segments contributing to the mgnetic field t P : one of rdius b, nd one of rdius. The currents re in the opposite directions for the two loops, so their fields re in opposing directions. The outer loop of rdius b hs its field pointing into the pge, nd the inner loop of rdius hs its field pointing out of the pge, so if we (rbitrrily) cll in plne direction positive, we cn subtrct the field of the smller loop from tht of the lrger. Wht bout the stright bits of wire? For those segments, the direction field is zero. Since the mgnetic field circultes round the wire, long the wire xis it must be zero. Even if it were not, by symmetry the two stright bits would hve to give equl nd opposite contributions nd cncel ech other nywy. There is no field contribution t P from the stright segments! Thus, the totl field is just tht due to the qurter circle bits, B = µ o 8b µ o 8 = µ ( o 1 8 b 1 ) (b) There is sneky wy to solve this problem using symmetry. Qulittively, we cn immeditely observe tht the field must point out of the plne of the pge. Think of the hirpin s being broken up into three sections: n upper semi-infinte wire, hlf circle, nd lower semi-infinite wire. All three segments give the sme direction of field t P. Further, if we were to rotte the entire hirpin in the plne of the pge, this will not chnge. Do tht in your hed once... rotte the entire setup, sy, 90 clockwise, nd you will find tht the mgnetic field t P will not chnge. f this is the cse, let us consider the prticulr cse where we hve the sme rrngement of wires rotted full 180, nd dd this to the existing setup, s shown below: Since both the norml nd rotted hirpins give the sme field, this rrngement just gives
P b us twice the field of the originl rrngement. With this rrngement, we cn brek the problem down into two infinite wires nd single circulr loop of rdius r. We hve lredy clculted the field due to stright wires nd loops; t point P, these three fields superimpose, nd the clcultion is trivil: 2B = µ o 2πb + µ o 2πb + µ o 2b = B = µ ( o 1 + 2 ) 4πb π 3. Find the force on squre loop (side ) plced s shown below, ner n infinite stright wire. Both loop nd wire crry stedy current. d We know how to find the force between prllel segments of wire. f the prllel segments re of length l, we know F = µ o 1 2 l 2πd (1) Here we hve top prllel segment of length distnce d+ wy, nd bottom prllel segment distnce d wy. The bottom segment hs its current ntiprllel to the min wire, mking this repulsive force, while the top segment hs its current prllel to the min wire, giving n ttrctive force. n finding the net force, this mens we should subtrct the force on the top segment from
tht on the bottom, since they ct in opposite directions. Wht bout the sides? We hve to think crefully bout the directions now. Chrges in the left segment move t right ngle to the field from the min wire (which pokes out of the pge t the position of the left segment), nd will feel force to the right. Chrges in the right segment, by the sme logic, will feel force to the left. These forces will be equl nd opposite, nd the net effect on the loop, if it is rigid, will cncel. (Were the loop flexible, it would be squished long the horizontl direction. Overll, the net force is just the difference between tht on the top segment nd tht on the bottom: F net = F bott F top = µ o 2 2πd µ o 2 2π (d + ) = µ o 2 ( 1 2π d 1 ) = µ o 2 2 d + 2π d (d + ) (2) 4. Wht is the induced EMF between the ends of the wingtips of Boeing 737 when it is flying over the mgnetic north pole? The internet hs most of the numbers you require. The induced voltge ii cn be found by considering the motion of the conducting metl plne in perpendiculr mgnetic field, nd mking few seemingly outlndish (but justifible) ssumptions. First, t the south mgnetic pole, the mgnetic field will be essentilly stright down. f the 737 is flying level over the ground, this mens tht its metl (conducting) skin is in motion reltive to mgnetic field. This in turn mens tht there will be motionlly-induced voltge. f the field is stright down, nd the 737 trvels stright forwrd, then positive chrges will experience force in the port (left) direction, nd negtive chrges towrd the strbord (right). This mens tht the wingtips will hve potentil difference between them due to the mgnetic force on the chrges in the conducting skin. f the wingspn is l meters, the irplne s velocity v nd the verticl mgnetic field B, then we know the potentil difference due to motion in mgnetic field is V =Blv. The wingspn of 737 is roughly 30 m, nd its cruising speed is bout 200 m/s. iii erth s mgnetic field iv t the south mgnetic pole v is bout 60 µt. Putting this together, Currently, the V = Blv = (60 µt) (30 m) (200 m/s) 0.36 V ii try to void using the term EMF nd usully just use voltge insted. EMF is bit ntiquted nd tends to confuse students in my opinion. f you see EMF just red it s potentil difference or voltge. iii http://en.wikipedi.org/wiki/boeing_737 iv http://www.ngdc.no.gov/geomg/mgfield.shtml v http://en.wikipedi.org/wiki/south_mgnetic_pole
5. Show tht, if the condition R 1 R 2 =L/C is stisfied by the components of the circuit below, the difference in voltge between points A nd B will be zero t ny frequency. V0 R1 L A C B R2