Physics 102 Confeence 6 Magnetic Field Confeence 6 Physics 102 Geneal Physics II Monday, Mach 3d, 2014 6.1 Quiz Poblem 6.1 Think about the magnetic field associated with an infinite, cuent caying wie. Using this mental image, sketch the magnetic field lines associated with a loop you will find the field along the cental axis in the next poblem, but you can get a sense of the field eveywhee. The pogession fom an infinite line of cuent to a loop is shown in Figue 6.1. Figue 6.1: An infinite line has magnetic field that ciculates aound it so does a loop of wie. 1 of 6
6.1. QUIZ Confeence 6 Poblem 6.2 A cicle of adius R has steady cuent I flowing aound it. What is the magnetic field (diection and magnitude) a height z above the cente of the loop (the Biot-Savat law, fo the contibution of a segment of cuent to the magnetic field at is: db = µ 0 4 π Idl [ ]) 3 ). ẑ ˆx I d ŷ Refeing to the figue: ẑ ˆx I d ŷ (side view) z diection of magnetic field R the contibution fom the shown will have hoizontal component cancelled by the patch of cuent opposite it only the vetical components will suvive the integation. So: db z = µ 0 I R dφ 4 π (R 2 + z 2 ) sin θ = µ 0 I R 2 dφ, (6.1) 4 π (R 2 + z 2 3/2 ) and then, integating aound the cicle gives: µ 0 I R 2 B = ẑ. (6.2) 2 (R 2 + z 2 3/2 ) 2 of 6
6.1. QUIZ Confeence 6 3 of 6
6.2. BIOT-SAVART LAW Confeence 6 6.2 Biot-Savat Law Poblem 6.3 Helmholtz coils ae a pai of cicula loops of adius R caying steady cuent I (in the same diection) they ae typically made using N tuns of wie, so that the net cuent caied by each loop is N I. The loops ae aanged a distance R apat. Find the magnetic field midway between the loops on the line connecting thei centes. ẑ R NI ˆx ŷ R Figue 6.2: A pai of cicula loops cay cuent N I. Find the magnetic field at the location shown. Conside a centeed pai of coils, so that the midpoint between the loops is at y = 0. The magnetic field due to the loop on the left is: B l = ( 2 µ 0 I R 2 R 2 + R2 4 ) 3/2 ŷ = 4 µ 0 I 5 5 R The magnetic field due to the loop on the ight is: B = ( 2 µ 0 I R 2 R 2 + R2 4 ) 3/2 ŷ = 4 µ 0 I 5 5 R so the total field at the point of inteest is (supeposition): ŷ. (6.3) ŷ, (6.4) B = B l + B = 8 µ 0 I 5 ŷ. (6.5) 5 R 4 of 6
6.3. LORENTZ FORCE LAW Confeence 6 6.3 Loentz Foce Law The ole of the magnetic field in chage motion comes fom the Loentz foce given a magnetic field B and a paticle with chage q moving with velocity v, the foce on the paticle, due to the field, is: F = q v B. (6.6) Poblem 6.4 A chaged paticle (of mass m, chage q) entes a solenoid of adius R though a hole along the x axis. It exits though a hole in the y axis. a. Assuming the velocity of the paticle as it entes the solenoid is v = v ˆx, find the magnitude and diection of the magnetic field that caused the motion inside the solenoid. R ŷ ˆx Figue 6.3: A top down view of the solenoid with the paticle moving in along the ˆx axis, and exiting along the ŷ axis. Once inside the solenoid, the paticle is in a egion of unifom field. The diection of the field must be ẑ (out of the page) in ode to get a foce, as the paticle entes, pointing to the ight. We know that the moving chaged paticle will tavel along a cicula ac, in this case, of adius R. Then the magnetic field is poviding the centipetal foce, and we have: m v 2 R = q v B 0 B 0 = m v q R, (6.7) 5 of 6
6.3. LORENTZ FORCE LAW Confeence 6 so the magnetic field inside the solenoid must be B = B 0 ẑ = m v q R ẑ. b. Assuming the solenoid in the pevious poblem is made up of wie caying cuent I, with n tuns pe unit length fo the solenoid, find the cuent I that must be flowing to induce the motion of the chaged paticle. Using Ampee s law, we know that the magnetic field associated with a solenoid is: B = µ 0 I n ẑ, (6.8) and given ou taget magnitude, we must have: µ 0 I n = m v q R I = m v µ 0 n q R. (6.9) 6 of 6
6.3. LORENTZ FORCE LAW Confeence 6 Poblem 6.5 A squae loop of wie (mass m with side length l) caies unifom cuent I flowing in the diection shown below. It is placed in a magnetic field, whee its magnetic dipole moment makes a small angle θ with espect to the magnetic field. The moment of inetia of the loop (calculated about a od though its cente) is: I l = m l2 6, (6.10) use this, and the small angle appoximation, to find the peiod of the oscillatoy motion of the loop. I B = B0 ŷ ẑ µ ˆx ŷ B = B0 ŷ top down view Figue 6.4: A squae loop of wie has magnetic moment that makes an angle of θ with espect to a constant magnetic field pointing to the ight. Fo small θ, find the peiod of the esulting oscillatoy motion. The toque on the loop is given by: τ = µ B = I l 2 B 0 sin θ ( ẑ). Using the small angle appoximation (sin θ θ), togethe with I α = τ gives: so that: A typical solution to this ODE is: I l d 2 θ(t) dt 2 = I l 2 B 0 θ(t), (6.11) d 2 θ(t) dt 2 = I l2 B 0 m l 2 /6 θ(t) = 6 I B 0 θ(t). (6.12) m θ(t) = θ 0 cos ( 6 I B0 ) m t, (6.13) so the peiod is: 6 I B0 m m T = 2 π T = 2 π. (6.14) 6 I B 0 7 of 6
6.3. LORENTZ FORCE LAW Confeence 6 8 of 6
6.3. LORENTZ FORCE LAW Confeence 6 We saw that the Loentz foce law implied a foce pe unit length of: F = µ 0 I 2 2 π d (6.15) fo two paallel wies caying the same cuent, and sepaated by a distance d. We can use this expession to get a sense of the magnitude of the magnetic foce compaed to the electic foce. Poblem 6.6 Take two infinite line chages with unifom λ, and pull them upwads with constant speed v. Assume the line chages ae sepaated by a distance d. v v d Figue 6.5: Two infinite line chages pulled upwad these lines ae sepaated by a distance d. a. What is the electostatic foce pe unit length acting between the wies? (Think of the left-hand wie as the souce wie, and the ight-hand wie as the foced wie). The electostatic field set up by the wie on the left has magnitude: E = λ 2 π ɛ 0 s (6.16) whee s is the distance fom the left-hand wie. The foce exeted on a segment of wie of length l on the ight is: F = λ2 l 2 π ɛ 0 d (6.17) 9 of 6
6.3. LORENTZ FORCE LAW Confeence 6 so the foce pe unit length, elevant fo compaison is: F E = λ2 2 π ɛ 0 d. (6.18) b. What is the cuent I (fo use in (6.15)) in this case? Wite the magnitude of the magnetic foce in tems of λ and v. The cuent associated with eithe wie is I = λ v, so that the magnetic foce pe unit length is: F B = µ 0 λ 2 v 2 (6.19) 2 π d c. With what speed v must you pull the wies in ode to get the electostatic foce to have the same magnitude as the magnetostatic one? What does this tell you about the elative size of the electostatic and magnetostatic foces? When the foces ae equal: µ 0 λ 2 v 2 2 π d F B = F E = λ2 2 π ɛ 0 d (6.20) we can solve fo v: v = 1 µ 0 ɛ 0 3 10 8 m/s. (6.21) The magnetic foce is, numeically, much smalle than the electic one. 10 of 6
6.4. FIRST ORDER ODES Confeence 6 6.4 Fist Ode ODEs Poblem 6.7 Suppose you want to model a population you denote the numbe of animals at time t by N(t). It is easonable to imagine that the ate of gowth is popotional to the population size, so that dn(t) dt N(t). Call the popotionality constant α, then the diffeential equation descibing the population is: dn(t) dt = α N(t) (6.22) and we must be given the size of the population at time t = 0: N(0) = N 0. a. Fo ease of gaphing, take N 0 = 2 and α = 2 1/s. Then dn dt t=0 = 4, so the slope of N(t) at t = 0 is fou. Daw a line with that slope that stats at t = 0 and goes to t = 1/2. What is the appoximate value of N(1/2)? Using that value, detemine the slope of N(t) at t = 1/2 and daw a line with that slope fom t = 1/2 to t = 1, that will allow you to estimate the value of N(t) at t = 1. Continue with t = 3/2, what sot of function ae you getting? This is a coase appoximation to the exponential, and it is off by quite a lot because of ou choice to take 1/2 s steps but the cuve is clealy gowing... quickly. b. One impotant featue of population models is the notion that thee is a natual maximum to the population, N max modify the stating equation dn dt = α N(t) so that if N(t) < N max you get dn dt > 0 and if N(t) > N max, you have dn dt < 0 (causing decay in the population). Take the simplest extension that will lead to an equation that is quadatic in N(t) on the ight, and will educe to dn(t) dt = α N(t) fo N(t) N max. The simplest fix is to take dn(t) dt ( = α 1 N(t) ) N(t) (6.23) N max 11 of 6
6.4. FIRST ORDER ODES Confeence 6 32 24 16 8 1 2 12 of 6
6.4. FIRST ORDER ODES Confeence 6 32 24 16 8 1 2 13 of 6