COMMONWEALTH OF AUSTRALIA Copyright Regulations 1969 Warning - Do not remove this notice This material has been reproduced and communicated to you by or on behalf of the University of New South Wales pursuant to Part VB of the Copyright Act 1968 (The Act). The material in this Communication may be subject to copyright under the Act. Any further reproduction or communication of this material by you may be the subject of copyright protection under the Act. I have read the above statement and agree to abide by its restrictions.
TilE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF PHYSICS FINAL EXAMINATION NOVEMBER 2014 PHYS2210 Electromagnetism and Thermal Physics- PAPER 1 PHYS2050 Electromagnetism Time Allowed - 2 hours Total number of questions - 4 Answer ALL questions Questions ARE NOT of equal value This exam is worth 30% of the final grade for PHYS221 0 students This exam is worth 60% of the final grade for PHYS2050 students Candidates must supply their own university approved calculator. Answers must be written in ink. Except where they are expressly required, pencils may only be used for drawing, sketching or graphical work. Candidates may keep this paper.
Ph YS2210 PHYS2050 Gradient Divergence Curl Laplacian Identities Volume element Gradient theorem Divergence Theorem Stokes' Theorem Coulomb's Law Electric Field Definit ions and Formulae 'Vf 8 f A () f A a f k ='j5"""" l +i:) J +"'2\ vx vy vz..., A 8Ax 8Ay 8A. v =-+-+ - 8x 8y 8z 'V X A= [&A.- oay] i + [8Ax- &A.] j + [ClAy- 8Ax ] k 8y 8z 8z ax 8x 8y 82! 82! 82! 'V. ('V!) = 72 J = 8x2 + 8y2 + 8z2 '\1 X ('\1 f)= 0 'V.('V X A)= 0 'V X ('V x A ) = 'V('V A) - 7 2 A dx dy dz (Cartesian coordinates) rdrd dz (Cylindrical coordinates) r 2 sin 8 dr d8 d (Spherical coordinates) 1b ('V f) dl = f(b) - f(a) is A ds = fv ('V A ) dv h A dl = h ('V x A) ds F= Q1Q2 f 41TEor 2 F= Q E E = _ l _ { Pv f dv 47rfo Jv r 2 Gauss' Law Electric Potential { E ds = Q ls Eo V(b)- V(a) = -1b E d.e 'il E =Pv. fq E (r) = - 'VV(r) V= -- l 1 -dv Pu 47TEo v r Current Density Charge Conservation Stored Energy J = Pvv = PuJ. E = a E 'V. J =-8p at W = ~ Lq;V(ri) W = 4 fv V(r)p(r) dv = 4 fv EE 2 dv = t fv D E dv Electric Diplacement Linear Media Capacitance D =EoE + P fs n -ds=q 1 Q C = V D = Eo( l + Xe)E = ee p.l
DC Circuit s: Ohm's Law:.6. V = I R resistance, R = plj A [D] Kirchhoff's Laws: (1) EI = 0 at a junction (2) Et' - 'f.! R = 0 around each loop Joule heating: power dissipa ted, P = I.6. V = 1 2 R = (.6. V) 2 J R Ohm's law: J = ue power dissipated/ unit volume= J E = ue 2 [vv] Current density in a. metal: J = - nevd Magnetism: ~~Iag ne t ic force on a moving charge q is F = q(v x B ) Ivfagnetic force on a current element I d is df = I d x B There are no 'magnetic charges', so for a. closed surface S Is B ds = 0 or ~ 'V BdV = 0 so: 'V B = 0 I3iot-Savart Law: B field from a moving charge q', with velocity v ': B = J.Lo :{ (v 1 x f) 47r.,.2 element of magnetic field produced by a current element I d is: db = J.Lo Id x f 411' r- 2 or ldbi = J-Lo I d sinfj 47r 1' 2 11-o II' de' x (d x r) Force between two current elements - - 411' r-2 (like currents attract, unlike currents repel) so: force/unit length between two long parallel current-carrying wires is F 11-o II' e = 21r --:;:- Force on wire of length, perpendicular to magnetic field: F = B I e [NJ Particle of mass m, charge q moving perpendicular to magnetic field: cyclotron radius, r = mvj (qb) [mj; cyclotron frequency, f = qbj(27rm) [Hz] Hall efect: Hall coefficient, RH = l jnq, charge mobility vd/ E = u RH. Ampere's law: f B d = p,ol (I is current linked by the closed path) Differential form: \7 x B = p..oj J.Loi, Magnetic field at distance r from a long straight wire: B = - 27rr (B is in circles around wire) Magnetic field on axis of a circular wire loop of radius R carrying current I is: B _ 11-o I 1rR 2 z - 27r (z2 + R2)~ Magnetic dipole moment of a small circular current loop ism = I1rR 2 general formula is m = I A [Am 2 J p.2
Axial magnetic field inside a long solenoid is B = f.l, J.LonJ, where n is the number of turns per unit lenglh Faraday's Law for EMF by induction: Or: EMF = rate of cutting magnetic flux. 8B Differential form: 'VxE = - - 8t f C' d<i> 8 d r E dl =c.-= -dt =- dt Js B-dS Mutual Inductance: 1 - Flux in circuit 2 -'lz - Current in coil 1 Self Inductance: L = ; V = -Ldl dt Nz Self Inductance of a solenoid: L = f.lr f.lo T A Magnetic energy: U = ~LJ 2 Energy density in magnetic field: u = ~ = ~ B H - f.l, f.lo Magnetic media: B = f.lo(h + M ) = f.lo(l+xm) H = f.lrf.lo H ie H = ~ -M f.lo \7 B = 0, so \7 H + \7 M = 0 Ampere's law becomes: \7 x H = J free At a boundary, B~ = 8 1. and H IJ = Hu Maxwell's Equations In a vacuum: 'V E p to 'VxE ab at 'V B 0 'V xb Lorentz force law: F = q (E + v x B ) EM Waves: 82E Wave equation for E in free space: 'V 2 E = f.loto otz ie c = 1/..jiiOf:O (in a medium: v = 1/ JJ.LrJ..Lotr 0 = cfn, n = refractive index) Solution: Ex = Eo sin(kx- wt) for monochromatic wave travelling in +vex-direction. E, B and the direction of propagation k are mutually perpendicular: :E x:b = k k-e = 0 k B = 0 cb = k xe p.3
The direction of E is the direction of polarization of the E-M wave. 1 2 Poynting vector: N = ExH = - (ExB) =- J.LO j.lqc Useful informat ion Eo= 8.854 X w- 12 Frn- 1 (l/4uo) = 8.99 x 10 9 mf- 1 J.Lo = 4n X w- 7 Hm- 1 speed of light, c = 3 x 10 8 ms- 1 elementary charge, e = 1.60 X lq- 19 C lev= 1.60 X w- 19 J electron mass = 9.11 X lq- 3! kg proton mass = 1.67 X 10-27 kg Avogadro's number, NA = 6.022 x 10 23 mol- 1 Boltzmann's const, ks = 1.381 X w- 23 JK- 1 p.4
Question 1 (20 marks) Consider two very long, coaxial conducting cylinders with radii a and b respectively. T he inner cylinder carries a charge per unit length +.X, while the outer cylinder carries - >.. +i. -I. (a) Calculate the electric fie ld in t he space between t he two cylinders. (b) Calculate t he corresponding electric potential difference between the two cylinders. (c) Calculate t he capacitance per unit length of t his configuration. s
Question 2 (25 marks) Consider a very long cylindrical conductor carrying a uniform current density J = Ji. (a) Show that the magnetic field inside the conductor is given by Hint: if> = - sin x + cos y. B- = -J( xya - yx A). 2 (b) Now suppose an equally long, cylindrical cavity is cut into the conductor in an off-axis position as shown: y where the z-direction points out of the page. The conductor still carries a uniform current density J = Jz. The cavity, however, is completely source-free. Find the magnitude and direction of B at t he point P. Hint: Use the principle of superposition, and observe that adding up two current densities J and - J gives exactly zero.
Question 3 (35 marks) A small loop of wire of radius a is held a distance z above the centre of a large loop of radius b. (a) Suppose a current I flows in the big loop. Show that the magnetic field at a distance z directly above its centre is (b) Find the flux through the small loop. You may assume that t he small loop to be so small that the field due to the big loop is essentially constant. (c) Suppose a current I flows in the small loop. Find the flux through the big loop. Here, you may treat the small loop as a magnetic dipole. Hint: the magnetic vector potential due to a magnetic dipole is - ~-toiiixf Adip(i) = --- 2-. 47T r (d) Find the mutual inductance, and confirm that M12 = M21 -
Question 4 (20 marks) Consider an electromagnetic wave propagating in free space. Its electric field component is given by E(x, t) =Eo sin(kx- wt) y. (a) Show that the corresponding magnetic field component is given by where w = kc. B(x, t) = Eo sin(kx- wt) z, c (b) Compute the corresponding magnetic vector potential. (c) Show that, at any one time, half of the energy in the wave is stored in t he electric field, and the other half in the magnetic field.