Foundation Course Unit SEF006 Fields and Waves. SOLUTIONS Examiners: Dr. A. G. Polnarev Prof. F. Di Lodovico c Queen Mary University of London, 2016
Page 2 SEF006 (2016) SECTION A SOLUTIONS TO SECTION A Question A1 Solution A1(a): Gravitational field strength, g, is the negative of the potential gradient: g = dv dr. 2/2 marks Solution A1(b): The small mass, m, will escape to infinity if its K.E. exceeds its P.E. (P.E. = V m),i.e. Hence, v esc = mv 2 esc 2 = GMm R, 2GM R = 2GM R 2 R = 2gR. Question A2 Solution A2(a): The simple harmonic motion is oscillation described by the following equation: a = ω 2 s, where ω is a constant. 1/1 marks Solution A2(b): 2/4 marks hence 2/4 marks T 1 = 1/10 min, T 2 = 60/300 min = 1/5 min, T 1 /T 2 = 1/2 l 1 l 2 = 1/2 l 1 l 2 = 1/4 = 0.25. Question A3 Solution A3(a): In transverse waves the disturbances are at right angles to the wave direction. 1/2 marks Example: electromagnetic waves. 1/2 marks Solution A3(b): In longitudinal waves the disturbances are parallel to the line of the wave direction. No, only transverse waves can be polarized, because there are two directions perpendicular to the line of the wave direction, and only one direction parallel to this line. Example: sound waves.
SEF006 (2016) Page 3 Question A4 Solution A4(a): ω 1 = ω 2 = ω, e 1 = e 2 = e. 1/1 marks Solution A4(b): Nodes: positions of zero amplitude. Antinodes: positions of greatest amplitude. = 2A e sin[ E = A e {sin[ω(t x/c)] + sin[ω(t + x/c)]} = ω(t x/c) + ω(t + x/c) ω(t x/c) ω(t + x/c) ] cos[ ] = 2A e sin(ωt) cos(ωx/c). 2 2 ωx n /c = π/2 ± mπ, ωx an /c = ±mπ, x n x an = cπ 2ω = λ/4 = λ 1/4 = 0.25 cm = 2.5 mm. Question A5 Solution A5(a): Faraday s law of electromagnetic induction states that ɛ is opposite to the rate of change of Φ: 1/2 marks 1/2 marks ɛ = dφ dt. Solution A5(b): dφ = BdA = Blvdt, hence ɛ = Blv = 10 3 T 5 10 2 m 0.5 m s 1 = 2.5 10 5 V = 25µV. Question A6 Solution A6(a): V is negative of work done by electric force to bring a unit charge from infinity to point A. 2/2 marks Solution A6(b): The sign of U is negative, since the electric force between the proton and electron is attractive. U = e V 9 109 (1.6) 2 10 38 10 10 J 2.3 10 18 J 14 ev. Turn over
Page 4 SEF006 (2016) Question A7 Solution A7(a): Assume that the particle has a size r, cross-section perpendicular its motion A ( r) 2 and volume V ( r) 3. The density of charge is q/ V, hence I ( A q q )v, hence f = BI r = B( A V V q )v r = B( r)2 v r = Bqv. ( r) 3 Solution A7(b): f = 80 10 3 3.2 10 19 1.6 10 7 N 4 10 13 N. 2/2 marks Question A8 Solution A8(a): v st = E ρ = πd 2 le m 3.14 16 10 4 2.5 2 10 11 24.5 m s 1 10 4 m s 1. 3/3 marks Solution A8(b): 2/2 marks 300 T/300 K = 1 10 104 T 300 (10/3) 2 3300 K. Question A9 Solution A9(a):The dipole is a vector, D formed by to charges which are negative to each other, i.e. q 1 = q = q 2, and separated by the vector d, whish is directed from the positive charge to the negative charge: D = q d. 2/2 marks Solution A9(b): Let Applying the principle of superposition, we have R = AO = αd. V D = 3/3 marks q 4πɛ 0 R q q(r + d R) = 4πɛ 0 (R + d) 4πɛ 0 R(R + d) = qd 4πɛ 0 d 2 α(1 + α) = q 4πɛ 0 αd(1 + α).
SEF006 (2016) Page 5 Question A10 Solution A10(a):Provided a constant phase difference is maintained between the sources of the waves, the points of cancelation and reinforcement do not move, so the two sets of waves are said to have produced an interference pattern. 2/2 marks Solution A10(b): w = λ θ = Dλ d λ = wd D = 6 10 3 5 10 4 5 λ = 600 nm corresponds to the red light. m = 6 10 7 m = 600 nm. Turn over
Page 6 SEF006 (2016) SECTION B SOLUTIONS TO SECTION B Question B1 Solution B1(a): 2/6 marks [A] = [T ][L] 1/2, (e.g. s m 1/2 ). There is no [M], (e.g. kg. Hence, we can propose the following hypothesis: A does not depend on mass, m. 2/6 marks (iii) We can propose the experiment with the same l and some different mass. T should be the same. 2/6 marks Solution B1(b): 3/8 marks N = 60. 2/8 marks (iii) Indeed, 3/8 marks T = t/n, A = t N l 2 s m 1/2. 2π/ 1 g 6.28 9.8 m s 2 2 s m 1/2 A. c ) Solution B1(c): 3/11 marks 2/11 marks 3/11 marks 3/11 marks g M = GM M R 2 M = 4πGρ MR 3 M 3R 2 M = g ρ M ρ E R M R E = 4πGρ MR M 3 9.8 m s 2 3.9 5.5 3.4 6.4 3.8 m s 2. gm 3.8 A M A g 2 9.8 s m 1/2 1.24 s m 1/2. =
SEF006 (2016) Page 7 Question B2 Solution B2(a): The spreading of a wave as it passes a small object or passes through a small aperture. 2/10 marks The total electric and magnetic fields at a point is equal to the sum of the electric and magnetic fields in individual waves. 2/10 marks (iii) When a wave crest meets a wave trough, the result is that they cancel one another out. 2/10 marks (iv) Coherent waves are waves which maintain a constant phase difference. 2/10 marks (v) We should have at least two coherent waves. 2/10 marks Solution B2(b): From Fig.B1 we have 4/9 marks θ BQ BA = λ/2 w/2 = λ OM, on other hand, θ w OA = d, hence, dw λd. D 2/9 marks hence 3/9 marks D = D = dw λ = d λd λ = d D λ, d λ, D = d2 λ = 10 6 m 2 6 10 7 1.6 m. m Solution B2(c): c ) From Fig.B2 we have 3/6 marks θ S 1Q = λ/2 = λ OP, on other hand, θ S 1 S 2 s 2s OM = d λd, hence, s D 2d. D = 2sd λ = 2 10 3 m 10 3 m 6 10 7 m 3.3 m. 3/6 marks Turn over
Page 8 SEF006 (2016) Question B3 Solution B3(a): 3/7 marks Taking into account that 4/7 marks Solution B3(b): v 2 r = GM r 2 2πr, hence, T = v = 2πr, GM/r M = 4π 3 ρr3, we have: T = ( 3π 1 + h ) 3/2. Gρ R T 1 = t ( 1 3π 50 = Gρ 1 + h 1 R ) 3/2 T 2 = t ( 2 3π 50 = Gρ 1 + h 2 R ) 3/2, t 1 t 2 = ( R + h1 R + h 2 ) 3/2, R = h ( ) 2α h 2/3 1 1 α, α = t1 = (100/119) 2/3 0.8, R = 10 3 2α 1 7000 km. t 2 1 α ρ = 3π G (1 + 1/7)3 (50/t 1 ) 2 = 3 3.14 52 8 3 6.67 7 3 (3.6) 2 1011+2 4 4 2 4.1 10 3 kg m 3. (iii) M = 4π 3 4.1 103 (7 10 6 ) 3 5.9 10 24 kg. (iv) U = GM/R = 6.0 10 7 m 2 s 2. (v) g = GM/R 2 = 6.67 10 11 5.9 10 24 (7 10 6 ) 2 8.6 m s 2 ; v = 2gR = 2 8.6 7 10 6 11.0 km s 1.
SEF006 (2016) Page 9 Question B4 Solution B4(a):This law states that the direction of the induced current is always so as to oppose the change which causes the current. This law expresses the principle of conservation of energy. 4/4 marks Solution B4(b): 4/8 marks ɛ = dφ dt = d(ab) = Bl dx dt dt = Blv. v = ɛ Bl = 5 5 1 m s 1 = 1 m s 1, where A is the area and B is the strength of magnetic field. 4/8 marks c ) Solution B4(c): The flux of magnetic field For one ring is Φ 1 = AB, hence for N rings it should be equal to NAB and this is called the magnetic flux linkage, hence, Φ = BAN. 4/13 marks According yo Faraday s law, 5/13 marks ɛ = dφ dt = NAdB dt = NAB t = BAN/t. (iii) t = BAN ɛ = 5 T 40 10 4 m 2 2 10 3 5 V = 8 s. 4/13 marks End of Paper Turn over