advertisement

[1]. A low pass filter has an inductor, a resistor and a capacitor in series with an ac emf E = Vin cos ωt. The output is the voltage, Vout is that across the capacitor. (a) Draw the circuit indicating the voltage which is defined as Vout . (b) Derive an expression for Vout /Vin . (c) Find the limiting expressions for ω large and small. Are these results consistent with the fact that this is called a low pass filter? Why? (a) Draw figure (b) The impedance of this is Z= p R2 + (1/ωC)2 and so the current in the circuit is i= Vin cos ωt Vin cos ωt =p . Z R2 + (1/ωC)2 The voltage across the capacitor is then, (using the same definition as for Vin ) Vout cos ωt = iXC = so that i Vin =p cos ωt 2 ωC R + (1/ωC)2 ωC Vout 1 . =p Vin R2 + (1/ωC)2 ωC (c) If ω is large the Z ≈= R and Vout 1 . = Vin RωC which is small when R > (1/ωC) which is what large ω means. When ω is small Z ≈ (1/ωC) and Vout = 1. Vin So yes, these results are consistent with the name since the output is small of ω > 1/RC and is equal to the input when the inverse is true. 1 ~ = Bx î + By ĵ + Bz k̂, at the centre of a semi[2]. (a) By direct integration, calculate the magnetic field, B circular segment carrying a current I in the clockwise sense. The radius of the segment is a, i.e., has coordinates x, y such that x2 + y 2 = a2 . The semi-circle is such that y > 0, i.e., is centered at the origin ad lies above the x-axis. y a x ~ is Consider small d~` in the clockwise direction. The contribution to B ~ =I dB µ0 d~` × r̂ . 4π r2 ~ is into of the paper, i.e., in the negative z direction, so The direction of dB ~ = −I dB µ0 d` k̂. 4π a2 independent of which d~` is involved. Thus Z Z ~ = dB ~ = −I µ0 ( d`)k̂ = I µ0 (πa)k̂ = − Iµ0 k̂ B 4πa2 4πa2 4a (b) A particle with charge q moves with velocity ~v = vx î + vy ĵ + vz k̂ in the resulting magnetic field. Calculate the force F~ (magnitude and direction) as the particle passes through the origin. Have ~ = q~v × B ~ = −q Iµ0 (vx î + vy ĵ + vz k̂) × k̂ = −q Iµ0 (vy î − vx ĵ) F~ = q~v × B 4a 4a (c) Determine the rate at which the magnetic field is doing work on the particle. (Hint: P = F~ · ~v .) Iµ0 Iµ0 P = F~ · ~v = −q (vy î − vx ĵ) · (vx î + vy ĵ + vz k̂) = −q (vy vx − vx vy ) = 0 4a 4a as it should be. 2 [3]. (a) An electromagnetic wave has an electric field given by ~ E(y, t) = −(5.10 × 106 V/m)k̂ sin[(2.25 × 1012 rad/s)t − ky] (a) In which direction is the wave moving? Given sin[(2.25 × 1012 rad/s)t − ky] the wave is traveling in the positive y direction. ~ (b) Re-write E(y, t) filling in the missing parameter k. Have ω = ck or k = ω/c = (2.25 × 1012 rad/s)/(3.00 × 108 m/s) = 7.5 × 103 m−1 . ~ E(y, t) = −(5.10 × 106 V/m)k̂ sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y] (c) What is the wave length, λ, of the wave? λ= 2π 2π = (7.5 × 103 m−1 ) = 0.84 × 10−3 m = 8.4 × 10−4 m k k ~ ~ as part (b). (d) Write the vector equation for B(y, t), i.e., write the equivalent to the above for E ~ ×B ~ is in the direction of propagation which must be in the y direction. Have ~ = 1E The vector S µ0 ~ must be in the −î direction, i.e., k̂ × î = ĵ so B (5.10 × 106 V/m) ~ B(y, t) = − î sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y] (3.00 × 108 m/s) or ~ B(y, t) == −(1.7 × 10−2 T)î sin[(2.25 × 1012 rad/s)t − (7.5 × 103 m−1 )y] 3 [4] A parallel plate capacitor has horizontal plates of area A and the distance between the plates is d. (a) Find the capacitance C in terms of A, d and fundamental constants. Ignore edge effects. The electric field close to a metallic surface is E = σ/0 , while σ = Q/A and V = Ed so that V = Ed = so that C= Qd σ d= 0 0 A Q 0 A = . V d (b) If there is a potential difference V across the plates what is the magnitude of the electric field E between the capacitor plates? Again ignore edge effects. As noted above E= V . d (c) Ignoring edge effects, what is the energy density uE due to this electric field and what is the total energy UC stored. Does this agree with 12 CV 2 using C from part (a)? Have uE = (1/2)E 2 so the energy stored in the space between the plates is U = AduE = (1/2)0 E 2 Ad = (1/2)) E 2 (A/d)V 2 which is to be compared with U = (/12)CV 2 with th result C= 0 A , d as in part (a). A cork ball of mass M and charge q is placed between the plates and remains at rest. (d) Find the voltage V across the capacitor in terms of M , q and g. Your solution must include a free body diagram. (Draw a diagramme.) The electric force is qE = q(V /d) while the gravitational force is M g and for a = 0, this implies V q = Mg d or M gd V = . q (e) If the space is filled with a dielectric with a dielectric constant κ what is the new value of V ? If the potential difference was V , when the dielectric is included it becomes V /κ so V = M gdκ . q 4 [5] (a) Find the emf’s E1 and (b) E2 in the circuit shown in the figure. (c) Find the potential difference between the points a and b. 20 V 1.00Ω 1.00A + 4.00Ω 6.00Ω 1.00Ω X1 b a 2.00A 1.00Ω + + 2.00Ω X2 For E1 , note the potential difference between a and b is obtained from the top leg of the diagramme, i.e., is 20 − 7.00Ω!.00A = 13V. The current in the middle brach is 1.00A in the direction b → a so 13 = E1 − 5.00Ω1.00A or E1 = 13 + 5 = 18.0V. Then 13 = E2 + 3.00Ω2.00A so that E2 = 13 − 6 = 7V. It has already been determine that Vab = 13V. 5 [6] The current in the long straight wire AB shown in the figure is upwards and increasing at a steady rate di/dt. ~ at a distance (a) At an instant when the current is i, what are the magnitude and direction of the field B r to the right of the wire? Consider the wire in isolation and use Ampere’s law, I ~ · d~` = 2πrB = µ0 i that isB = µ0 i B 2πr and is into the paper. (b) What is the flux dΦB through the narrow shaded strip? dΦB = BLdr = µ0 i Ldr 2πr (c) What is the total flux ΦB through the loop? By integration Z ΦB = µ0 i dΦB = L 2π b Z a dr µ0 iL b = ln r 2π a (d) What is the emf E induced in the loop? Farday’s law states µ0 L dΦB =− E =− dt 2π b ln a di dt (e) Evaluate the numerical value of E if a = 12.0cm, b = 36.0cm, L = 24.0cm and di/dt = 9.60A/s. Have the |E| = µ0 × 0.24 2π ln 36 12 6 9.6 = 5.06 × 10−7 V