2.1 The electric field outside a charged sphere is the same as for a point source, E(r) =

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Baldwin Gregory
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1 Chapter Exercises. The electric field outside a charged sphere is the same as for a point source, E(r) Q 4πɛ 0 r, where Q is the charge on the inner surface of radius a. The potential drop is the integral The capacitance is therefore a Q V b 4πɛ 0 r dr C Q 4πɛ 0 Q ( ) ab V 4πɛ 0 b a ( a b ). This is inversely proportional to the resistance found in Exercise.4.. The planar capacitor formula is Solving for A, A Cd κɛ 0.3 The solenoid inductor formula is C Aκɛ 0 d. ( F)(0 9 m) 0( C /N m ). m. L Aµ 0N. l The loop area is A π(5 cm ) 76 cm m. The density of coils is N/l (6 4 )/(.05 )(π(.05 m)) /m. Solving for l, l L Aµ 0 (N/l) H ( m )(4π 0 7 N/A 7 cm. )( /m).4 Integrate the flux over a rectangular section between the wires of length l. The magnetic field from a wire is B µ 0I πr. 4

2 The flux from one wire is a µ 0 I Φ l b πr dr µ 0Il π ln(a/b). The second wire contributes flux in the same direction, so the total flux is The voltage drop is Φ tot µ 0II π V Φ tot t ln(b/a). µ 0l π ln(a/b) I t. Therefore L µ 0l π ln(a/b). [Note: Technically we should account for the magnetic field inside each wire. The current inside the radius r is Ir /b. B(πr) µ 0 Ir /b, so B µ 0rI πb inside the wire. The integral of the flux is inside the wire is b µ 0 ri Φ l 0 πb lµ 0I 4π. The contribution to the inductance L from both wires is then L µ 0l π. which implies L/L ln(a/b), which means the field inside the wire is negligible if a b. ].5 The Biot-Savart law is db( r) µ 0 Id l. 4π 3 For a loop of radius L centered at (0,0), the unit vector along d l is ˆθ ˆx sin θ + ŷ cos θ. The distance from a point of the circle to a point r on the x-axis is L(cos θ, sin θ) (r, 0) (L cos θ r) + L sin θ The direction of is L(cos θ, sin θ) (r, 0) ˆ (ˆx(L cos θ r) + ŷl sin θ). 5

3 The cross product is ˆθ ˆ ẑ (r cos θ L). The integral of the Biot-Savart element is then B µ 0Iẑ 4π π 0 r cos θ L Ldθ [(L cos θ r) + L sin θ] 3/ This function involves elliptic integrals. Doing it numerically for r 0 (the center of the loop), gives B µ 0Iẑ 4L, while at r L/, it is B (3.9/π) µ 0Iẑ 4L. The magnitude of B rises sharply near r L..6 This has the solution I C t (V ) C t. (t) V e t/c..7 Kirchhoff: This has the general solution Setting this to V at t t 0 gives 0 C t +. (t) V e t/c + V. The current through the capacitor is V V e t 0/C + V. I(t 0 ) V C t (V e t/c + V ) C tt0 ( V ) C e t 0/C which implies V V e t 0/C. Plugging this into the above gives V 0, so the solution is (t) V e (t t 0)/C. This is decay to zero with the same time constant C. 6

4 .8 The integrator circuit of Figure.6(a) is governed by equation (.4.3), For sin ωt, this implies + C. sin ωt + ωc cos ωt. When ωc, the sin ωt term is negligible, and is proportional to, i.e., is proportional to the antiderivative of. The differentiator circuit of Figure.6(b) is governed by equation (.4.7), For sin ωt, this implies + C. ω cos ωt + C sin ωt. When ωc, the cos ωt term is negligible, and is proportional to, i.e., is proportional to the derivative of..9 This is the circuit shown in Fig..8(b). We have.0 i/ωc + Z C i/ωc + i/ωc + /ω C + /ω C. This is a high-pass filter.. From Exercise.9, we have Z L + Z L iωl + iωl + /ω C Solve for ω: ω ( C / ) / 7

5 a) b) c) 3 db 0 log 0 / / ω /C / 0 0. ω /3C / ω /0C. a) 0 dbm 0 log 0 (V/V 0 ) V V 0 0 / V 3.V 0 V P V 0 mw. We could also have done this just by noting that 0 dbm is 0 times greater than 0 dbm. b) 35 db 0 log 0 V /V V V 0 35/ From (.7.), sin ωt eiωt e iωt i F (ω ) π δ(ω ω) δ(ω + ω) i.4 f(t) is given by (note typo in book) f(t) {, (n )T < t < (n /)T 0, (n /)T < t < nt. 8

6 c n T T T/ e iπnt/t dt e iπn e iπn iπn e 3iπn/ (e iπn/ e iπn/ ) iπn πn e 3iπn/ sin(πn/) f(t) n c n e iπnt/t + n c n n 0 ± iπ ± 0 ±3 i 3 π ±4 0 ±5 i 5 π ±6 0 ±7 i 7 π i πn (eiπnt/t e iπnt/t ) πn sin(πnt/t ) The sum of the terms up through n 7 (first five nonzero terms) is shown in Figure 6. n Figure 6: Fourier sum for Exercise.4. 9

7 .5 The Fourier transform is f(t) is given by The transform is F (ω) n f(t) (n /)T (n )T F (ω) f(t)e iωt dt. {, (n )T < t < (n /)T 0, (n /)T < t < nt. e iωt dt n i ω (eiωt/ e iωt ) ) i ω (e iω(n /)T ) e iω(n )T ) ) n e iωnt. The sum will be equal to infinity for ω πn /T and zero otherwise (this is equivalent to a δ-function). Thus we have i πn F (ω) /T i πn, ω πn /T 0, else. This is the same as the result of Exercise.4..6 The Fourier transform is F (ω) Θ(t)e st e iωt dt The response function (.5.7) is and the product is The reverse Fourier transform is f(t) π F o (ω) 0 i/ωc i/ωc For t > 0, ω +i converges, so this becomes e st e iωt dt s iω i/c ω i/c ( i/c ) ( i ) ω i/c ω is ( ) ( ) i/c i e iωt dt. ω i/c ω is i ω is i f(t) i( i/c) i/c is ei(i/c)t + i ( i/c) is i/c ( i)ei(is)t. 0

8 Setting s 0 gives f(t) e t/c..7 Loops: I + I c Z C I + I L Z L Node: I I C + I L Solution:.8 (.8.) is (.8.) is I C Z L Z C + Z L + Z C Z L V C I C Z C iωl i/ωc + iωl + L/C L/C i/ωc + iωl + L/C i/ωc + iωl. I + I C + LÏ. For (0)e iωt and I I 0 e iωt+φ, this becomes iω (0)e iωt iωi 0 e i(ωt+φ) + I 0 C ei(ωt+φ) Lω I 0 e i(ωt+φ). The output is across, so I 0 e i(ωt+φ). The above equation therefore becomes Solving for / gives which is the same as (.8.). iω iω + C L ω. iω iω + /C ω L/

9 .9 We want a high-pass filter like that shown in Fig..8(b), which has response (see Exercise.9) + /ω C We solve for C: C ω π(00 Hz) /(.95) / Ω 0.6 µf..0 Solve for 0% value: For 90% value, On the positive side, t 0 t 90 e t 0 /τ. t 0 τ ln(0.). t 90 τ ln(0.9). τ ln(0.9) τ ln(0.) τ ( ln(0.) ln(0.9)). τ.. The electric field is purely radial. For charge +Q on the inner sphere and Q on the outer sphere, the electric field is E(r) Q 4πɛ 0 r and the potential drop is r V (r) E(r)dr Q Q. r 4πɛ 0 r 4πɛ 0 r Comparing to the definition Q C V, this gives C 4πɛ 0 ( r r ).. The electric field from a line charge is found from Gauss s law, E σ πrɛ 0,

10 where σ Q/l is the charge density. The voltage drop from wire of radius b to a distance a is a V E(r)dr Q/l ln(a/b). b πɛ 0 By superposition, the other wire contributes the same, so we multiply by. This implies C πɛ 0l ln(a/b)..3 ω 0 LC C Aɛ 0 d lwɛ 0 d L µ 0ld w µ0 ld w lwɛ 0 d µ0 ɛ 0 l c l..4 The high-pass filter of Fig..8(b) has response (see Exercise.9) From (.5.9), i/ωc ( + i/ωc) + /ω C. tan φ Im / e / /ωc At high frequency, φ 0. At low frequency, φ π/. 3

11 .5 iωl iωl + iωl/ iωl/ +. Figure?? shows the plot with frequency in units of /L. Figure 7: esponse function for Exercise.5..6 is increased by (3/).5. 0(log 0.5) 3.5 db.7 From Exercises.9 and.8: + /ω C Solving for ω, ω ( ) / C / 0. C.8 The low-pass filter response (.5.7) is i/ωc i/ωc. 4

12 Power response is i/ωc i/ωc i/ωc + i/ωc /ω C + /ω C ω (C) +. Solve for ω: 0%: ω 3 C. ω C / 90%: ω 3C range: ω.667 C..9 db 0 log 0 V /V V V 0 db/0 0 3/0 4.5 The amplitude signal-to-noise ratio increases by a factor of 4.5, i.e. from to 9. It is also common to talk in terms of the signal-to-noise power ratio..30 The circuit is shown in Figure 8. Figure 8: Circuit for Exercise.30. 5

13 Loops: I + I C Z C I + I + I Z C Node: I I C + I Set, C C C. Solution: Z C. + 3Z C + ZC + 7C ω + C 4 4 ω 4 In the Figure 9, the lower curve is this response function, while the upper curve is the single low-pass response, from Exercise.7. Figure 9: esponse functions for Exercise Assuming perfect detector; no current flows into output; I o is the current flowing from top to bottom in the right side of the circuit. Loops: I C Z C + I 3 6

14 I + I C3 Z C I + I o + I o Z C + I 3 Nodes: I I C3 + I o I C + I o I 3 Solution: ( + C ω ) 4iCω + C ω This is plotted in Figure 0. When ω /C, this equals 0. When ω 0 or ω, it approaches unity. Figure 0: esponse function for Exercise.3..3 The circuit in the book is missing a 50-Ω resistor between and the rest of the circuit as drawn, the circuit will give for all inputs. Putting a resistor there gives Loops: I + I (Z C + Z L ) 7

15 I (Z C + Z L ) Node: I I + I Solution: ( + C L ω )( + C L ω ) + ic ω + ic ω C L ω C L ω ic C L ω 3 ic C L ω 3 + C C L L ω 4 This is a double notch filter, with zeroes where the two terms in the numerator vanish. Figure shows a plot for the values given: Figure : Double notch response function, for Exercise a) Capacitor relation: Loops: I C V C t V s I + I V C + I C C + I These become 8

16 V s I + I iωv e iωt I C C + iωv ACe i(ωt+φ) V DC + V AC e i(ωt+φ) I Node: I C + I I. Write I I DC + I AC and I I DC + I AC (note that there is no DC component through the capacitor), and set DC terms equal and AC terms equal: V s I DC + I DC 0 I AC + I AC iωv e iωt I C C + iωv ACe i(ωt+φ) V DC I DC V AC e i(ωt+φ) I AC I C + I AC I AC I DC I DC We solve these for I C, I DC, I AC, I DC, I AC, V DC, and V AC e iφ, which gives V DC V s +, where V AC e iφ C ω i i + C ω V C eff ω i + C eff ω V eff We want a low-pass filter that eliminates AC frequency of 60 Hz. We use the circuit shown in Fig..8(a) with a polarized capacitor with the negative side grounded. Formula (.5.) gives us / + ω C. We would like low series to prevent DC droop of the voltage supply. Pick Ω, which is small compared to a typical 50 Ω load impedance. Solve for C: C / / ω 9

17 To eliminate 99% of the ripple, pick /.0. Setting ω π(60 Hz) 377 s, we then have C F 530 µf. (377 s )(50 Ω) 30

Handout 11: AC circuit. AC generator

Handout 11: AC circuit. AC generator Handout : AC circuit AC generator Figure compares the voltage across the directcurrent (DC) generator and that across the alternatingcurrent (AC) generator For DC generator, the voltage is constant For