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# Pracfinl.Key

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### Pracfinl.Key

1. 1. Final Exam Tuesday, April 21st 1:00 pm 300 Richards
2. 2. 1. (35 points) The resistors in the circuit shown have values R1 = R2 = R3 = 2.2 kΩ and R4 = 1.1 kΩ, and the battery has a 18V emf. (You may want to do c and/or d first but this is not necessary.) (a) (10 points) Rank the four resistors in order of the current that they carry (for example R3 = R2 > R4 > R1), and briefly justify your answer. I I1 I2 E I3 (b) (10 points) Rank the four resistors in order of the potential difference across them, and briefly justify your answer. (c) (10 points) Calculate the equivalent resistance of the four resistors. I d) (5 points) Calculate the total current produced I y the battery, neglecting its internal resistance. E I I ΔVloop = + E - V4231 = 0 V = IR I
3. 3. R1 = R2 = R3 = 2.2 kΩ R4 = 1.1 kΩ E R42 = R4 + R2 } R42 = 1.1 kΩ + 2.2 kΩ I I1 I42 E R42 = 3.3 kΩ I3 V3 = V42 Iin = Iout ΔVloop = 0 = +V3 – V42 I1 = I3 + I42 V3 = V42 (V = IR) I42 = I1 - I3 Plug in I3R3 = I42R42 Solve for I3 I3R3 = (I1 - I3)R42 I3 = I1R42/(R3 + R42) = (3/5)I1 I3R3 = I1R42 - I3R42 Rearrange Solve for I42 I3R3 + I3R42 = I1R42 I42 = I1 - I3 = I1 - (3/5)I1 = (2/5)I1 Factor out I42 = I4 = I2 = (2/5)I1 I3(R3 + R42) = I1R42 1 1 1= 1+ 1 = + R423 R42 R2 3.3 kΩ 2.2 kΩ E 1= 2 3 5 + = R423 6.6 kΩ 6.6 kΩ 6.6 kΩ R432 = 1.32 kΩ E R4231 = R423 + R2 R4231 = 1.32 kΩ + 2.2 kΩ = 3.52 kΩ
4. 4. 2. (35 pt.s) Three long wires, two carrying a current I = 5A out of the page and the third carrying a current 3I = 15 A into the page, are placed at the vertices of an hexagon, each 0.12 m away from point P at the center of the hexagon. 2 (a) (13 pt.s) Calculate the magnitude of the B- field at point P due to the 15 A current. d µ 0I B= 2πr 3 F 1 B1 d r2 = d = .12 m I2 = 15 A d B2 x axis (b) (12 pt.s) Calculate the magnitude and direction of the d d total B-field at point P. B3 y axis Btot = B2 + B1 + B3 I1 = I3 = 5 A r1 = r3 = d Magnitude for B1 and B3 B1 counterclockwise about wire 1, B3 counterclockwise about wire 3, and B2 clockwise about wire 2, by (modified) right-hand rule (rhr). Directions for B1 , B2 and B3 shown in figure, with B1 and B3 both 600 from +x axis, (tangent lines perpendicular to radii). The y-components of B1 and B3 cancel, while the x-components add. In negative x direction (c) (10 pt.s) A fourth wire carrying a 10 A current into the page is now placed at point P. What is the magnitude and direction of the magnetic force acting on a 0.8 m length of this wire? F = I(L x Btot) Direction for F shown, ⊥ to Btot and L, by rhr. F = ILBtotsin900 = 13.3 x 10-5 N Magnitude of F
5. 5. 3. (35 points) A capacitor consisting of two 6.3 cm2 parallel conducting plates separated by a 0.95 mm air gap is connected to a battery and charged to 18 V. (a) (8 points) Calculate the magnitude of the electric field between the plates. E = 18 V A = .00063 m2 d = .00095 m Relationship between voltage and (uniform) electric field V = Ed for parallel capacitor plates separated by distance d. V=E Voltage across capacitor equals EMF of battery. (b) (9 points) Calculate the surface charge density on the positively charged plate. Treat plates as infinite sheets of charge to find the electric field between them, far from the edges. Electric field between two infinite sheets of charge E = σ/ε0 with opposite charge densities ±σ. (c) (9 points) Calculate the total energy stored in the capacitor. Energy stored in a capacitor with charge ±Q on the U = ½ QV two opposing plates, and a voltage V between them. V=E σ ≡ Q/A, so Q = σA (d) (9 points) If the capacitor is disconnected from the battery and then discharged through a 1.7 kΩ resistor, what is the initial current through The voltage across the the resistor? capacitor must be the same as VC = VR the voltage across the resistor VR = IR +Q I0 VC = E = I0R True at initial time. −Q
6. 6. 4. (20 points) The magnetic field component of a traveling electromagnetic wave has the form B(x ,t) = j (0.82mT) sin[(2.4 x 106 rad/m) x – (7.2 x 1014 rad/s) t] a.) (10 points) Calculate the wavelength and speed of the wave. Asin(k x – ωt] Form of equation for a traveling wave ω = 2πf = 7.2 x 1014 rad/s k = 2π/λ = 2.4 x 106 rad/m (10 points) What is the magnitude and direction of the maximum Electric field? S = Eµ B Direction of propagation determined by Poynting Vector. x 0 5. (20 points) An underwater stationary sonar system emits sound of frequency 16.2 kHz to detect and measure the speed of nearby submarines. The speed of sound in water is 1460 m/s. a.) (10 pts) What is the wavelength of the sound in the water? v = λf b.) (10 pts) If a sub is approaching the system at a speed of 11 m/s, what will be the frequency of the sound detected on the sub? Observer moves, so correction factor in numerator (on top). Observer moves towards, so correction factor makes f larger. Sub (observer) hears f'.
7. 7. 6. (35 pt.s) In the figure to the right, the two open circles represent charges of +6.7 nC. They are separated by distances as indicated. 3 cm a. (15 pt.s) What is magnitude and direction of the electric field at the lower left point (the dark circle)? Q1 Total E-field is vector sum of Etot = E1 + E2 4 cm r1 E-field due to each charge. E2 Q2 E1 = −(kQ1/r12)ĵ= −(37646 N/C)ĵ r2 E2 = −(kQ2/r22)î = −(66926 N/C)î E1 Magnitude from kQ/r2 , (Q as a magnitude). Direction from applying the rule “opposites charges attract, like charges repel” to a positive test charge at lower left point. Etot2 = E12 + E22 tanθ = E1/E2 b. (10 pts) What is the electric potential at this point? Total electric potential (voltage) is (scaler) sum of Vtot = V1 + V2 voltage due to each charge. Sign of charge matters. V1 = kQ1/r1= +1506 V V2 = kQ2/r2= +2008 V c. (10 pts) If a proton is placed at this point, what is its potential energy relative to a point very far away? (Effectively at infinity.) ΔU = qΔV V∞ = 0
8. 8. 7. (5 points) A dielectric slab is placed midway between the parallel plates of a capacitor, which has had a charge Q placed on it. Does the voltage across the capacitor: (a) increase? (b) decrease (c) remain the same? Why? C = Q/V C = εA/d ε = κε0 effective κ > 1, so V decreases 8. (5 points) A computer monitor uses a grid to accelerate and a magnetic field to deflect an electron beam (“cathode ray”) vertically. In the region of the magnetic field, the electrons move in a circular arc. The direction of the magnetic field is: (a) upwards along the page (b) to the left (c) downwards along the page (c) into the page (d) to the right (e) out of the page Explain. Assume electron velocity is in the + x-direction. F = qv x B Force y is in the + y-direction. v x B in the - y-direction for electron, (q = -e). y x z
9. 9. 9. (5 points) A 5000-Hz sound wave is emitted by a stationary source toward an object moving 3.5 m/s toward the source. What is the frequency of the wave reflected by the moving object as detected by a detector at rest near the source? (a) 5051 Hz (b) 4948 Hz (c) 5103 Hz (d) 4898 Hz Show your work here. Double Dopler shift: Sound waves travel to object, echo off the object, then travel back to the original source. For the 1st leg of wave’s journey, and the object is the observer, which is moving with speed u towards the stationary source. frequency f' heard by object. For 2nd leg of journey, the object is the source, sending out (reflected) waves of frequency f', and moving with speed u towards the observer, which is the original (stationary) source. frequency f'' heard by stationary observer. 10. (5 points) The circuit to the right contains two identical bulbs. What happens to bulb 1 when the switch is closed? (a) goes out (b) gets dimmer E (c) stays the same 2 1 (d) gets brighter Explain briefly. Light bulb are just a kind of resistor. The bulbs shown are wired in parallel. When the switch is closed, the equivelent resistance of the circuit will be half the resistance of one bulb, so the current out of the battery will double, but will split (evenly) at the junction, so the current to bulb 1 will be the same, so the power P = I2R will be the same.