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Physics 221 β Exam 2 π = 1.6 × 10β19 πΆ π = 9.0 × 109 πβ π2 πΆ β) Lorentz Force Law: π(πΈβ + π£ × π΅ βπ = βππΈππππ π Q = πΆΞπ πΌ= = βπΈβ β βπ π2 πΆ2 ππ = 8.9 × 10β12 πβ π2 β β π΄ Ξ¦π΅ = π΅ point charge: π = 1 1 ππΈ = 2πΆ = 2 πβπ = 2 πΆ(βπ)2 Ξπ Ξπ‘ Ξππ = πΌπ ΞπΌ ΞππΏ = πΏ Ξπ‘ For AC Circuits: π = πΌπ Ξ¦ = πΏπΌ ΞππΏ = πΌππΏ |π| = βπ 2 + (ππΏ β ππΆ )2 ΞππΆ = πΌππΆ π πΌ π Long, Straight Wire: π΅ = 2ππ 1 Volt = 1 J/C πΏ πΆπ ππ = (β πΆ ) π π π ππ = β π π 1 β1 πΏπππ = (β πΏ ) π ππΆ = 1 2πππΆ 1 Volt/m = 1 N/C 1 β1 πΆπππ = β πΆπ π = ππ΄ πΏπ ππ = β πΏπ ππ π πβ ππ ππ = 4π × 10β7 π 2 β π΄2 1 β1 π πππ = (β π ) π ππ 2 π΄ π Solenoid: πΏ = ππΏ = 2πππΏ |π| = β(Ξππ )2 + (ΞππΏ β ΞππΆ )2 πΆ = π πΆπ£ππ πΌπ ππ = πΌππππ ππΏ βππΆ ) π πΏ = ArcTan ( PHYS 221 Exam 2 THIS PAGE LEFT BLANK 2 10 July 2015 PHYS 221 Exam 2 10 July 2015 1. An RLC series circuit consists of a 1 kΞ© resistor, a 1 ΞΌF capacitor and a 0.2 H inductor connected to a 150 V A/C source. What is the current delivered by the source at resonance? a. 20.3 mA b. 150 mA At resonance, the impedance is equal to the resistance which means the current is just c. 10.3 mA 150 π 1π Ξ© = 150 ππ΄ d. 15.0 mA e. 54.3 mA 2. Consider the circuit shown below with V = 9.0 V, R1 = 3000 Ξ©, R2 = 7000 Ξ© and L = 7.7 mH. What is the energy stored in the inductor? a. 3.5 x 10-8 J b. 6.9 x 10-8 J c. 1.2 x 10-5 J d. 3.0 x 10-3 J e. 0 J The energy stored in the inductor is based on the current flowing through it. Since the inductor and R1 are in the same branch, finding the current through R1 will do the trick. Using Kirchoffβs rules, we can take the outer loop. After a long time, the voltage across the inductor goes to zero which means the full all 9.0 V are dropped across R1. This produces a 3 mA current in the branch with the inductor. 1 1 ππΏ = πΏπΌ 2 = (7.7 ππ»)(3 ππ΄)2 = 3.5 × 10β8 π½ 2 2 3 PHYS 221 Exam 2 10 July 2015 3. A 24 V DC battery is connected to the primary of a transformer as shown. The transformer has N1 = 5 turns and the transformer secondary has N2 = 10 turns. Assuming a that the magnetic flux linkage is 100%, what is the voltage across R? a. 0 V b. 6 V c. 48 V d. 12 V Transformers rely on Faradayβs law of induction to operate. Faradayβs law of induction requires a magnetic field that changes over time, but a DC voltage source creates a steady current which makes a constant magnetic field. There is no induction, so there is no voltage across R. 4. The series RLC circuit shown below has R = 10 Ξ©, L = 1.0 H, and C = 1.0 ΞΌF. The AC generator is applying a peak voltage given by β° = 14.14 V. What is the RMS voltage across the inductor at resonance? a. 0 V b. 160 V c. 225 V d. 1000 V e. 1414 V The RMS voltage across the inductor is simple this: ΞππΏ,π ππ = πΌπ ππ ππΏ . As before, we are at resonance, so that means the impedance is just the resistance giving a peak current of 1.414 A. However, we need the RMS current, so we convert and obtain a 1 A. Next we need the inductive reactance which is: ππΏ = ππΏ = ( that π = 1 βπΏπΆ 1 βπΏπΆ πΏ ) πΏ = β = 1000Ξ© where we used the fact πΆ at resonance. The final result is that the RMS voltage is 1000 V. This is greater than the voltage source, but remember that the voltage across the capacitor cancels this out since they are equal, but out of phase by 180 degrees. 4 PHYS 221 Exam 2 10 July 2015 5. Each chamber in the figure below has a unique magnetic field. A particle with charge q = 25 mC and a mass of 10-10 kg enters the left chamber where B = 1.0 T directed into the page with a velocity of 75 m/s. If the magnetic field in the second chamber is 0.5 T directed out of the page, at what velocity does the particle leave chamber 2? a. 37.5 m/s b. 75 m/s c. 150 m/s d. 5625 m/s e. Insufficient information to determine Magnetic forces do no work so the incoming speed is the outgoing speed. 6. Two charged particles are traveling in circular orbits with the same speed in a region of uniform magnetic field that is directed into the page. The magnitude of the charge on each charge is identical, but the signs are unequal. π£ π£ πΉ β π£×π΅ β π£×π΅ πΉ Which one of the entries in the table is correct? a. b. c. d. e. Mass Relationship m1 = m2 m1 > m2 m1 < m2 m1 > m2 m1 < m2 Sign of Q1 + + + Sign of Q2 + + - The relative direction of the cross product and the force gives us the sign of the charge. For the first particle they are opposite to each other meaning the charge is negative. The second particle is then a positive charge. If we use Newtonβs second law with the understanding that we are undergoing uniform circular motion, then we get the following equation: ππ£π΅ = π first particle has a larger mass. π£2 π 5 . Solving for r shows us that the radius is proportional to the mass meansing the PHYS 221 Exam 2 10 July 2015 7. A negative ion is moving east near the equator where the Earthβs magnetic field is horizontal to the Earthβs surface pointing north. The direction of the magnetic force on the ion is: β π΅ a. North b. South π£ c. Up d. Down e. There is no force Marking out the velocity and magnetic field vectors makes it easier to use the right hand rule. Notice how to rotate the velocity into the direction of the magnetic field we make a CCW rotation. This means that the cross product is into the page. However, the charge is negative. This means the force is in the opposite directionβ¦ i.e. it is up. 8. A positively charged particle moves through a magnetic field with a velocity that makes an angle of 32° with the magnetic field. Describe the kind of motion the particle undergoes. a. Parabolic b. Uniform circular motion c. Spiral d. Straight line e. Helical Essentially a uniform magnetic field causes the component of the velocity that is perpendicular to it to rotate. In a uniform field, this causes uniform circular motion perpendicular to the magnetic field. So if the entire velocity is along the magnetic field, we just have a straight lineβ¦ it itβs completely perpendicular then we have uniform circular motion. Anything in between is a combination of the twoβ¦ helical motion. 6