PHYS 211 Exam 2, Spring 2007

PHYSICS 211 Quiz II TIME: 50 minutes April 16, 2008 Part A (30%) Part B: Multiple choice questions ( 25%): 1. (5%) Gauss’s Law states that the net electric flux, · dA, through any closed surface is proportional to the charge enclosed: · dA = . The analogous formula for magnetic fields is: a. · dA = 0. b. · dA = . c. · dA = . d. · dA = . e. · dA = – . 2. (5%) Which diagram correctly shows the magnetic field lines created by a circular current loop in which current flows in the direction shown? 3. (5%) Equal charges, one at rest, the other having a velocity of 10 4 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field? a. The charge that is at rest. b. The charge that is moving, if its velocity is parallel to the magnetic field direction when it is released. c. The charge that is moving if its velocity makes an angle of 45 o with the direction of the magnetic field when it is released. d. The charge that is moving if its velocity is perpendicular to the magnetic field direction when it is released. e. All the charges above experience equal forces when released in the same magnetic field. Score:________ Check if solution is continued on the back. 1 4. (5%) A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3  10 –27 kg, q = 1.6  10 –19 C) a. 19 mm b. 16 mm c. 20.3 mm d. 10 mm e. 9.0 mm 5. (5%) At what rate is thermal energy generated in the 30- resistor? a. 20 W b. 27 W c. 60 W d. 13 W e. 30 W 6. (5%) At t = 0 the switch S is closed with the capacitor uncharged. If C = 50 µF,  = 20 V, and R = 4.0 k, what is the charge on the capacitor when I = 2.0 mA? a. 360 µC b. 480 µC c. 240 µC d. 600 µC e. 400 µC Score:________ Check if solution is continued on the back. 2 Part C- Problems (70 %) 1.a. Demonstrate that if 3 Resistors are set in parallel then the equivalent resistor is given by Using Ohm’s law we have DV=IR. The current in the equivalent circuit should be the same as the total current in the three loops leading to Ieq=I1+I2+I3 Ieq=( DV1/R1) + ( DV2/R2) + ( DV3/R3) = ( DV/Req) In parallel DV is the same leading to DV1=DV2=DV3 This leads to b.(8%) Demonstrate that if 3 resistors are set in series, then the equivalent resistor is given by Req = R1 + R2 + R3 In series we DV1 + DV2 + DV3 = DV whereas I is unchanged. Replacing DV1=IR1 We obtain V=IReq= IR1+IR2+IR3 leading to Req = R1 + R2 + R3 2. Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kV and enter a uniform magnetic field of 1.20 T directed perpendicular to their velocities (take proton mass to be 1.66 × 10−27  kg and the electric charge = 1.6x10-19 C). (a) Assuming energy is conserved, determine the velocity, and the radius of their circular path. so (a) (b) Repeat for uranium-235 ions and indicate how does the ratio of these path radii depend on the accelerating voltage and on the magnitude of the magnetic field? Score:________ Check if solution is continued on the back. 3 (b) The ratios of the orbit radius for different ions are independent of and B. 3. Given the circuit below with 3 A of current running through the 4 Ω resistor as indicated … Determine … a. the current through each of the other resistors, b. the voltage of the battery on the left c. the power delivered to the circuit by the battery on the right  a. Let's identify the currents through the resistors by the value of the resistor (I1, I2, I3, I4) and the currents through the batteries by the side of the circuit on which they lay (IL, IR). Start with the 2 Ω resistor. Apply the loop rule to the circuit on the lower right. 20 V = I2(2 Ω) + (3 A)(4 Ω) I2 = 4 A Proceed to the 3 Ω resistor. Apply the junction rule to the junction in the center of the circuit. I2 = I3 + I4 4 A = I3 + 3 A I3 = 1 A The current through the 1 Ω resistor most certainly runs from right to left. If we apply the loop rule to the top circuit, we'll have to run against that current. This changes what is normally considered a potential drop into a potential increase. (Kind of like skiing up a mountain instead of down.) I1(1 Ω) = (4 A)(2 Ω) + (1 A)(3 Ω) I1 = 11 A Score:________ Check if solution is continued on the back. 4  b. Apply the loop rule to the outer circuit to get the voltage of the battery on the left (continuing with the assumption that the current is running counterclockwise). We find ourselves running through the left battery backwards. This changes what is normally considered a potential increase into a potential decrease. (Kind of like using the ski lift to travel down a mountain instead of up.) 20 V = (11 A)(12 Ω) + V2 VL = 9 V Let's verify this result by repeating the procedure for the bottom circuit. 20 V = (4 A)(2 Ω) + (1 A)(3 Ω) + V2 VL = 9 V Good, we get the same answer by two methods. We must be doing the right thing.  c. The power delivered to the circuit by the battery on the right is the product of its voltage times the current it drives around the circuit. We already have the voltage (it's given in the problem) all that remains is to determine the current. Apply the junction rule to the junction on the left … IL = I1 + I3 IL = 11 A + 1 A IL = 12 A and again to the junction at the bottom … IR = IL + I4 IR = 12 A + 3 A IR = 15 A to find the power of the battery on the right … P = VI P = (20 V)(15 A) P = 300 W Score:________ Check if solution is continued on the back. 5