PHY 303L-Final

# PHY 303L-Final - final 01 – VARELA CHRISTOPHER – Due...

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Unformatted text preview: final 01 – VARELA, CHRISTOPHER – Due: Dec 17 2007, 4:00 pm 1 Question 1, chap 32, sect 4. part 1 of 1 10 points E S C R L I Let the resistor have a resistance of 1 MΩ, the capacitor to have a capacitance of 2 μ F, and the inductor to have an inductance of 5 mH. Finally, let the battery have a voltage of 18 . 1 V. What is the charge on the capacitor after the switch has been closed for a long time? Correct answer: 3 . 62 × 10 − 5 (choice number 6). Explanation: Let : E = 18 . 1 V and C = 2 μ F = 2 × 10 − 6 F . After a long time t the capacitor will reach its maximum charge q max . When this occurs the current will drop to zero. Setting I = 0 in the loop equation from Part 1 gives E − q max C = 0 , so q max = C E = (2 × 10 − 6 F) parenleftbigg 1 × 10 − 6 F μ F parenrightbigg (18 . 1 V) = 3 . 62 × 10 − 5 C . Question 2, chap 32, sect 3. part 1 of 1 10 points An inductor has a 47 . 1 Ω impedance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 138 . 5 V rms voltage? Correct answer: 4 . 99028 (choice number 3). Explanation: Let : X L = 47 . 1 Ω , f = 60 Hz , f ′ = 50 Hz , and V rms = 138 . 5 V . The maximum voltage of the circuit is V max = V rms √ 2 , and the inductive reactance is X L = ω L, so X ′ L = ω ′ ω X L = f ′ f X L , and the maximum current is I max = V max X ′ L = √ 2 V rms f f ′ X L = √ 2 (138 . 5 V) (60 Hz) (50 Hz) (47 . 1 Ω) = 4 . 99028 A . Question 3, chap 30, sect 1. part 1 of 1 10 points A device (“source”) emits a bunch of charged ions (particles) with a range of ve- locities (see figure). Some of these ions pass through the left slit and enter “Region I” in which there is a vertical uniform electric field (in the − ˆ direction) and a B uniform mag- netic field (aligned with the ± ˆ k-direction) as shown in the figure by the shaded area. q m Region of Magnetic Field B + V d x y z r Region I Region II final 01 – VARELA, CHRISTOPHER – Due: Dec 17 2007, 4:00 pm 2 Figure: ˆ ı is in the direction + x (to the right), ˆ is in the direction + y (up the page), and ˆ k is in the direction + z (out of the page). The ions that make it into “Region II” are observed to be deflected downward and then follow a circular path with a radius of r . The charge on each ion is q. What is the mass of the ions? 1. m = q B r E 2. m = q E r B 3. m = q B r E 2 4. m = q B 2 r E 5. m = q E 2 r B 6. m = q B r E 7. m = q E r B 8. m = q E r B 2 9. m = q B 2 r E correct 10. m = q E 2 r B Explanation: Since the electric and magnetic forces on the ion are equal, q E = q v B v = E B . The radius of a circular path taken by a charged particle in a magnetic field is given by r = mv q B ....
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PHY 303L-Final - final 01 – VARELA CHRISTOPHER – Due...

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