Midterm%20Exam%202%20Solutions

Midterm Exam 2%2 - Physics 11b Midterm Exam II with Solutions Name Please place a check mark X next to your section leader Ognjen Ilic Subhaneil

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Unformatted text preview: Physics 11b: Midterm Exam II with Solutions April 17, 2008 Name: Please place a check mark ( X ) next to your section leader: Ognjen Ilic Subhaneil Lahiri Josh Lapan Daniyar Nurgaliev Roberto Martinez Problem Score Total Points 1 15 2 20 3 15 4 10 5 20 6 10 Total Raw Score: / 90 1 Problem 1 A metal rod is forced to move with constant velocity ~v along two parallel metal rails, connected with a strip of metal at one end as in the figure below. A magnetic field of magnitude B = 0 . 75 T points out of the page. (a) If the rails are separated by L = 15 . cm and the speed of the rod 55 . cm/s, what is the emf generated? (b) If the rod has a resistance of 18.0 Ω and the rails and connector have negligible resistance, what is the current in the rod? Explain your work. (Solution) a) Over small time dt the rod moves the distance vdt increasing the area of the loop by dS = Lvdt which leads to change of magnetic flux through this loop by d Φ = BdS = BLvdt . The emf generated is given by | ε | = d Φ dt = BLvdt dt = BLv = 6 . 2 × 10- 2 V . b) According to Omh’s law i = ε R = 3 . 4mA . Problem 2 A variable capacitor with a range from 10 to 365 pF is used with a coil to form a variable-frequency LC circuit to tune the input to a radio. Sketch the circuit. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain fre- quencies from . 54 MHz to 1 . 60 MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance should be added and (c) what inductance should the coil have? Explain your work. (Solution) L C var Figure 1: Circuit with inductor L and variable capacitor 10 pF ≤ C var ≤ 365 pF. (a) Figure 1 shows the circuit diagram. We know that ω = 2 πf = 1 √ LC , so f (0) max = 1 2 π q LC (0) min , f (0) min = 1 2 π q LC (0) max . where C (0) max = 365 pF and C (0) min = 10 pF, so the ratio is f (0) max f (0) min = v u u t C (0) max C (0) min ≈ 6 . 04 . (1) L C var C Figure 2: Circuit with inductor L in series with a constant capcitor C and a variable capacitor...
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This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

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Midterm Exam 2%2 - Physics 11b Midterm Exam II with Solutions Name Please place a check mark X next to your section leader Ognjen Ilic Subhaneil

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