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Exam3_2010Spr_Solutions

# Exam3_2010Spr_Solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2010 8.02 Exam Three Spring 2010 Solutions Problem 1: (25 points) Five Concept Questions. Please circle your answers. Question 1 (5 points): A very long solenoid consisting of N turns has radius R and length d ( d >> R ). Suppose the number of turns is halved keeping all the other parameters fixed. The self inductance a) remains the same. b) doubles. c) is halved. d) is four times as large. e) is four times as small. f) None of the above. Solution e. The self-induction of the solenoid is equal to the total flux through the object which is the product of the number of turns time the flux through each turn. The flux through each turn is proportional to the magnitude of magnetic field. By Ampere’s Law the magnitude of the magnetic field is proportional to the number of turns per unit length or hence proportional to the number of turns. Hence the self- induction of the solenoid is proportional to the square of the number of turns. If the number of turns is halved keeping all the other parameters fixed then he self inductance is four times as small.

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` 2 Question 2 (5 points): The sketch below shows three wires carrying currents 1 I , 2 I and 3 I , with an Ampèrian loop drawn around 1 I and 2 I . The wires are all perpendicular to the plane of the paper. Which currents produce the magnetic field at the point P shown in the sketch (circle one)? a) 3 I only. b) 1 I and 2 I . c) 1 I , 2 I and 3 I . d) None of them. e) It depends on the size and shape of the Amperian Loop. Solution c. All there currents 1 I , 2 I and 3 I contribute to the magnetic field at the at the point P .
` 3 Question 3 (5 points): A circuit consists of a battery with emf V, an inductor with inductance L, a capacitor with capacitance C, and three resistors, each with resistance R, as shown in the sketch. The capacitor is initially uncharged and there is no current flowing anywhere in the circuit. The switch S has been open for a long time, and is then closed, as shown in the diagram. If we wait a long time after the switch is closed, the currents in the circuit are given by: a) i 1 = 2 V 3 R i 2 = V 3 R i 3 = V 3 R . b) i 1 = V 2 R i 2 = 0 i 3 = V 2 R . c) 12 3 0 33 VV ii i R R == = . d) 123 0 22 iii RR === . e) None of the above. Solution b.: If we wait a long time after the switch is closed, the capacitor is completely charged and no current flows in that branch, 2 0 i = . Also the current has reached steady state and is not changing in time so there is no effect from the self- inductance. Hence the inductor acts like a resistance-less wire. (Note that real inductors do have finite resistance as you saw in your lab.) Therefore the same current flow through resistors 1 and 3 and is given by 13 /2 iiVR = = .

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` 4 Question 4 (5 points): At the moment depicted in the LC circuit the current is non-zero and the capacitor plates are charged (as shown in the figure below). The energy in the circuit is stored
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Exam3_2010Spr_Solutions - MASSACHUSETTS INSTITUTE OF...

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