exam 3 example - Physics 1112 Spring 2009 University of...

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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler Solution to PHYS 1112 In-Class Exam #3A Thu. April 9, 2009, 11:00am-12:15pm Conceptual Problems Problem 1: If a wire of some length L and with a circular cross-section of diameter D has a resistance of 50 kΩ, what will be the resistance of a wire made from the same material, at the same temperature, of length 15 L and diameter 5 D ? (A) 10 kΩ (B) 30 kΩ (C) 60 kΩ (D) 83 . 3 kΩ (E) 150 kΩ Answer: (B) : Using R = ρL/A and A = π ( D/ 2) 2 , we get R L/D 2 . Increasing L L ± = 15 L and D D ± = 25 D thus changes R R ± = 15 × R/ 5 2 = (3 / 5) × 50Ω = 30Ω. Problem 2: Three diFerent circuits, X , Y and Z , are built with the same three resistors, R 1 > 0, R 2 > 0, and R 3 > 0, and the same battery of battery voltage E , as shown in ±ig. 3.04. Compare and rank the magnitude of the voltage drops V 1 across R 1 , observed in the three diFerent circuits. (Hint: V 1 = R 1 I 1 .) Fig. 3.04 (X) R 1 I 2 R 3 R 2 I 3 I 1 E R 3 I 2 R 2 R 1 I 1 I 3 E I 3 R 3 R 1 I 1 R 2 I 2 E (Z) (Y) I o I o I o 1
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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler (A) V 1 ( X ) >V 1 ( Y ) 1 ( Z ) (B) V 1 ( X ) 1 ( Z ) 1 ( Y ) (C) V 1 ( Y ) 1 ( X ) 1 ( Z ) (D) V 1 ( Z ) 1 ( Y ) 1 ( X ) (E) V 1 ( Y ) 1 ( Z ) 1 ( X ) Answer: (E) In circuit Y : R 1 is directly connected to battery, hence I 1 ( Y )= E /R 1 . In circuit Z : R 1 and R 2 in series are connected to battery. Hence I 1 ( Z E / ( R 1 + R 2 ) < I 1 ( Y ), since R 1 + R 2 >R 1 . In circuit X : R 1 , R 2 and R 3 in series are connected to battery. Hence I 1 ( X E / ( R 1 + R 2 + R 3 ) <I 1 ( Z ) 1 ( Y ), since R 1 + R 2 + R 3 1 + R 2 . Since V 1 = R 1 I 1 in all 3 circuits (with the same R 1 in all 3 circuits!), I 1 ( X ) 1 ( Z ) 1 ( Y ) implies V 1 ( X ) <V 1 ( Z ) 1 ( Y ). That is the same as saying V 1 ( Y ) 1 ( Z ) 1 ( X ). Problem 3: A molecular ion beam containing four diFerent types of ions, called P , Q , R and S here, enters a uniform magnetic ±eld, as shown in ²ig. 3.15, with B ± = 0 above the lower horizontal line, and 8 B perpendicular to, and pointing out of, the plane of the drawing. The incident beam, below the lower horizontal line, is in the plane of the drawing. B (out) Fig. 3.15 P S Q R B=0 ²rom the semicircular ion trajectories in the 8 B -±eld in ²ig. 3.15, ±nd the signs (+ or - ) of the charges q P , q Q , q R and q S of each of the four diFerent ion types in the beam. 2
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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler (A) P :+, Q :+, R : - , S : - , (B) P :+, Q : - , R : - , S :+, (C) P : - , Q :+, R :+, S : - , (D) P :+, Q : - , R :+, S : - , (E) P : - , Q : - , R :+, S :+, Answer: (B) For simplicity consider the magnetic force 8 F acting on the ions only at, or immediately after, their point of entry into the 8 B -±eld. At/immediately after their point of entry into the 8 B -±eld, ions R and Q are beginning to be de²ected to the left, ions P and S are beginning to be de²ected to the right. Also, from the mechanics of circular motion, we know that for (semi-)circular trajectories the respective forces 8 F acting on the ions must always point towards the center of that (semi-)circle . And for R and Q that center of the circle in Fig. 3.15 is to the left of the point of entry; whereas for P and S that center of the circle is to the right of the point of entry. Thus, at/immediately after the point of entry we ±nd: for R and Q , 8 F points to the left; for P
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This note was uploaded on 04/16/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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exam 3 example - Physics 1112 Spring 2009 University of...

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