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Unformatted text preview: nguyen (jmn727) homework 21 Turner (59070) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A metal strip 1 . 8 cm wide and 0 . 09 cm thick carries a current of 28 A in a uniform magnetic field of 1 . 8 T. The Hall voltage is measured to be 4 . 35 V. The charge on an electron is 1 . 6 10 19 C. vector B 1 . 8 cm . 09 cm I a b Calculate the drift velocity of the electrons in the strip. Correct answer: 0 . 134259 mm / s. Explanation: Let : V H = 4 . 35 V = 4 . 35 10 6 V , B = 1 . 8 T , and w = 1 . 8 cm = 0 . 018 m . The Hall voltage as a function of the drift velocity of the electrons in the strip is V H = v d B w , so v d = V H B w = 4 . 35 10 6 V (1 . 8 T) (0 . 018 m) 10 3 mm 1 m = . 134259 mm / s . 002 (part 2 of 3) 10.0 points The charge on the electron is 1 . 6 10 19 C. Find the number density of the charge car riers in the strip. Correct answer: 8 . 04598 10 28 m 3 . Explanation: Let : I = 28 A , t = 0 . 09 cm = 0 . 0009 m , and q = 1 . 6 10 19 C . The current is I = n A q v d n = I A q v d = I w t q v d = 28 A (0 . 018 m) (0 . 0009 m) 1 (1 . 6 10 19 C) (0 . 000134259 m / s) = 8 . 04598 10 28 m 3 . 003 (part 3 of 3) 10.0 points Is point a or point b at the higher potential? 1. V a < V b 2. V a > V b correct 3. V a = V b Explanation: Apply a righthand rule to I vector and vector B to conclude that positive charge will accumulate at a and negative charge at b , so V a > V b . 004 (part 1 of 2) 10.0 points A current I = 3 A flows through a wire perpendicular to the paper and towards the reader at A and back in the opposite direc tion at C . Consider the wires below the plane at A and C to be semiinfinite. In the figure, L 1 = 5 m, R = 5 m, and L 2 = 6 m and there is a B = 9 . 73 T magnetic field into the paper (not including the field due to the current in the wire). Caution: It may be necessary to take into account the contribution from the long straight wire which runs up to and down from the underneath side of the page. nguyen (jmn727) homework 21 Turner (59070) 2 5m 3A 3 A 6 m 3 A 5 m A C O B = 9 . 73 T What is the magnitude of the force on the wire due to the external magnetic field B ? Correct answer: 433 . 941 N. Explanation: Let : R = 5 m , I = 3 A , L 1 = 5 m , L 2 = 6 m , and B = 9 . 73 T . By the BiotSavart law, d vector B = 4 I dvectors r r 2 . The contribution from the long straight wire which runs into and out of the page is zero since the external field and the current are parallel. The force on a current carrying wire from point A, at vectorr 1 , to point C, at vector r 2 , in a uniform field is vector F = I integraldisplay vectorr 2 vectorr 1 ( dvectors vector B ) ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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