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HW 21 - terry(ect328 homework 21 Turner(59130 This...

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terry (ect328) – homework 21 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice 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 . 2 cm wide and 0 . 13 cm thick carries a current of 22 A in a uniform magnetic field of 1 . 4 T. The Hall voltage is measured to be 4 . 22 μ V. The charge on an electron is 1 . 6 × 10 19 C. vector B 1 . 2 cm 0 . 13 cm I a b Calculate the drift velocity of the electrons in the strip. Correct answer: 0 . 25119 mm / s. Explanation: Let : V H = 4 . 22 μ V = 4 . 22 × 10 6 V , B = 1 . 4 T , and w = 1 . 2 cm = 0 . 012 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 . 22 × 10 6 V (1 . 4 T) (0 . 012 m) · 10 3 mm 1 m = 0 . 25119 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: 3 . 50893 × 10 28 m 3 . Explanation: Let : I = 22 A , t = 0 . 13 cm = 0 . 0013 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 = 22 A (0 . 012 m) (0 . 0013 m) × 1 (1 . 6 × 10 19 C) (0 . 00025119 m / s) = 3 . 50893 × 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 correct 2. V a < V b 3. V a = V b Explanation: Apply a right-hand 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 = 7 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 semi-infinite. In the figure, L 1 = 5 m, R = 3 m, and L 2 = 7 m and there is a B = 5 . 75 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.
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terry (ect328) – homework 21 – Turner – (59130) 2 5 m 7 A 7 A 7 m 7 A 3 m A C O B = 5 . 75 T What is the magnitude of the force on the wire due to the external magnetic field B ? Correct answer: 515 . 452 N. Explanation: Let : R = 3 m , I = 7 A , L 1 = 5 m , L 2 = 7 m , and B = 5 . 75 T . By the Biot-Savart law, d vector B = μ 0 4 π I dvectors × ˆ 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 ) . Since vector B is a constant, it can be taken out of the integral and we can write (recalling that if we change the order of the cross product, we need to change the overall sign) vector F = I vector B × integraldisplay vectorr 2 vectorr 1 dvectors = I vector B × ( vectorr 2 vectorr 1 ) .
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