18. Currents and the b field-solutions

# 18. Currents and the b field-solutions - gilvin(jg47854 18...

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gilvin (jg47854) – 18. Currents and the b field – meyers – (21235) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This homework is due Tuesday, April 19, at midnight Tucson time. 001(part1of2)10.0points Consider the application of the Biot-Savart law Δ vector B = μ 0 4 π I Δ vector × ˆ r r 2 . Consider the magnetic field at O due to the current segments B B +semicircle+ CC . The length of the linear segments, B B = CC = d . The semicircle has a radius R . x y I I I 180 O R B B C C A The magnetic field B O at the origin O due to the current through this path is 1. undetermined, since B 0 = 0 . 2. out of the page. 3. into the page. correct Explanation: BasicConcepts: Biot-Savart Law Solution: ˆ r is pointing from A to O ; i.e. , from the location of the source element to the location of the field of concern. So ˆ is the direction. 002(part2of2)10.0points At O , the magnitude of the magnetic field due to the current segments described above is given by 1. B O = μ 0 I R 2. B O = μ 0 I π R 3. B O = μ 0 I 2 R 4. B O = μ 0 I π 2 R 5. B O = μ 0 I π 4 R 6. B O = μ 0 I 2 π R 7. B O = μ 0 I 4 π R 8. B O = 0 9. B O = μ 0 I π R 10. B O = μ 0 I 4 R correct Explanation: Since the Biot-Savart equation has the cross product ˆ ı × hatwide R in the numerator, this term implies that the contribution of B B and CC to the field at O is zero. The magnetic field at at the center of an arc with a current I is B = μ 0 I 4 π integraldisplay dvectors × hatwide R R 2 = μ 0 I 4 π R 2 integraldisplay ds = μ 0 I 4 π R 2 integraldisplay R dθ = μ 0 I 4 π R integraldisplay π 0 = μ 0 I 4 π R θ vextendsingle vextendsingle vextendsingle vextendsingle π 0 = μ 0 I 4 R . 003 10.0points In Niels Bohr’s 1913 model of the hydro- gen atom, an electron circles the proton at a distance of 7 . 09 × 10 11 m with a speed of 1 . 84 × 10 6 m / s. The permeability of free space is 1 . 25664 × 10 6 T · m / A .

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gilvin (jg47854) – 18. Currents and the b field – meyers – (21235) 2 Compute the magnetic field strength that this motion produces at the location of the proton. Correct answer: 5 . 8566 T. Explanation: Let : μ 0 = 1 . 25664 × 10 6 T · m / A , q = 1 . 6 × 10 19 T , v = 1 . 84 × 10 6 m / s , and R = 7 . 09 × 10 11 m . Current generated by the circulating electron is I = q v 2 π R = (1 . 6 × 10 19 T) (1 . 84 × 10 6 m / s) 2 π (7 . 09 × 10 11 m) = 0 . 000660863 A . Thus the magnetic field at the center of the circular loop is B = μ 0 I 2 R = (1 . 25664 × 10 6 T · m / A) 2 (7 . 09 × 10 11 m) × (0 . 000660863 A) = 5 . 8566 T . 004 10.0points The closed loop shown in the figure carries a current of 16 A in the counterclockwise direc- tion. The radius of the outer arc is 60 cm , that of the inner arc is 40 cm . Both arcs ex- tent an angle of 60 . The permeability of free space is 4 π × 10 7 T m / A.
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