Currents and the b field-solutions

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

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I ?? O = ? 0 I O = ? 0 I O = ? 0 I ? O = ? 0 I ? O = 2? 0 I O = 4? 0 I O = ? 0 I 0 gilvin (jg47854) - 18. Currents and the b field - meyers - (21235) 1 This print-out should have 25 questions. 2. B 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. O = π 0 RI 3. B 2 R 4. B 2 R 001 (part 1 of 2) 10.0 points 5. B Consider the application of the Biot-Savart 4 R law 6. B ℓ × r π R Δ B = µ 0 4 π r 2 7. B Consider the magnetic field at O due to the current segments B B+semicircle+CC . The 8. B O = 0 length of the linear segments, B B = CC = π R d. The semicircle has a radius R. 9. B O = µ 0 RI π y 10. B 180 A I I R 4 R correct Explanation: Since the Biot-Savart equation has the cross product ˆı × R in the numerator, this term implies that the contribution of B B and CC to the field at O is zero. B B O x C I C The magnetic field at at the center of an arc with a current I is The magnetic field B O at the origin O to the current through this path is 1. undetermined, since B 0 = 0 . 2. out of the page. 3. into the page. correct Explanation: Basic Concepts: Biot-Savart Law due B = µ 0 I 4 π = µ 0 I 4 π R 2 = µ 0 I 4 π R 2 = µ 0 I 4 π R ds × R ∫ R 2 ds R dθ π v & π Solution: 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 −ĵ is the direction. = µ 0 I 4 π R θ = µ 0 I 4 R . v & 0 002 (part 2 of 2) 10.0 points At O, the magnitude of the magnetic field due to the current segments described above is given by 1. B 003 10.0 points 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 R 1.25664 × 10 −6 T · m/A .

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12 r i ? 12 r o ? r o ? 1 ? ? 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 . Find the component of the magnetic field at point O along the x axis. Correct answer: −1.39626 × 10 −6 T . Explanation: Let : r i = 40 cm = 0.4 m , r o = 60 cm = 0.6 m , I = 16 A , and µ 0 = 4 π × 10 −7 T m/A . The magnetic field due to a circular loop at
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Currents and the b field-solutions - gilvin(jg47854 18...

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