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ohw19-solutions - howard (cah3459) ohw19 turner (56705)...

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ohw19-solutions - howard (cah3459) ohw19 turner (56705)...

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howard (cah3459) – ohw19 – turner – (56705) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A wire carries a current oF I = 44 A along the x -axis From x 1 = - 5 cm to x 2 =1 . 6 cm. I x x P y 1 x 2 ±ind the magnitude oF the resulting mag- netic feld at the point r = 4 cm on the y axis. Correct answer: 126 . 785 μ T. Explanation: Let : x 1 = - 5 cm , x 2 . 6 cm , and r = 4 cm = 0 . 04 m . ±rom the Biot-Savart Law d & B = μ 0 4 π I d&s × ˆ r r 2 , the magnetic feld at a point From a straight line segment is B = μ 0 I 4 π r (cos θ 1 - cos θ 2 ) . In this case, θ 1 = arctan ± y - x 1 ² = arctan ³ 4 cm - ( - 5 cm) ´ = 38 . 6598 , and θ 2 = 180 - arctan ± y x 2 ² = 180 - arctan ± 4 cm 1 . 6 cm ² = 111 . 801 , so the resultant magnetic feld is B = μ 0 I 4 π d (cos θ 1 - cos θ 2 ) = μ 0 (44 A) 4 π (0 . 04 m) (cos 38 . 6598 - cos 111 . 801 ) = 126 . 785 μ T . 002 (part 1 of 2) 10.0 points Imagine a very long, uniForm wire that has a linear mass density oF 0 . 0033 kg / m and that encircles the Earth at its equator. Assume the Earth’s magnetic dipole mo- ment is aligned with the Earth’s rotational axis. The Earth’s magnetic feld is cylindri- cally symmetric (like an ideal bar magnetic). The acceleration oF gravity is 9 . 8m / s 2 and the magnetic feld oF the earth is 9 × 10 - 5 T . What is the magnitude oF the current in the wire that keeps it levitated just above the ground? Correct answer: 359 . 333 A. Explanation: Let : g =9 . / s 2 , B Earth × 10 - 5 T , and μ =0 . 0033 kg / m . The Earth’s geographical North pole is a mag- netic South pole and vice versa . The North pole oF a compass is attracted to the South magnetic pole oF the Earth, the place where Santa Claus lives. The magnetic and gravitational Forces must balance. ThereFore, it is necessary to have (For unit length oF wire) F = IB = μg I = B = (0 . 0033 kg / m) (9 . / s 2 ) 9 × 10 - 5 T = 359 . 333 A .
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howard (cah3459) – ohw19 – turner – (56705) 2 003 (part 2 of 2) 10.0 points The current in the wire goes in the 1. same direction as the Earth’s spinning motion (West to East). correct 2. opposite direction as the Earth’s spinning motion (East to West). Explanation: Since the Sun rises in the East, the Earth is turning towards the East and the Earth’s angular velocity is parallel to the Earth’s axis, pointing from South to North. S N v E F B I v Earth The total force on the wire is zero and the gravitational force is downward (inward), so the magnetic force must be upward (out- ward). If the current goes from West to East, in the same direction as the Earth’s spinning motion, & I × & B will be vertically upward (out- ward). 004 (part 1 of 2) 10.0 points A long, solid cylindrical conductor of radius 0 . 8 m carries a current of 2 . 4 nA parallel to its axis and uniformly distributed over its cross- section. The conductor has a resistivity of 0 . 006 Ω · m.
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