2UFhp8988pro_1210016696_jwk572

2UFhp8988pro_1210016696_jwk572 - × 10-6 T m A The radius...

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inclasstest 06 – KIM, JI – Due: Mar 3 2008, 1:00 pm 1 Question 1, chap -1, sect -1. part 1 of 1 1 points This is a question for showing your atten- dance. Please choose the correct answer be- low, or you will lose 25% of this weekly test. 1. This is the correct answer. correct 2. This is the wrong answer. Explanation: It is correct because the choice does not lie. Question 2, chap 29, sect 5. part 1 of 1 3 points An inFnite wire is bent as shown in the Fgure. The current is I . It consists of three segments, AB , BCD , and DE . AB and DE are parallel and both inFnite in length. BCD is a semi-circular arc, with radius OB = OC = OD = r . The following questions are concerned with the magnetic Feld vector, v B , at O . x y I II III IV z D B C O E A The permeability of free space is 1 . 25664
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Unformatted text preview: × 10-6 T · m / A. The radius of curvature is 0 . 5 m . ±ind the magnitude of v B at O due to a 5 A current through the entire wire ABCDE . Correct answer: 5 . 14159 × 10-6 . Explanation: Let r = 0 . 5 m I = 5 A and μ = 1 . 25664 × 10-6 T · m / A . B C O A I x Δ x s r θ Note: The magnetic Feld due to DE is equal in magnitude and direction to the mag-netic Feld due to AB . The semicircular arc BCD subtends π ra-dian. Thus the total magnetic Feld is B total = B AB + B DE + B BCD = 2 p μ I 4 π r P + μ I 4 r = μ I 4 r p 2 π + 1 P = (1 . 25664 × 10-6 T · m / A) (5 A) 4 (0 . 5 m) × p 2 π + 1 P = 5 . 14159 × 10-6 T ....
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