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Unformatted text preview: × 106 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 × 106 . Explanation: Let r = 0 . 5 m I = 5 A and μ = 1 . 25664 × 106 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 magnetic Feld due to AB . The semicircular arc BCD subtends π radian. 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 × 106 T · m / A) (5 A) 4 (0 . 5 m) × p 2 π + 1 P = 5 . 14159 × 106 T ....
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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