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Unformatted text preview: GenericStudent – Homework 8 – savrasov – 39823 – Jan 02, 2008 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 1 points Consider the application of the BiotSavart law Δ ~ B = μ 4 π I Δ ~ ‘ × 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 3 m. The permeability of free space is 1 . 25664 × 10 6 T · m / A . x y 7 A 7 A 7 A 180 ◦ O 3 m B B C C A What is the path of the magnetic field B O at the origin O due to the 7 A current? 1. into the page 2. out of the page correct 3. undetermined, since B = 0 Explanation: Basic Concepts: BiotSavart 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 (part 2 of 2) 1 points Find the magnitude of the magnetic field due to the current segments described above at O . Correct answer: 7 . 33038 × 10 7 T. Explanation: Let : μ = 4 π × 10 7 T · m / A = 1 . 25664 × 10 6 T · m / A , I = 7 A , and r = 3 m . Since the BiotSavart equation has the cross product ˆ ı × ˆ 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 I 4 π Z d~s × ˆ r r 2 = μ I 4 π r 2 Z ds = μ I 4 π r 2 Z r dθ = μ I 4 π r Z π dθ = μ I 4 π r θ fl fl fl fl π = μ I 4 r = (1 . 25664 × 10 6 T · m / A) (7 A) 4 (3 m) = 7 . 33038 × 10 7 T . keywords: 003 (part 1 of 3) 1 points A conductor in the shape of a square, whose sides are of length 0 . 802 m, carries a clockwise 14 . 7 A current as shown in the figure below. The permeability of free space is 1 . 25664 × 10 6 T · m / A. . 802 m 14 . 7 A P GenericStudent – Homework 8 – savrasov – 39823 – Jan 02, 2008 2 The permeability of free space is 1 . 25664 × 10 6 T · m / A. What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire? Correct answer: 20 . 7371 μ T. Explanation: Let : μ = 1 . 25664 × 10 6 T · m / A , ‘ = 0 . 802 m , and I = 14 . 7 A . By the BiotSavart law, dB = μ 4 π I d~s × ˆ r r 2 . Consider a thin, straight wire carring a con stant current I along the xaxis with the y axis pointing towards the center of the square, as in the following figure. y x O r P x I ˆ r ds a θ Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element d~s is at a distance r from P . The direction of the field at P due to this element is out of the paper, since d~s × ˆ r is out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the field at...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Winter '09 term at UC Davis.
 Winter '09
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