This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version One – Homework 8 – Savrasov – 39821 – May 14, 2007 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. BiotSavart Law 30:01, trigonometry, numeric, > 1 min, nor mal. 001 (part 1 of 1) 2 points The segment of wire in the figure carries a cur rent of 5 A, where the radius of the circular arc is 3 cm. The permeability of free space is 1 . 25664 × 10 6 T · m / A . 5 A 3 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 26 . 1799 μ T. Explanation: Let : I = 5 A and R = 3 cm = 0 . 03 m . For the straight sections d s × ˆ r = 0. The quarter circle makes onefourth the field of a full loop B = μ I 8 R into the paper. Or, you can use the equation B = μ o I 4 π θ , where θ = π 2 . Thus the magnetic field is B = μ I 8 R , = (1 . 25664 × 10 6 T · m / A) (5 A) 8 (0 . 03 m) = 26 . 1799 μ T , into the paper . keywords: Magnetic Field from a Segment 01 30:01, calculus, numeric, > 1 min, wording variable. 002 (part 1 of 2) 1 points Consider two radial legs extending to infinity and a line segment carrying a current I as shown below. r I r I 5 7 π I O What is the magnitude of the magnetic field B at the origin O due to the current through this path? 1. μ I 2 π r tan 5 14 π correct 2. μ I 2 π r tan 9 14 π 3. μ I 2 π r tan 5 7 π 4. μ I 2 π r sin 5 14 π 5. μ I 2 π r sin 9 14 π 6. μ I 2 π r sin 5 7 π 7. μ I 2 π r cos 5 14 π 8. μ I 2 π r cos 9 14 π 9. μ I 2 π r cos 5 7 π Version One – Homework 8 – Savrasov – 39821 – May 14, 2007 2 10. None of these Explanation: Let : α = 5 7 π , and θ = π 2 α 2 = 1 2 5 14 π = 9 14 π , and α 2 = π 2 θ = 5 14 π . 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 vertex point O , as in the following figure. y x O r P x I ˆ r ds a θ α 2 Figure: Is not drawn to scale. 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 P . In fact, taking the origin at O and letting P be along the positive y axis, with ˆ k being a unit vector pointing out of the paper, we see that d s × ˆ r = ˆ k  d s × ˆ r  = ˆ k ( dx sin θ ) . Substituting into BiotSavart law gives d B = ˆ k dB , with dB = μ I 4 π dx sin θ r 2 . (1) In order to integrate this expression, we must relate the variables θ , x , and r . One approach is to express x and r in terms of θ . From the geometry in the figure and some simple differ entiation, we obtain the following relationship...
View
Full
Document
This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas.
 Spring '10
 SHIH
 Work

Click to edit the document details