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 print-out should have 7 questions. Multiple-choice 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 one-fourth 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 = µ 1 2- 5 14 ¶ π = 9 14 π , and α 2 = π 2- θ = 5 14 π . By the Biot-Savart 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 x-axis 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 Biot-Savart 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
- Spring '10
- Magnetic Field