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: Nguyen, Thanh – Homework 13 – Due: Mar 7 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 15 questions. Multiple-choice 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) 10 points Consider a current configuration shown be- low. A long (effectively infinite) wire segment is connected to a quarter of a circular arc with radius a . The other end of the arc is connected to another long horizontal wire segment. The current is flowing from the top coming down vertically and flows to the right along the pos- itive x-axis. a I O O y x What is the direction of the magnetic field at O due to this current configuration? 1. along the positive + x-axis 2. 135 ◦ counterclockwise from the + x-axis 3. 45 ◦ counterclockwise from the + x-axis 4. 315 ◦ counterclockwise from the + x-axis 5. perpendicular to and into the page 6. perpendicular to and out of the page cor- rect 7. along the negative y-axis 8. along the positive y-axis 9. 225 ◦ counterclockwise from the + x-axis 10. along the negative x-axis Explanation: From the Biot-Savart law we know that d ~ B ∝ I d ~ l × ˆ r . One should first verify that the magnetic field at O contributed by any infinitesimal current element along the current configuration given is perpendicular to the page, which is coming out of the page. Therefore the resulting mag- netic field must also be pointing out of the page. 002 (part 2 of 2) 10 points Let I = 6 . 9 A and a = 0 . 79 m. What is the magnitude of the magnetic field at O due to the current configuration? Correct answer: 3 . 1188 × 10- 6 T. Explanation: Let : I = 6 . 9 A , and a = 0 . 79 m . Consider first the one-quarter of a circular arc. Since each current element is perpendic- ular to the unit vector pointing to O , we can write B arc = μ 4 π I Z π/ 2 adθ a 2 = μ 4 π I π 2 a = μ I 8 a . For the straight sections, we apply the formu- las derived from the figure below, θ 1 θ 2 a I ~ B = ˆ z μ I 4 π a Z θ 2 θ 1 dθ sin θ = ˆ z μ I 4 π a (cos θ 1- cos θ 2 ) where ˆ z is the unit vector perpendicular to the plane of the paper that points to the reader. Nguyen, Thanh – Homework 13 – Due: Mar 7 2008, 7:00 pm – Inst: Weathers 2 For the downward y-directed current, θ 2 = π 2 and θ 1 = 0. For the horizontal x-directed current θ 2 = π and θ 1 = π 2 . Thus we find the contribution at the point O for the magnetic field from the long vertical wire is same as in the long horizontal wire, and the sum is equal to ~ B = ˆ z μ I 4 π a (1 + 1) = ˆ z μ I 2 π a . (Note that the sum of the two is the same as a long straight wire) Adding the contributions from the straight sections and the arc, B tot = μ I 8 a + µ μ I 2 π a ¶ = μ I 2 a µ 1 4 + 1 π ¶ = (1 . 25664 × 10- 6 T · m / A) (6 . 9 A) 2 (0 . 79 m) × µ 1 4 + 1 π ¶ = 3 . 1188 × 10- 6 T ....
View Full Document
This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
- Spring '00