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: gilvin (jg47854) – 19. Ampere’s law – meyers – (21235) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This homework is due Tuesday, april19, at midnight Tucson time. 001 10.0 points A lightening bolt may carry a current of 4200 A for a short period of time. The permeability of free space is 1 . 25664 × 10 − 6 T · m / A. What is the resulting magnetic field 68 m from the bolt? Suppose that the bolt extends far above and below the point of observation. Correct answer: 12 . 3529 μ T. Explanation: Given : L = 68 m and I = 4200 A . Treat the lighting bolt as a long, straight conductor. Then the magnetic field is B = μ I 2 π L = ( 1 . 25664 × 10 − 6 T · m / A ) (4200 A) 2 π (68 m) × parenleftbigg 1 . × 10 6 μ T 1 T parenrightbigg = 12 . 3529 μ T . 002 10.0 points A long, straight, thinwalled cylindrical shell of radius R carries a current I . Find the magnetic field inside the cylinder and outside the cylinder. 1. B = 5 μ I 2 π r , r > R ; B = 0 , r < R 2. B = μ I 2 π r , r > R ; B = 0 , r < R correct 3. B = 3 μ I 2 π r , r > R ; B = 0 , r < R 4. B = 0 , r > R ; B = 3 μ I 2 π r , r < R 5. B = 0 , r > R ; B = μ I 2 π r , r < R 6. B = 0 , r > R ; B = 5 μ I 2 π r , r < R Explanation: Applying Amp` ere’s law for r > R , contintegraldisplay C vector B · dl = B (2 π r ) = μ I B r>R = μ I 2 π r . applying Amp` ere’s law for r < R , contintegraldisplay C vector B · dl = B (2 π r ) = 0 B r<R = . 003 (part 1 of 3) 10.0 points A wire of radius 1 cm carries a current of 90 A that is uniformly distributed over its crosssectional area. The permeability of free space is 4 π × 10 − 7 T · m / A. Find the magnetic field at a distance of . 1 cm from the center of the wire. Correct answer: 0 . 00018 T. Explanation: Let : I = 90 A , R = 1 cm = 0 . 01 m , r = 0 . 1 cm = 0 . 001 m , and μ = 4 π × 10 − 7 T · m / A . Using Amp` ere’s law, we have contintegraldisplay C vector B · d vector l = μ I C . Because r < R , I C = r 2 R 2 I , so gilvin (jg47854) – 19. Ampere’s law – meyers – (21235) 2 B (2 π r ) = μ r 2 R 2 I B = μ r 2 π R 2 I = (4 π × 10 − 7 T · m / A) (0 . 001 m) 2 π (0 . 01 m) 2 × (90 A) = . 00018 T ....
View
Full
Document
This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.
 Spring '11
 doc
 Magnetism, Work

Click to edit the document details