Old Exam 3 Sol

# Old Exam 3 Sol - Platt David – Oldquiz 3 – Due 4:00 am...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 32 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 4) 10 points Consider the current segment shown where a current i is flowing through a straight segment from C to A , then an arc from A to D , and a straight segment from D to F . 13 24 π D F ∞ i A C ∞ i r O Both segments AC and DF extend to infinity. The circular arc AD is 13 48 of a circle with a radius a . The arc is centered at O . What is the magnitude ( k ~ B CA k ≡ B CA ) of the magnetic field at O due to the CA segment alone? 1. B CA < μ i 8 π a 2. B CA = μ i a 3. B CA = μ i 8 π a 4. B CA = μ i 4 a 5. B CA = μ i 4 π a correct 6. B CA > μ i a 7. B CA = μ i 6 a 8. B CA = μ i 2 π a 9. B CA = μ i π a 10. B CA = μ i 2 a Explanation: Note: 13 48 360 ◦ = 97 . 5 ◦ = 13 24 π . This problem has a direct way for its solu- tion (based on the Biot-Savart law) and a fast insightful way, based on Ampere’s law and the principle of superposition. Fast Way: Choose the orientation of the x-axis so that CA is along the negative x-axis with origin at A (see figure). θ P P A a O C C From C to A a current I flows, and imagine this current continuing from A to C . At O , the magnetic field B from the current in the extended wire ( CAC ) can be found from Ampere’s law, which gives 2 π a B = μ I , or B = μ I 2 π a , with B coming out of the paper. This follows from the standard right hand rule associated with Ampere’s law. Note: Contribution to B from the current in the segment CA equals the contribution from the current in the segment AC . Reason: This follows by noticing (from the Biot-Savart law) that the contribution to ~ B from a current through the line element d ‘ centered about P is the same as the contribu- tion from an equal incremental line element d ‘ centered about P with P and P equally distant from point A . Thus each half of the extended wire gives the desired magnetic field B , and B + B = B = μ I 2 π a . Therefore B = μ I 4 π a . Direct Way: Set up a coordinate system with the origin at A , such that the segment CA goes along the negative x-direction as in the figure. The distance from the field point O to the wire is a . Consider a certain point P along CA with x-coordinate x . The distance from O to this particular point is called r . Denote the (counterclockwise) angle between Platt, David – Oldquiz 3 – Due: Nov 13 2005, 4:00 am – Inst: Ken Shih 2 the line OP and the horizontal by θ . We find sin θ = a r , cos θ =- x r . Dividing gives x =- a cot θ , then dx = a sin 2 θ dθ = r 2 a dθ ....
View Full Document

## This note was uploaded on 07/26/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

### Page1 / 16

Old Exam 3 Sol - Platt David – Oldquiz 3 – Due 4:00 am...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online