oldhw19 - oldhomewk 19 – LEE BENJAMIN – Due 4:00 am 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: oldhomewk 19 – LEE, BENJAMIN – Due: Oct 16 2007, 4:00 am 1 Question 1, chap 29, sect 5. part 1 of 2 10 points The wire is carrying a current I . x y I I I 180 ◦ O R Find the magnitude of the magnetic field vector B at O due to a current-carrying wire shown in the figure, where the semicircle has radius r , and the straight parts to the left and to the right extend to infinity. 1. B = μ I 3 π r 2. B = μ I 7 π r 3. B = μ I 2 π r 4. B = μ I 3 r 5. B = μ I 4 π r 6. B = μ I 8 π r 7. B = μ I π r 8. B = μ I 2 r 9. B = μ I r 10. B = μ I 4 r correct Explanation: By the Biot-Savart Law, vector B = μ I 4 π integraldisplay dvectors × ˆ r r 2 . Consider the left straight part of the wire. The line element dvectors at this part, if we come in from ∞ , points towards O; i.e. , in the x- direction. We need to find dvectors × ˆ r to use the Biot-Savart Law. However, in this part of the wire, ˆ r is pointing towards O as well, so dvectors and ˆ r are parallel meaning dvectors × ˆ r = 0 for this part of the wire. It is now easy to see that the right part, having a dvectors antiparallel to ˆ r , also gives no contribution to vector B at O . Let us go through the semicircle C. The element dvectors , which is along the wire, will now be perpendicular to ˆ r , which is pointing along the radius towards O . Therefore | dvectors × ˆ r | = ds using the fact that ˆ r is a unit vector. So the Biot-Savart Law gives for the magnitude B of the magnetic field at O B = μ I 4 π integraldisplay C ds r 2 Since the distance r to the element dvectors is con- stant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is integraldisplay C ds r 2 = 1 r 2 integraldisplay C ds = 1 r 2 L C , where L C = π r is the length of the semicircle. Thus the magnitude of the magnetic field is B = μ I 4 π 1 r 2 π r = μ I 4 r . Question 2, chap 29, sect 5. part 2 of 2 10 points Note: ˆ i is in x-direction, ˆ j is in y-direction, and ˆ k direction is perpendicular to paper to- wards reader. Determine the direction of the magnetic field vector B at O due to the current-carrying wire. 1. hatwide B = − ˆ k correct 2. hatwide B = +ˆ 3. hatwide B = − ˆ ı oldhomewk 19 – LEE, BENJAMIN – Due: Oct 16 2007, 4:00 am 2 4. hatwide B = 1 √ 2 parenleftBig ˆ i − ˆ parenrightBig 5. hatwide B = 1 √ 2 parenleftBig ˆ k +ˆ parenrightBig 6. hatwide B = + ˆ k 7. hatwide B = +ˆ ı 8. hatwide B = − ˆ 9. hatwide B = 1 √ 2 parenleftBig ˆ k − ˆ parenrightBig 10. hatwide B = 1 √ 2 parenleftBig ˆ i +ˆ parenrightBig Explanation: We know from Part 1 that the only contri- bution to the magnetic field at O comes from the semicircle. Furthermore, we need only consider the direction of dvectors × ˆ r for one typi- cal segment dvectors . If we go along the semicircle from left to right, and we know that ˆ r is point- ing in towards O , the right hand rule tells us that the field resulting from this element is...
View Full Document

This note was uploaded on 04/07/2008 for the course PHY 303K taught by Professor Turner during the Fall '07 term at A.T. Still University.

Page1 / 6

oldhw19 - oldhomewk 19 – LEE BENJAMIN – Due 4:00 am 1...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online