PS6 Solutions

PS6 Solutions - Physics 217 Problem Set 6 Solutions by...

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Unformatted text preview: Physics 217: Problem Set 6 Solutions by Jolyon Bloomfield and Yao Liu Due Friday, March 13th, 2009 1 Problem 6.03 Purcell The magnetic field along the axis of the ring is given by ~ B = μ I a 2 2( a 2 + z 2 ) 3 / 2 ˆ z in SI notation where a is the radius of the ring. Let us construct an Amperian loop which travels from- R → R in the z direction, then forms a large semicircle of radius R to close the loop. We’ll look at the limit of R → ∞ . We can then write I ~ B · ~ dl = Z R- R μ I a 2 2( a 2 + z 2 ) 3 / 2 dz + Z r = R ~ B · ~ dl as our loop integral, and take the limit as R → ∞ . Let’s look at the first of these terms, the integral along the z-axis. We can take the limit here without qualms. Z ∞-∞ μ I a 2 2( a 2 + z 2 ) 3 / 2 dz = μ I You may evaluate the integral with a trigonometric substitution (not shown here). Let’s look closely at the integral over the arc now. The ~ B field along the axis is proportional to 1 /z 3 for large z , but the field far out in the z = 0 plane falls off as 1 /r 2 . At any rate, we can safely say that ~ B falls off as 1 /r 2 or faster. At r = R , we may write ~ B ∼ 1 /R 2 . The length of the semicircle is πR , so this component of the loop integral falls off as 1 /R or faster, and so will vanish as R → ∞ . Therefore, we have verified Amp´ ere’s law, which says that H ~ B · ~ dl is μ times the total current enclosed by the loop, which in this case is just I . More rigorously, in the style of a typical argument in analysis, | ~ B | = Z μ 4 π Id ~ l × ˆ r r 2 ≤ Z μ 4 π Id ~ l × ˆ r r 2 = μ I 4 π Z | d ~ l × ˆ r | r 2 ≤ μ I 4 π Z dl ( R- a ) 2 = μ I 4 π 2 πa ( R- a ) 2 The first inequality is a sort of infinitesimal version of triangle inequality (also valid for real integral, complex integral, and even more general definition of integration), and in the second inequality we take the maximum in the numerator, and the minimum in the denominator. Then Z r = R ~ B · d ~ l ≤ Z r = R | ~ B | dl ≤ Z μ I 2 a ( R- a ) 2 dl = μ I 2 aπR ( R- a ) 2 → as R → ∞ A similar procedure is used to calculate certain definite integrals that could not be done easily by normal approach. You will learn about contour integration and method of residues in a course on complex analysis. 1 2 Problem 6.04 Purcell There are a few ways to do this. One is to use the Biot-Savart law to integrate along the entire wire, breaking up the integral into three components, one for each segment of wire. However, we’ve already done most of that work, so why not use it? We’re going to use the principle of superposition. There are two half-infinite straight wires, whose contribution will add together. Note that you should be able to take either of these half-wires and rotate it about P and it will still give the same contribution to the superposition. Let’s take the top wire and rotate it by 180 degrees around P , to find that the contribution is identical to that of an infinite straight wire. So,, to find that the contribution is identical to that of an infinite straight wire....
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PS6 Solutions - Physics 217 Problem Set 6 Solutions by...

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