8.022pset7solution

8.022pset7solution - Massachusetts Institute of Technology...

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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 7: Magnetic fields and Induction. TA: Yi Mao, [email protected] 1. Purcell 6.4: Practice with B-S law. I r e dl P P r I A B C C D B = B points normal to and out of the page B r I Figure 1: Magnetic Field of a hairpinlike shaped long wire: decomposed into two semi-infinite long wires AB and CD and a half-circle BC. Refer to Figure 1. To calculate the magnetic field, we decompose the wire into two semi-infinite long wires AB and CD and a half-circle BC. First we check the direction of magnetic field from each sub-wire, and we find all contribute ~ B in the direction normal to and out of the page. By Ampere’s law we can calculate a magnetic field of a whole infinite long wire B × 2 πr = (4 π/c ) I , or B = 2 I/cr . Since all element wires contribute ~ B in the same direction, ~ B given by the half-infinite long wire AB is just B AB = (1 / 2) × (2 I/cr ) = I/cr . So is ~ B given by CD. For ~ B given by the half-circle CD, B CD = Z Idl cr 2 = I × πr cr 2 = πI cr . (1) So the total magnetic field at point P is B P = (2 + π ) I cr . (2) I z I a a b z B P Figure 2: Helmholtz Coil: the magnetic field in the vicinity of the origin is of interest to us. 2. Purcell 6.13: Helmholtz coil. Let’s consider the following configuration: two coaxial current rings of radius a axially seperate by distance b , each carrying a current I of the same direction. We set up a coordinate in which the origin locates on the axis midway between the two coils. We hope the magnetic field in the vicinity of O is as nearly uniform as possible; that means, for the magnetic field as a function of locations around O, we want to make the function have vanishing derivatives at O of as many orders as possible. (You can see this point by Taylor expanding the function around z = 0.) For simplicity, consider a point P on the axis and in the vicinity of O. The magnetic field at the axis of a single current ring is given by eq.(41), Purcell p.227. So given seperation b , the magnetic field at point P is B ( z ) = 2 πIa 2 c f b ( z ) , (3) where f b ( z ) = ( a 2 + ( b 2 + z ) 2 )- 3 / 2 + ( a 2 + ( b 2- z ) 2 )- 3 / 2 . (4) The calculation of derivatives is lengthy but eventually we get f b ( z = 0) = , f 00 b ( z = 0) =- 768( a 2- b 2 ) (4 a 2 + b 2 ) 7 / 2 , (5) f 000 b ( z = 0) = , f (4) b ( z = 0) = 92160(2 a 4- 6 a 2 b 2 + b 4 ) (4 a 2 + b 2 ) 11 / 2 , (6) ··· The vanishing of eq.(5) sets the condition b = a, (7) under which we have vanishing derivatives of three orders, B ( z ) = B (0) + O ( z 4 ); or in other words, a uniform magnetic field is around z = 0 within the order of O ( z 3 )....
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8.022pset7solution - Massachusetts Institute of Technology...

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