8.022pset7solution

# 8.022pset7solution - Massachusetts Institute of Technology...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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: 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 )....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

8.022pset7solution - Massachusetts Institute of Technology...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online