non_coulomb - 1/3 p NON-COULOMB E FIELD CZENDLJ<1 TIE/V...

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Unformatted text preview: 1/3 p NON-COULOMB E FIELD CZENDLJ <1 TIE/V), We begin by considering a uniform §~field represented by a set of parallel lines, Next, , i imagine an area A A Whose normal _). is along fi . Then, as in the case of E we definevflux of Bthrough AAto be _ = _ s »—> —> - —> - Ad) 3 = BaAA = BAA C0s(B,fi) —> —+ —) As before, flux is maximum when E H n (i to area) and, zero if J. fi( 3 lines lie in plane of —> —> AA ).. . ' . . . . . AC1) B V . ' Faraday’s dlscovery was that if 1th) B varies w1th tune, that IS, M at 0 , there will be an emf a _ ' induced in every closed loop surrounding the region where Ad) 3 is changing. For example, if in the icture B varies with the time, Oran le of "B with fi _ 1 p a. . g 07%. ; F/cczc- CCU" r N W Vang; mr‘H-i 05944-9, Ltu._ or:- ' . and ‘91 m a, Md”! 6.24. 4g 1:” e ., varies w1th tlme, At a6 0 , and an emf W111 appear 1n the closed loop, represented by the dotted Iine,surrounding AA . We know that if there is an e‘mf in a circuit (closedyloop) there must be an E -field present at every point of the circuit such that a = Z c E aAl ‘ —) , —> —> .2/3 ‘ Where the sum Is over the closedloop and Al represents displacement along the loop. ._.) . Next, we add Lenz’s principle to Faraday’s discovery to write EOE-AZ=—A®B a a At The minus sign on the right is crucial, it implies that the emf or equivalently E field, is such that —> it opposes the change in the flux of B Which generates the emf. . , g Another remarkable point to note is that the E ~field generated by a time varying (I) B is *—> represented by lines which close on themselves — there is no beginning and no end. This is totally different from the E -field implied by Coulomb’s law: E : 4”: r2 19 which “started” at +ive charges and “ended”at —ive charges. r , ' We call the E —field generated by the time varying flux of B a NON-COULOMB E ~FIELD: —> —> —> Ad) ZCENC-Ah— B ._) . a AL . Notice that for E NC Gauss’ law will always give 4 ECEIC'A'AE 0. Total flux 'of E No through a closed surface isalways equal to zero. Note: A charge q placed in E NC experiences a force —> -- fihei7fimc exactly as for a Coulomb E —field. APPLICATIONS Calculate the E NC for a solenoid at a distance r from its axis when the flux of B is varied by a a . - . - . Ai AB AC1) time variation of the current in the solenord. That is, — ¢ 0 :> — at 0 :> B At At At solenoid wound on a tube of radius a. If there are n turns per meter and the current flow is as shown, there is a uniform B inside it _) V = 0. Consider a . B=flonifi and Ag A2 A 3/3 The problem has cylindrical symmetry about axis of solenoid, ENC at r is a function of r only ' —> and must be azimuthal. Letj increase with time. Then flux of B along + f) is increasing with ' —) time. Take a circular 100p. As shown direction of E NC must be clockwise ( as viewed frOm —> above) to oppose increase of (I) 3, Next, _A_z' At A . If r>a ENC 27:7 = ~ ,uomraz X; [LOOP OUTSIDE SOLENOID] Ifr<a ENC 2727 = — #0117172 [LOOP INSIDE SOLENOID] Magnitude ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 122 taught by Professor Bhagat during the Fall '11 term at Maryland.

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non_coulomb - 1/3 p NON-COULOMB E FIELD CZENDLJ<1 TIE/V...

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