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Unformatted text preview: 1/3 p NONCOULOMB 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‘Hi 059449, 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 EOEAZ=—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 NONCOULOMB E ~FIELD:
—> —> —> Ad)
ZCENCAh— 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.
 Fall '11
 Bhagat
 Physics

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