Homework Assignment 9 Solution Set

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1 Homework Assignment 9 Solution Set PHYCS 44 3 Mrch, 4 Problem (Griffiths 77) The mgnitude of the current in the loop is loop = ε induced = Φ B = A B = π = π µ n (µ n) = π µ nk According to Lense s Lw this current flows in such wy s to try to restore the chnging field Thus, for positive k the current in the circuit will flow in the opposite direction from the current in the solenoid Thus, s it is drwn in the text the current will flow left to right b We hve loop (t) = Φ B To find the totl chrge we just integrte loop (t)dt = Q = Φ B dt dφ B = π (µ n ( µ n)) = π µ n

2 Problem (Griffiths 79) The mgnetic field inside the torroid is nd the cross-sectionl flux is Br, t = µ N(t) πr Φ B (t) = +w µ N(t) hdr = µ hn(t) πr π ln + w Since w nd h re much smller thn we cn tret the torroid s skinny wire of cross-sectionl flux Φ B µ hn(t) w π (expnding in Tylor s series) n this sense, the electric field t point z bove the center of this loop of (chnging) flux should behve the sme s the mgnetic field distnce z bove the center of loop of current (chnging chrge) From Exmple 56 in the book we get the mgnetic field due to such loop of current f points in the ˆθ direction then the field is B(z) = µ ẑ ( + z ) 3 Now, we hve the dulity B d l = µ encl nd E d l = dφ B t So, we should be ble to use the result for the current loop nd just exchnge E for B, for, nd dφ B t for µ Thus, for mgnetic field in the ˆθ direction we should get E(z) = Φ B Problem 3 (Griffiths 74) = µ hn π ẑ ( + z ) 3 w (t) = µ hnkw 4π( + z ) 3 ẑ ( + z ) 3

3 Agin we hve Thus, the induced emf is induced (t) = Φ B = = Nµ h π rout r in (t) = Nµ hω π N µ (t) πr hdr ln r out r in ln r out r in sin ωt ()4π 7 ()(6)(5) ln sin π(6)t = 6 4 sin(377t) nd the current in the resistor is sin(377t) = 5 7 sin(377t) b For the bck emf we just need to know the self-inductnce This we hve from the book, so ε = L = µ N h ln r out π r in (5 7 sin(377t)) = 7 ln (5 7 )(337) cos(337t) = 44 7 cos(337t) The rtio of their mplitudes is bck emf induced emf 3 This is s it should be; the bck emf is never comprble to the externlly induced emf, except in specil cses (see Griffiths Problem 8) Problem 4 (Griffiths 78) n this peculir rrngement there is no self-generted flux tht psses through ny closed loop of current So we cn t use the stndrd L = Φ B However, using wht we know bout mgnetic energy we cn write U = B = V µ L ( ) µ r µ π lπrdr = L 3

4 = L l µ l 8π = µ 8π = L Problem 5 (Griffiths 7 - optionl) Let s refer to the smller of the two loops s loop nd to the lrger s loop Φ = B d A = B A (since loop is so smll) = µ b π (b + z ) 3 = µ π b (b + z ) 3 b We know nd so Φ = B (dipole) = µ (3( m ˆr)ˆr m) 4πr3 B d A = = µ = µ = µ µ 4πr 3 (3( m ˆr)ˆr m) πr dr ẑ (where r is in the plne of loop ) = µ π = µ π (3( m ˆr)ˆr ẑ m ẑ)r dr (r + z ) 3 (3m cos θ m)r dr (r + z ) 3 (r + z ) 3 π z (3( r + z ) )r dr = µ π b (b + z ) 3 (r + z ) 3 r z r 3 dr (r + z ) 5 r z r 3 r + z dr 4

5 c M = Φ = µ π b (b + z ) 3 = Φ = M Problem 6 (Griffiths 73) The constnt current in the two wires cuses stedy chnge in the chrge density on the circulr surfces tht form the boundry of the gp Since w we cn tret the gp s prllel plte cpcitor with electric field E(t) = σ(t) ẑ ɛ where ẑ is the direction of the current We cnnot pply the old mgnetosttics version of Ampere s Lw becuse no current is ctully pssing through the gp The full form of Ampere s lw, s fixed by Mxwell, is given in Eqution 738 in the text Thus, B d E l = µ encl + µ ɛ d A ( ( ) ) = B(r)πr σ(t) = µ ɛ πr ˆθ ɛ = B(r) = µ ( ) r π ˆθ = µ r π ˆθ which is the sme s the field inside the wires t distnce r < Fortuntely, Ampere is rescued We get the sme field s we would expect if the current hd flowed right through the gp This lso stisfies our boundry condition tht the prllel component of H be continuous in the bsence of ny surfce currents Problem 7 (Griffiths 75) With the definitions of V nd V s given we cn write the closed integrl of E round the circuit in terms of the voltges cross ech resistor E d l + E d l = b V V = α E d l = α Also, the current must be the sme everywhere in the circuit (given tht the voltmeters drw negligible current), so from Ohm s lw we hve V = V 5

6 Combining these two equtions gives the desired result V = α + V = α + This relly shouldn t be too surprising t just demonstrtes the fct tht the induced emf cn be thought of s n idel voltge source situted nywhere in the circuit These voltges re the sme s wht would result from two resistors in series with bttery supplying voltge α Problem 8 (Griffiths 758) Since the ribbons re close together we cn tret them s prllel plte cpcitor C = ɛ A d = C = ɛ w l h b Agin, since the sheets re close together the field between them is the sme s tht from pir of infinite surfce currents Thus, c L = Φ B = µ Klh = L = µ h l w For the ribbon configurtion we hve L C = µ ɛ = l l c nd for the cylindricl trnsmission line in Exmple 73 we get L C l l ( ( )) ( ) µ b = π ln πɛ ln ( ) = µ b ɛ d n medium (s opposed to vcuum) we just replce µ nd ɛ with µ nd ɛ respectively Thus, L C = µɛ l l nd v = < c µɛ 6

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