P812 Midterm Examination February Solutions

Transcription

1 P8 Midterm Exmintion Februry s. A one dimensionl chin of chrges consist of e nd e lterntively plced with neighbouring distnce. Show tht the potentil energy of ech chrge is given by U = ln. 4πε Explin qulittively why NCl crystl melts if plced in wter. Assume =5.5 m. The dc permittivity of wter is ε =8ε. The potentil energy is U = e 3 4 µ + + 8πε = e ln, 4πε where the fctor is due to contributions from both sides. Note tht the potentil energy of chrge q i is X q j U i = q i,i6= j. 4πε r ij If the chin is plced in wter, the potentil energy is reduced to U = e 4πε ln = e ln. 4π 8ε If this mgnitude is smller thn the therml energy 3 k BT, the chin is broken. For NCl crystl plced in wter, the potentil energy is U = ln =.33 ev. Therml energy t room temperture is 3 k BT = ev =.39 ev. Therefore, the crystl should melt if plced in wter. This rgument is still qulittive becuse ctul crystl is 3D.. The surfce potentil of long cylinder of rdius is specified s +V, <φ<π/, V, π/ <φ<π, Φ s (φ) = +V, π < φ < 3 π, 3 V, π<φ<π. Using the two dimensionl Green s function, G(, )= 4π ln + cos φ φ ( /) + cos φ φ, e

2 determine the potentil Φ(, φ) both inside ( <) nd outside ( >) the cylinder in the following form, ³ m sin mφ, <, m Φ(, φ) = µ m m sin mφ, >. Hint: The following expnsion should be useful, x +x x cos θ =+ X The outer potentil is given by Z π G Φ(, φ) = + where x = /. Using we find = π = π Z π Z π = dφ m= x m cos(mθ). + cos φ φ Φ s φ dφ x +x x cos φ φ Φ s φ dφ, x +x x cos θ =+ X Φ(, φ) = π m= x m cos(mθ), X Z π x m Φ s (φ )cos[m(φ φ)]dφ, m= where ½ /, <, x = /, >. The surfce potentil hs periodicity π nd thus hs Fourier components of m =, 6,,. Then the first nonvnishing m is m =followed by m =6. Note tht m =4term vnishes. Φ(, φ) = 4V π Φ(, φ) = 4V π " µ ³ sin (φ)+ 3 µ 6 sin (6φ)+ #,>, ³ 6 sin (φ)+ sin (6φ)+,< A chrge q is plced t r =(b, θ,φ =)(in the x z plne) ner grounded (Φ =)conducting plte which hs hemisphericl bob of rdius (< b) s shown. Using the method of imge, determine the potentil Φ(r) t n rbitrry position r =(r, θ, φ). (The solution is essentilly the Green s function for the geometry.)

3 The imges re µ b q t r = b,θ,, q t r = b, π θ,, nd + µ b q t r = b,π θ,. Then the potentil is given by Φ(r) = µ q 4πε r r b q r r q r r + b q r r = q 4πε p r + b br cos γ /b r where r +( /b) b r cos γ q 4πε p r + b br cos γ /b r, r +( /b) r cos γ b cos γ = cosθ cos θ +sinθsin θ cos φ, cos γ = cos θ cos θ +sinθsin θ cos φ. For n rbitrry zimuthl loction of the chrge φ, this my be generlized s cos γ = cosθ cos θ +sinθsin θ cos(φ φ ), cos γ = cos θ cos θ +sinθsin θ cos(φ φ ). 4. () A thin ring of mjor rdius nd minor rdius b ( ) crries current I. Show tht the norml component of the mgnetic field t the ring surfce ( = b) is pproximtely given by B n = B ' µ I ln 4π µ 8 sin θ. b 3

4 Hint: In the region, it is convenient to use the qusi toroidl coordintes (, θ, φ) with metric coefficients h =,h θ =, h φ =+ cos θ. (b) A thin toroidl superconducting ring hving mjor rdius nd minor rdius b ( ) is plced in n externl mgnetic field B = B e z with its xis prllel to the field. Find the current to be induced in the ring nd sketch qulittively the mgnetic field profile when b/ = () The vector potentil due to ring current is A φ = µ I π µ r + +r sin θ k K k k E k, where k 4r sin θ = r + +r sin θ, nd θ is the polr ngle in the sphericl coordintes. Ner the ring,, r + +r sin θ = + + cos θ +( + cos θ) ' 4 +4 cos θ, 4

5 nd Then k = = r + +r sin θ ' 4r sin θ r + +r sin θ 4( + cos θ) 4 +4 cos θ + = 4( + cos θ) cos θ + ³ '. ³ + cos θ / ' µ 8 µ k K k k E k ' ln nd the vector potentil ner the ring my be pproximted by A φ = µ µ I µ 8 π cos θ ln, µ cos θ,, vlid to order /. The rdil component of the mgnetic field t the surfce ( = b) is B = µ + b µ+ b cos θ θ cos θ A φ ' µ µ I 8 ln sin θ. 4π b If the externl mgnetic field is downwrd, the current in the ring is counterclockwise if seen from top which is in φ direction of the qusi-toroidl coordintes. Then, B ' µ µ I 8 ln sin θ. 4π b (b) Since the ring is superconducting, the totl rdil mgnetic field (externl plus the field due to the current in the reing) should vnish. The externl mgnetic field cn be decomposed s B e z = B sin θe B cos θe θ. Then from B n = B =t = b, we find B µ µ I 8 ln =, 4π b or When /b =8, 4πB I = µ 8 µ ln b µ I =.4 4πB.. 5

6 The xil mgnetic field due to the ring current is B z (z) = µ I (z + ) 3/ =.4 π 3 (z + ) 3/ B, which opposes the externl field B. Stgntion points cn be found from (z + ) 3/ =.4 π 3, z = ±.954. Field profile is qulittively shown for =5cm, b =5cm. (Mtlb is used.) A semi-infinite thin solenoid crries liner mgnetic moment density of κ =κe z (A m) in the region <z<. () Show tht the vector potentil is given by A φ (r, θ) = µ κ 4πr cos θ,θ6= π. sin θ (b) Then determine the mgnetic field B. (ThisistomimicthefieldofDirc smgnetic monopole.) 6

7 () The vector potentil is Since A φ = µ κ 4π tn θ = sec θ dθ = Z sin θ r dz. r sin θ r cos θ z r sin θ (r cos θ z ) dz, we find A φ = = Z θ µ κ sin θ dθ 4πr sin θ µ κ ( cos θ),θ6= π. 4πr sin θ (b) The mgnetic field B = A is purely rdil, B r = r sin θ,θ6= π. r = µ κ 4π θ (r sin θa φ) 6. The Lorentz force J B in conductor vnishes if the current density J is prllel to the mgnetic field everywhere. Such configurtion is clled force-free nd plys key role in designing superconducting wire strnds nd certin plsm devices such s Reversed Field Pinch nd Spheromks (Compct Tori). The Ampere s lw in force-free configurtion reduces to B = λb (= µ J) where λ is constnt hving dimensions of m. Solve this eqution for cylindricl conductor of rdius ssuming B =[,B φ (),B z ()], φ = z =, nd totl xil current is I, i.e., π J z ( = ) =, Z J z ()d = I. Hint: Z J (x)xdx = xj (x). The first root of J (x) =is x '.448. From B = λb, we find µ d d + d d + λ B z () =, 7

8 nd Since B z () =, The totl current is B z () =B J (λ), B φ () =B J (λ). I = π λ =.45. Z J z () d Z = πλb J (λ) d µ = πλb µ λ J (λ) = πb µ J (λ). Since we find the reltion between I nd B, J (.448) =.59, I =.59 πb µ. 8