PHYS463 Electricity& Magnetism III ( ) Problems Solutions (assignment #3) r n+1

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Neil Doyle
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1 . (Problem 3.38, p.6) Solution: Use equation (3.95) PHYS463 Electricity& Magnetism (3-4) Problems Solutions (assignment #3) Φ 4π² X n ³ r n Pn ³cos ³ ϑ ρ r dτ r n+ Now λ Q/a a<z<+a for <z<a,r z, ϑ ϑ. for <z<, r z z, ϑ π ϑ and cos ϑ cos(π ϑ) cos ϑ So Φ λ 4π² X n a n ³ z Pn ³cos ϑ dz r n+ Phys. 463 E & M 3-4 / Solution3

2 evaluate a n ³ z Pn ³cos ϑ dz n ³ z Pn ³cos a ϑ dz + ³ z n Pn ³cos ϑ dz n a n ³ z Pn (cos ϑ) dz + ³ z Pn ( cos ϑ) dz n a ³ z Pn (cos ϑ) dz + n ³ z ( ) n P n (cos ϑ) dz for odd n a n ³ z Pn ³cos ³ ϑ dz P n (cos ϑ) z n a dz + ³z n ( ) n dz ³ ( ) n P n (cos ϑ) z n a ³ dz + z n dz a ³ ( ) n P n (cos ϑ) z n dz " n+ # a z ( ) n P n (cos ϑ) n + P n (cos ϑ) an+ n+ n,, 4, n, 3, 5,. Φ λ 4π² X Q 4π² a Q 4π² r Q 4π² r r n+ n X neven X neven + a ³ z n Pn ³cos ϑ dz r n+ P n (cos ϑ) an+ n + a n Pn (cos ϑ) n +³ r ³ a ³ a 4 P (cos ϑ)+ P4 (cos ϑ)+ r r S E B E r B θ e z S E r B θ V ln b a r πr V π ln b a r Phys. 463 E & M 3-4 / Solution3

3 Power through the cross-section a<r<b P b a b b S d A Sπrdr V a π ln b a V ln b ln b a V a Exactly the power delivered by the power supply. a V πrdr r ln b a 3. Problem 8. (ref. to Problem 7.3, p. 34) A fat wire, radius a,carries a constant current, uniformly distributed over its cross section. A narrow gap in the wire, of width w a, forms a parallel-plate capacitor. b a r dr (a) Find the electric and magnetic field in the gap, as functions of the distance s from the axis and the time t.(assumethechargeiszeroatt.) (b) Find the energy density u em and the Poynting vector S in the gap. Note especially the direction of S. Check that equation 8.4 is satisfied. Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap (eq. 8.9 in this case W, Because there is no charge in the gap). [f you worried about the fringing fields, do it for a volume of radius b<awell inside the gap.] Solution (a) Electric field: from the definition of the capacitance C Q V t Ew t Ew Phys. 463 E & M / Solution3

4 on the other hand, the capacitance of parallel plate is So or independent of s Magnetic fields: within the gap, j. So B Ã j + ² E A C ² w ² πa w t Ew ² πa w E π² a t E π² a t e z! ² E e π a z integrate over a circular cross section of radius s ³ B d A π a e z d A LHS: ³ B d B A d l πsbθ RHS: so or or π e a z d A π πsb θ π a πs B θ π a s B π a s e θ a πs Phys. 463 E & M / Solution3

5 (b) u em ² E + B ² π² a t + S E B π² π² πa st ( e 4 r ) radially inwards. Now check a t π π π² πa 4 st a s a s e z e θ e r n this case, u mech and (u mech + u em ) S u em ² t π² a π² πa t 4 therefore S s s (ss s) s s π² πa 4 s t π² πa t 4 (u mech + u em ) S (c) Total energy in the gap " s U em u em dτ π ² " π 4 ² π² a t ws + w 8 π² a t ws + ws 4 6π a π² a t + π a π s 4 # # a s wsds U em ws t π² a Phys. 463 E & M / Solution3

6 Also U em S s d A S s da S s (πsw) π² πa st (πsw) 4 π² a 4 ws t dw dt S d A π² ws t a π² a 4 ws t 4. c r sinϑ cos φ e x +sinϑ sin φ e y +cosϑ e z On the bowl d A da c r R sin ϑdϑdφ [sin ϑ cos φ e x +sinϑ sin φ e y +cosϑ e z ] T d A Q T zx ε E z E x ε 4πε R Q T zy ε E z E y ε 4πε R T zz ε E z E x E y cos ϑ sin ϑ cos φ cos ϑ sin ϑ sin φ ε Q cos ϑ sin ϑ 4πε R z T zx da x + T zy da y + T zz da z (T zx sin ϑ cos φ + T zy sin ϑ sin φ + T zz cos ϑ) R sin ϑdϑdφ " Q ε cos ϑ sin ϑ cos φ sin ϑ cos φ 4πε R Q +ε cos ϑ sin ϑ sin φ sin ϑ sin φ 4πε R + ε Q cos ϑ sin ϑ cos ϑ# R sin ϑdϑdφ 4πε R Phys. 463 E & M / Solution3

7 Q ε cos ϑ sin ϑ cos φ +cosϑsin ϑ sin φ 4πε R + cos ϑ sin ϑ cos ϑ R sin ϑdϑdφ Q ε 4πε R Q ε 4πε R ε Q 4πε R cos ϑ sin ϑ + cos ϑ sin ϑ cos ϑ R sin ϑdϑdφ R sin ϑ cos ϑdϑdφ cos ϑ +sin ϑ cos ϑ R sin ϑdϑdφ 5. Problem 8.4 (hints: read the example 8. for integral over the disk first. You need to find the electric field in the mid plane first) Solution: Create a semi hemisphere consisting of the mid-plane and the half spherical surface on the right hand side with radius The force on the charge on the left hand side is T F d A T d A + T d A mid plane spher. surface Calculate the field on the mid-plane now. consider a point on the x axis at (r,, ). Due to left-right symmetry, the fieldisonthex axis direction. Field due to one charge Field due to both charges E x E 4π² q R q 4π² R cos θ q π² R r R qr π² (r + a ) 3/ Due to symmetry about z axis. the field on the x-y plane in a cylindrical coordinate system is radially outwards.. E r (r) π² q R r R π² qr (r + a ) 3/ Phys. 463 E & M / Solution3

8 Since d A da e z but since E z and T d A Txz da e x + T yz da e y + T zz da e z T zz ² T xz ² E x E z, T yz ² E y E z ² E z Ex ² E y Ã qr π² (r + a ) 3/ The surface integral on the mid-plane is T d A mid plane e z mid plane E x + Ey ² E! q 8π ² r (r + a ) 3 T zz da e z mid plane q e 8π z ² q 8π² e z q r 8π ² (r + a ) 3 da r (r + a ) 3 (πrdr) r (r + a ) 3 dr r (r + a ) 3 dr t (t + a ) 3 dt td (t + a ) Phys. 463 E & M / Solution3

9 So t (t + a ) (t + a ) mid plane On the half spherical surface + a T d A q (t + a ) dt + 8π² e z (t + a ) dt q e a 6π² a z T, da r r4 So T d A Total force spher. surface T q F d A T d A mid plane 6π² a e z Direct calculation reveals the force on the charge on the left hand side 6. Problem 8.6 Solution F 4π² q (a) q 6π² a in the e z direction. A charged parallel plate capacitor (with uniform electric field E E e z ) is placed in a uniform magnetic field B B e x as shown. Phys. 463 E & M / Solution3

10 (a) Find the electromagnetic momentum in the space between the plates. p momentum ² S ² E B ² EB e y Assume the distance between the plates is d, surface area of A, the total momentum is p total ² EBAd e y (b) Now a resistive wire is connected between the two plates, along the z-axis, so that the capacitor slowly discharges. The current through the wire will experience a magnetic force; What is the total impulse ( R Fdt) delivered to the system, during the discharge? Assume the current on the wire is (t). t should flow from the bottom plate to thetopplate.theforceonthewireis But Also So and The impulse h F (t)d l B (t)db e y p (t) dq dt E σ q ² A² (t) A² de dt de F AdB² e y dt F dt e y ² BEAd e y AdB² de dt This is exactly the initial EM momentum stored between the plates. (c) Determine all nine elements of the stress tensor, in the region between the plates. Displayyouranswerasa3 3 matrix. T ij ² E i E j δ ije + B i B j δ ijb Phys. 463 E & M 3-4 / Solution3

11 For i 6 j T ij ² E i E j + B i B j Note E E e z, B B e x T ij, when i 6 j When i j T ii ² Ei E + Bi B T ² E x E + B x B ² E + B T ² E y E + B y B ² E B T 33 ² E z E + B z B ² E B T So ² E + B ² E B ² E B 7. Problem 8.5 Solution Now dw dt ³ E J dτ H J f + D E J E Ã H D ³ E H E D h H ³ ³ E i E H! E D Phys. 463 E & M 3-4 / Solution3

12 where Ã! ³ E H + H B E D Ã ³ E D H E + H! B ³ S u em S E H and u em E D + H B For linear media, ² and are independent of E and B respectively So or E D + H B ² E E + B B ² ³ E E + ³ B B ² E E + B B ³ E D + H B u em ³ E D + B H u em ³ E D + B H Phys. 463 E & M 3-4 / Solution3

dt Now we will look at the E&M force on moving charges to explore the momentum conservation law in E&M.

dt Now we will look at the E&M force on moving charges to explore the momentum conservation law in E&M. . Momentum Conservation.. Momentum in mechanics In classical mechanics p = m v and nd Newton s law d p F = dt If m is constant with time d v F = m = m a dt Now we will look at the &M force on moving charges