Solution From Ampere s law, the magnetic field at point a is given by B a = µ 0I a

Transcription

1 Prof. Anchordoqui Problems set # 8 Physics 69 April 4, 05. The unit of mgnetic flux is nmed for Wilhelm Weber. The prcticl-sie unit of mgnetic field is nmed for Johnn Krl Friedrich Guss. Both were scientists t Gottingen, Germny. Along with their individul ccomplishments, together they built telegrph in 833. It consisted of bttery nd switch, t one end of trnsmission line 3 km long, operting n electromgnet t the other end. (André Ampérè suggested electricl signling in 8; Smuel Morse built telegrph line between Bltimore nd Wshington in 844.) Suppose tht Weber nd Gusss trnsmission line ws s digrmmed in Fig.. Two long, prllel wires, ech hving mss per unit length of 40.0 g/m, re supported in horiontl plne by strings 6.00 cm long. When both wires crry the sme current I, the wires repel ech other so tht the ngle between the supporting strings is 6.0. (i) Are the currents in the sme direction or in opposite directions? (ii) Find the mgnitude of the current. Solution The seprtion between the wires is = 6.00cm sin 8.00 =.67 cm. (i) Becuse the wires repel, the currents re in opposite directions. (ii) Becuse the mgnetic force cts horiontlly, F B Fg = µ 0I l πmg = tn 8.00, yielding I = mgπ µ 0 l tn 8.00 nd so I = 67.8 A.. Figure is cross-sectionl view of coxil cble. The center conductor is surrounded by rubber lyer, which is surrounded by n outer conductor, which is surrounded by nother rubber lyer. In prticulr ppliction, the current in the inner conductor is.00 A out of the pge nd the current in the outer conductor is 3.00 A into the pge. Determine the mgnitude nd direction of the mgnetic field t points nd b. Solution From Ampere s lw, the mgnetic field t point is given by B = µ 0I πr, where I is the net current through the re of the circle of rdius r. In this cse, I =.00 A out of the pge (the current in the inner conductor), so B = 4π 0 7 T m/a A π m = 00 µt towrd top of the pge. Similrly t point b: B b = µ 0I b πr b, where I b is the net current through the re of the circle hving rdius r b. Tking out of the pge s positive, I b =.00 A 3.00 A =.00 A, or I b =.00 A into the pge. Therefore, B b = 4π 0 7 T m/a A π m = 33 µt towrd the bottom of the pge. 3. A long cylindricl conductor of rdius crries current I s shown in Fig. 3. The current density J, however, is not uniform over the cross section of the conductor but is function of the rdius ccording to J = br, where b is constnt. Find n expression for the mgnetic field B (i) t distnce r < nd (ii) t distnce r >, mesured from the xis. Solution Use Ampérè s lw, B d s = µ0 I. For current density J, this becomes B d s = µ 0 J da. r (i) For r <, this gives Bπr = µ 0 0 brπrdr nd B = µ 0br 3 (for r < or inside the cylinder. (ii) When r >, Ampérè s lw yields πr )B = µ 0 0 brπrdr = πµ 0b 3 3 or B = µ 0b 3 3r (for r > or outside the cylinder).

2 4. A toroid with men rdius of 0.0 cm nd 630 turns is filled with powdered steel whose mgnetic susceptibility χ is 00. The current in the windings is 3.00 A. Find B (ssumed uniform) inside the toroid. Solution: Assuming uniform B inside the toroid is equivlent to ssuming r (see Fig. 4); then B 0 = µ 0 H µ 0 NI s for tightly wound solenoid. This leds to B 0 = µ 0 π 0.00 = T. With the steel, B = ( + χ)µ 0 H = T = 0.9 T. 5. At sturtion, when nerly ll of the toms hve their mgnetic moments ligned, the mgnetic field in smple of iron cn be.00 T. If ech electron contributes mgnetic moment of A m (one Bohr mgneton), how mny electrons per tom contribute to the sturted field of iron? Iron contins pproximtely toms/m 3. Solution Consider torus mde of unmgnetied iron. If the current in the primry coil is incresed from ero to some vlue I, the mgnitude of the mgnetic field strength H increses linerly with I ccording to H = ni. Furthermore, the mgnitude of the totl field B lso increses with incresing current, s shown by the curve from point O to point in Fig. 5. At point O, the domins in the iron re rndomly oriented, corresponding to M = 0. As the incresing current in the primry coil cuses the externl field to increse, the ligned domins grow in sie until nerly ll mgnetic moments re ligned t point. At this point the iron core is pproching sturtion, which is the condition in which ll mgnetic moments in the iron re ligned. Next, suppose tht the current is reduced to ero, nd the externl field is consequently eliminted. The B-versus-H curve, clled mgnetition curve, now follows the pth b in Fig. 5. Note tht t point b, B is not ero even though the externl field H is ero. The reson is tht the iron is now mgnetied due to the lignment of lrge number of its mgnetic moments. At this point, the iron is sid to hve remnent mgnetition. Shortly fter removing the externl field we hve.00 T = µ 0 M. In ddition, M = xnµ B, where n is the number of toms per volume nd x is the number of electrons per tom contributing. Therefore B = µ 0 µ B xn, yielding x = B µ 0 µ B n =.00 T m N m/t 4π 0 7 T m/a = The mgnetic moment of the Erth is pproximtely A m. (i) If this were cused by the complete mgnetition of huge iron deposit, how mny unpired electrons would this correspond to? (ii) At two unpired electrons per iron tom, how mny kilogrms of iron would this correspond to? [Hint: Iron hs density of 7, 900 kg/m 3, nd pproximtely iron toms/m 3.] Solution (i) Number of unpired electrons = A m A m = Ech iron tom hs two unpired electrons, so the number of iron toms required is (ii) Mss = toms 7,900 kg/m toms/m 3 = kg. 7. Find the mgnetic field on the xis t distnce bove disk of rdius with chrge

3 density σ rotting t n ngulr speed (clockwise when viewed from the point P ). Solution Consider thin ring between r nd r + dr s shown in Fig. 6. The chrge per length on the ring is q = σdr nd ech point on it is moving t speed v = ω r, so the current is I = vq = σωrdr. By symmetry, the mgnetic field is in the direction. Using the result from problem 6 of the previous homework, the contribution from the ring is db = µ 0(σωr dr) The mgnetic field for the entire disk is B = µ 0σω substitution (dont forget to chnge limits) gives B = µ 0σω 4 [ ] µ 0 σω +. ( + ) / 0 r 3 r. (r + ) 3/ (r + ) 3/ dr. Let u = r, so du = rdr. This 0 u du = µ 0σω (u+ ) 3/ 4 (u+ ) (u+ ) / 0 8. A right-circulr solenoid of finite length L nd rdius hs N turns per unit length nd crries current I. Show tht the mgnetic induction on the cylinder xis in the limit NL is B = µ 0NI (cos + cos ), where the ngles re defined in Fig. 7. Solution We strt by computing the mgnetic field on xis for single loop of wire crrying current I shown in Fig. 8. This my be done by n elementry ppliction of the Biot-Svrt lw. By symmetry, only the component contributes B = µ 0I [d l ] 4π = µ 0I 4π = µ 0I 4π π = 3 3 dl sin α 3 µ 0 I. Substituting in = + µ yields B 3 = 0 I. We now use liner superposition ( + ) 3/ to obtin the field of the solenoid. Defining nd s in Fig. 8 (where + = L) we hve B = µ 0NI d. A simple trig substitution = tn α converts this integrl to ( + ) 3/ B = µ 0NI + tn ( /) tn ( /) cos αdα = µ 0NI sin α + tn ( /). A bit of geometry then demonstrtes tn ( /) tht this is equivlent to B = µ 0NI (cos + cos ). 9. Two circulr coils of rdius, ech with N turns, re perpendiculr to common xis. Ech coil crries stedy current I in the sme direction, s shown in Fig. 9 (i) Find the totl mgnetic field on the xis B tot s function of, using the midwy point s the origin ( = 0). (ii) Show tht db tot d = 0 t the midpoint. This mens tht, functionlly, t lest t tht one loction, the field is not chnging. (iii) If the seprtion b is picked correctly, then d B tot = 0 t the midpoint. This d configurtion is known s Helmholt coils. Determine the pproprite vlue of b. (iv) Show tht the mgnetic field on the xis ner the origin, s n expnsion in powers of (up to 4 inclusive) is given by B tot = µ 0NI [ d 3 + 3(b ) d 4 + 5(b4 6b + 4 ) 4 ] 6d 8 +, where d = + b /4. (v) Show tht the mgnetic field on the xis for lrge is given by the expnsion in inverse odd powers of obtined from the smll expnsion of prt (iv) by the forml substitution d. (vi) Wht is the mximum permitted vlue of / if the field is to be uniform to one prt in 0 4 (or one prt in 0 ) in the Helmholt configurtion? = Solution (i) The mgnetic field due to circulr hoop of rdius crrying current I, which µ lies in the x-y plne with its center t the origin, is B = 0 I. If the coils on the system ( + ) 3/

4 hve N turns, then the field from ech coil is just N times lrger. The mgnetic field from both coils is then given by B tot = B + B { } = µ 0NI [ + (b/ + ) ] 3/ + [ + (b/ ) ] 3/ () (ii) The derivtive is db tot d It is esy to see tht dbtot d { } = 3µ 0NI b/ [ + (b/ ) ] 5/ b/ + [ + (b/ + ) ] 5/. = 0. This is true regrdless of the seprtion b. (iii) Differentiting =0 { } 5(b/+) 5(b/ ) +. [ +(b/+) ] 7/ [ +(b/+) ] 5/ [ +(b/ ) ] 7/ [ +(b/ ) ] 5/ gin gives d B tot = 3µ 0NI d At the midpoint, the second derivtive becomes d B tot d { } = 3µ 0NI 0b /4 =0 [ + b /4] 7/ [ + (b/) ] 5/ [ ] = 3µ 0 NI + b /4 5b /4 ( + b /4) 7/ [ ] = 3µ 0 NI b ( + b /4) 7/. Therefore, the condition d B tot =0 = 0 is met if b =. (iv) All we must do now is to Tylor expnd d the terms to order 4. Noting tht we re seeking n expnsion in powers of /d, we my rewrite () s B tot = µ 0NI [(d b + ) 3/ + (d + b + ) 3/] = µ 0NI [ d 3 ( bζ + d ζ ) 3/ + ( + bζ + d ζ ) 3/] () = µ 0NI { [ bζ + ( d 3 + b /4)ζ ] 3/ [ + + bζ + ( + b /4)ζ ] } 3/, where we hve introduced ζ = /d. Expnding this in powers of ζ yields B tot = µ 0NI [ d (b )ζ + 5 ] 6 (b4 6b + 4 )ζ 4 +, which is the desired result. The mgnetic field long the xis chnges very slowly when is very smll. (v) For lrge we Tylor expnd B tot in inverse powers of, tht is B tot = µ 0NI 3 { [ b + ( + b /4) ] 3/ + [ + b + ( + b /4) ] 3/ }.

5 Compring this with the lst line of () shows tht the Tylor series is formlly equivlent under the substitution ζ, which my be ccomplished by tking d. (vi) For b = the field is of the form B tot = µ 0NI d 3 ( ) d 8 + = 4µ 0NI [ 5 3/ 3 44 ( ) ] Tking the ( /) 4 term s smll correction, the field non-uniformity is δbtot B tot 44 ( ) 4. 5 For uniformity to one prt in 0 4, we find / < 0.097, while for uniformity to one prt in 0, we insted obtin / < These numbers re ctully pretty good becuse of the fourth power. For exmple, the first vlue indictes we cn move ±0% of the distnce between the coils while mintining field uniformity t the level of 0.0%. Helmholt coils re very useful in the lb for cnceling out the Erth s mgnetic field. 0. Let us tret the motion of n electron (chrge e, mss m) in hydrogen tom clssiclly. Suppose tht n electron follows circulr orbit of rdius r round proton. Wht is the ngulr frequency ω 0 of the orbitl motion? Suppose now tht smll mgnetic field B perpendiculr to the plne of the orbit is switched on. Assuming tht the rdius of the orbit does not chnge, clculte the shift in the ngulr frequency ω = ω f ω 0 of the orbitl motion in terms of the mgnitude B of the mgnetic field, chrge e, mss m, nd rdius r. This is known s the Zeemn effect. Solution We first pply Newton s second lw to find the ngulr frequency ω ) of the orbitl motion. Coulombs lw describes the force cting on the electron due to the electric interction between the proton nd the electron F elec = 4πɛ 0 e r ˆr, where ˆr is unit vector in the plne of the circulr orbit pointing rdillly outwrd. Becuse we re ssuming the motion is circulr the ccelertion is = rω0 ˆr. Newtons second lw is then 4πɛ 0 e r ˆr = mrω 0 ˆr. (3) We cn now solve for the ngulr frequency e ω 0 = 4πɛ 0 mr 3. (4) When smll mgnetic field B perpendiculr to the plne of the orbit is switched on there is mgnetic force cting on the electron F mg = e v B. Let s first ssume tht the mgnetic force points rdilly inwrd nd tht the rdius of the orbit does not chnge. This will cuse the electron to speed up hence incresing the ngulr frequency to ω f. Choose coordintes s shown in Fig. 0. Assume tht the mgnetic field points in the positive -direction. In order to hve rdilly inwrd force, we require tht the velocity of the electron is v = rω f ˆ. Then the mgnetic force is given by F mg = e v B = erω f ˆ Bˆk = erωf Bˆr. Therefore Newton s second lw is now e 4πɛ 0 ˆr erω r f Bˆr = mrωf ˆr. Using (3) we cn rewrite this s mrω0 ˆr erω fbˆr = mrωf ˆr or 0 = ω f eω fb m ω 0. We cn solve this qudrtic eqution to obtin ω f = eb m ± e B + ω 4m 0. We choose the positive squre root to keep ω f > 0. We now

6 ents in the sme direction or in opposite dire Find the mgnitude of the current. Figure : Problem P or 0 = ωf Ampère s Lw + eω fb m y 6.00 cm 6.0 substitute (4) into the bove eqution yielding ω f = eb m + e B + e 4m 4πɛ 0. Then the chnge mr 3 in ngulr frequency is Ω = ω f ω 0 = eb m + e B + e e 4m 4πɛ 0 mr 3 4πɛ 0. Suppose we reverse the direction of the mgnetic field s shown in Fig. 0. Then the mgnetic force points mr 3 outwrd, Fmg = e v B = erω f ˆ Bˆk = +erωf Bˆr, resulting in smller ngulr frequency ω f. epeting the nlysis we just finished we hve tht mrω0 ˆr + erω fbˆr = mrωf ˆr ω 0. We cn solve this qudrtic eqution for ω f choosing the positive squre root to keep ω f > 0, tht is ω f = eb m ± e B + ω 4m 0. The chnge in ngulr frequency is now m + Ω = ω Four long, f ω 0 = prllel eb e conductors crry equl curr B 4m + e e 4πɛ 0 mr 3 4πɛ 0 mr A. Figure P30. is n end view of the cond current direction is into the pge t points A icted by the crosses) nd out of the pge t C icted by the dots). Clculte the mgnitud ction of the mgnetic field t point P, locted er of the squre of edge length 0.00 m. x

7 ., surrounded by nother rubber lyer. In prticulr ppliction, the current in the inner conductor is.00 A out of the pge nd the current in the outer conductor is 3.00 A into the pge. Determine the mgnitude nd of direction the cn of nd the I mgnetic the upwrd field current, t points uniformly nd b. distributed over its curved wll. Determine the mgnetic field () just inside the wll nd (b) just outside. (c) Determine the pressure on the wll. 8. Niobium metl becomes superconductor when cooled 3.00 A below 9 K. Its superconductivity is destroyed when the surfce mgnetic field exceeds T. b. Determine the mximum current.00 A.00-mm-dimeter niobium wire cn crry nd remin superconducting, in the bsence of ny externl mgnetic field. 9. A long cylindricl conductor of rdius crries current I s shown in Figure P30.9. The current density J, however, is not uniform over the cross mmsection of mmthe conductor but is function of the rdius ccording to J br, where b is Figure P30.3 constnt. Find n expression Figure : Problemfor. the mgnetic field B () t distnce r nd (b) t distnce r, 4. mesured The mgnetic from field the xis cm wy from long stright wire crrying current.00 A is.00 T. () At wht distnce is it 0.00 T? (b) Wht If? At one instnt, the two conductors in long household I extension cord crry equl.00-a currents in opposite directions. The two wires re 3.00 mm prt. Find the mgnetic field 40.0 cm wy r from the middle of the stright cord, in the plne of the two wires. (c) At r wht distnce is it one tenth s lrge? (d) The center wire in coxil cble crries current.00 A in one direction nd the sheth round it crries current.00 A in the Figure opposite P30.9 direction. Wht mgnetic Figure 3: Problem 3. field does the cble crete t points outside? In Figure A pcked P30.30, bundle both of currents 00 long, in stright, the infinitely insulted long wires wires re forms in the cylinder negtive of rdius x direction () cm. Sketch () If the ech mgnetic wire field crries pttern.00 A, in wht the re y plne. the mgnitude (b) At wht nd direction distnce of d the long p 8. N b su m c e 9. A s is is co ( m 30. In r fi th

8 Chpter 30 0 f f m 30. T j f Wb A m ma ent to ssuming olenoid. g 9 T B 09. T FIG. P30.4 fe J T j T K J 3 m M xn B. Then B 0 B xn where n is the of electrons per tom contributing. c 00. T O 0. f N m T 4 0 T m A je see tht the pplied field is described by B0 C d 0H, nd the definition of susceptibility T T Figure 30.3 Mgnetition curve Figure 5: Problem 5. for ferromgnetic mteril. K A Figure 4: Problem CHAPTE 3 0 Sources of the Mgnetic Field b B j e H the mgnitude the curve from re rndomly primry coil c until nerly ll pproching s iron re ligned Next, supp consequently e follows the pth externl field B lignment of the iron is sid If the curre mgnetic field unmgnetied

9 Find the mgnetic field on the xis t distnce bove disk of rdius with chrge density rotting t n ngulr speed (clockwise when viewed from the point P). Consider thin ring between r nd r + dr s shown below. The chrge per length on the ring is dr nd ech point on it is moving t speed v r, so the current is I v r dr. P where the ngles re defined in the figure. dr r We strt by computing the mgnetic field on xis for single loop current I. This my be done by n elementry ppliction of lw. By symmetry, the mgnetic field is Figure in the 6: Problem direction. 7. dl Using the result from Ex. 5.6 (Eq. 5.38), the contribution from the ring is 0 r dr r db r. B 3 α The mgnetic field for the entire disk is 0 r 3 dr B By symmetry, only the. component contributes r 3 0 Let u r FigureB 7: = µ Z 0I [d ~` ~ ] Problem 8., so du r dr. This substitution (don t 4 forget 3 = µ Z 0I d` sin to chnge 4 limits) gives 3 = µ 0I 4 3 here the ngles re defined in the figure. he ngles re defined in the figure. We strt by computing the mgnetic field on xis for single lo current I. This mysubstituting be done in by n elementry ppliction = + yields lw. dl strt by computing the mgnetic field on xis for single loop of wire crrying urrent I. This my be done by n elementry ppliction of the Biot-Svrt. α dl symmetry, only the component contributes B = µ Z 0I [d ~` ~ ] 4 3 By symmetry, only the component contributes α = µ Z 0I d` sin 4 3 B B = µ 0 I ( + ) 3/ We now use liner superposition to obtin the field of the solen nd B s follows (where + = L) we hve Figure 8: Solution of problem 8. B = µ 0I Z = µ 0I 4 3 = µ 0I 3 Nd ( + ) 3/ A simple trig substitution = tn converts this integrl to

10 I I > b Figure 9: Problem 9. Suppose we reverse the direction of the mgnetic field s in the figure below. Assume tht the mgnetic field points in the positive -direction. In order to hve Figure Then rdilly inwrd force, we require tht the velocity of the electron is v! the 0: mgnetic Problem force = r! f ˆ". Then the 0. points outwrd,! mgnetic force is given by F mg =!e v! " B! =!er# f ˆ$ "!Bˆk = +er# f B ˆr,! F mg =!e v! " B! =!er# f ˆ$ " Bˆk =!er# f B ˆr. resulting in smller ngulr frequency! f. Therefore Newton s Second Lw is now! ke r Using Eq. () we cn rewrite this s ˆr! er" f Bˆr =!mr" f ˆr. epeting the nlysis we just finished we hve tht or!mr" 0ˆr + er" f Bˆr =!mr" f ˆr 0 =! f + e! f B "! 0.