# Solução Halliday Volume 3 - Eletromagnetismo - ch32

(Parte 1 de 2)

1. (a) The flux through the top is +(0.30 T)πr2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb.

(b) The fact that it is negative means it is inward.

2. We use

3. (a) We use Gauss’ law for magnetism:z⋅= G G BdA0. Now, z⋅=++

Bd A C ΦΦΦ12, where Φ1 is the magnetic flux through the first end mentioned, Φ2 is the magnetic flux through the second end mentioned, and ΦC is the magnetic flux through the curved surface. Over the first end the magnetic field is inward, so the flux is Φ1 = –25.0 µWb. Over the second end the magnetic field is uniform, normal to the surface, and outward, so the flux is Φ2 = AB = πr2 B, where A is the area of the end and r is the radius of the cylinder. It value is

 2 350120 160 10 724 10 724=× = + × = +−−π m T Wb Wbb g c h µ

Since the three fluxes must sum to zero,

 =− − = − = −12 250 724 474 µ µ µWb Wb Wb

(b) The minus sign in c Φindicates that the flux is inward through the curved surface.

4. From Gauss’ law for magnetism, the flux through S1 is equal to that through S2, the portion of the xz plane that lies within the cylinder. Here the normal direction of S2 is +y. Therefore, r r

B r r i S S B x Ld x B x L dx Ld x iL µ µ

5. We use the result of part (b) in Sample Problem 32-1:

R dE B rR rd t µε=≥ to solve for dE/dt:

dE Br

6. From Sample Problem 32-1 we know that B ∝ r for r ≤ R and B ∝ r–1 for r ≥ R. So the maximum value of B occurs at r = R, and there are two possible values of r at which the magnetic field is 75% of Bmax. We denote these two values as r1 and r2, where r1 < R and r2 > R.

(a) Inside the capacitor, 0.75 Bmax/Bmax = r1/R, or r1 = 0.75 R = 0.75 (40 m)=30 m.

(b) Outside the capacitor, 0.75 Bmax/Bmax = (r2/R)–1 , or r2 = R/0.75 = 4R/3 = (4/3)(40 m)

= 53 m..

(c) From Eqs. 32-15 and 32-17,

iR i R d max

Tm A A

T. c hb g bg

7. (a) Noting that the magnitude of the electric field (assumed uniform) is given by E = V/d (where d = 5.0 m), we use the result of part (a) in Sample Problem 32-1 rd Edt rd dV dt rR== ≤µ ε µ ε0 022 forbg.

We also use the fact that the time derivative of sin (ωt) (where ω = 2πf = 2π(60) ≈ 377/s in this problem) is ω cos(ωt). Thus, we find the magnetic field as a function of r (for r ≤ R; note that this neglects “fringing” and related effects at the edges):

B r

Vt B rV

max max maxcosbg where Vmax = 150 V. This grows with r until reaching its highest value at r = R = 30 m:

0 max

(b) For r ≤ 0.03 m, we use the 00max max rV B µ εω = expression found in part (a) (note the

B ∝ r dependence), and for r ≥ 0.03 m we perform a similar calculation starting with the result of part (b) in Sample Problem 32-1:

max max max

for 2

R RdE dV BV t rd t rd dt rd

RV rR

 == = ¸

(note the B ∝ r–1 dependence — See also Eqs. 32-16 and 32-17). The plot (with SI units understood) is shown below.

8. (a) Inside we have (by Eq. 32-16) B = µoid r1 /2πR2 , where r1 = 0.0200, R = 0.0300, and the displacement current is given by Eq. 32-38: id = εo dΦE /dt = εo(0.00300), in SI units. Thus we find B = 1.18 × 10 −19 T.

(b) Outside we have (by Eq. 32-17) B = µoid /2πr2 where r2 = 0.0500 in SI units. Here we obtain B = 1.06 × 10 −19 T.

9. (a) Application of Eq. 32-3 along the circle referred to in the second sentence of the problem statement (and taking the derivative of the flux expression given in that sentence) leads to

B (2πr ) = εo µo (0.60 V. m/s)

Using r = 0.0200 (which, in any case, cancels out) and R = 0.0500 (SI units understood) leads to B = 3.54 × 10 −17 T.

(b) For a value of r larger than R, we must note that the flux enclosed has already reached its full amount (when r = R in the given flux expression). Referring to the equation we wrote in our solution of part (a), this means that the final fraction ( r

R ) should be replaced with unity. On the left hand side of that equation, we set r = 0.0500 m and solve. We now find B = 2.13 × 10 −17 T.

10. (a) Application of Eq. 32-7 with A = πr2 (and taking the derivative of the field expression given in the problem) leads to

 s)

B (2πr) = εo µo πr2 (0.00450 V/m.

With r = 0.0200 m, this gives B = 5.01 × 10 −2 T.

(b) With r > R, the expression above must replaced by

 s)

B (2πr) = εo µo πR2 (0.00450 V/m.

Substituting r = 0.050 m and R = 0.030 m, we obtain B = 4.51 × 10 −2 T.

1. (a) Here, the enclosed electric flux is found by integrating

ΦE = ³ r o E 2πr dr = t(0.500 V/m.s)(2π) ³ r o ()1 –

R r dr = t π ©§¹ r2 – with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3 leads to

B (2πr) = εo µo π ©§¹ · 1 r2 –

With r = 0.0200 m and R = 0.0300 m, this gives B = 3.09 × 10 −20 T.

(b) The integral shown above will no longer (since now r > R) have r as the upper limit; the upper limit is now R. Thus, ΦE = t π ©§¹ · 1

R2 –

3R =

 Consequently, Eq. 32-3

t π R2 becomes

B (2πr) = 1

6 εo µo πR2 which yields (for r = 0.0500 m) B = 1.67 × 10 −20 T.

12. Let the area plate be A and the plate separation be d. We use Eq. 32-10:

ddt d dt ddt Vd Ad dV dt or dVdt idA i C

 75 10

Therefore, we need to change the voltage difference across the capacitor at the rate of 57.510 V/s×.

13. The displacement current is given by0(/),d iAdEdtε= where A is the area of a plate and E is the magnitude of the electric field between the plates. The field between the plates is uniform, so E = V/d, where V is the potential difference across the plates and d is the plate separation. Thus

Now, ε0A/d is the capacitance C of a parallel-plate capacitor (not filled with a dielectric), so iC dV

14. We use Eq. 32-14: 0(/).d iAdEdtε= Note that, in this situation, A is the area over which a changing electric field is present. In this case r > R, so A = πR2 . Thus, dEdt i A i R

A m V msCNm22di b g

15. Consider an area A, normal to a uniform electric field G E. The displacement current density is uniform and normal to the area. Its magnitude is given by Jd = id/A. For this situation , 0(/)d iAdEdtε=, so

dEdt dE dt

16. (a) From Eq. 32-10,

E d d dE d iA A t A dt dt dt εε ε ε

(b) The negative sign in d iimplies that the direction is downward.

(c) If one draws a counterclockwise circular loop s around the plates, then according to Eq. 32-18

which means that G G

Bds⋅<0. Thus G B must be clockwise.

17. (a) We use G G

BdsI⋅=zµ0enclosed to find d d

JrI BJ r r µ µ −−

(b) From iJr d r dE

dt d dEdt

J d

Am Fm Vms2 .

18. (a) Since i = id (Eq. 32-15) then the portion of displacement current enclosed is

i R

(b) We see from Sample Problems 32-1 and 32-2 that the maximum field is at r = R and that (in the interior) the field is simply proportional to r. Therefore,

B r Rmax which yields r = R/4 = (1.20 cm)/4 = 0.300 cm.

(c) We now look for a solution in the exterior region, where the field is inversely proportional to r (by Eq. 32-17):

B R rmax which yields r = 4R = 4(1.20 cm) = 4.80 cm.

19. (a) In region a of the graph, i d

A dE dt d εε0

.Fm m NC NC s A.2c hc h

(b) id ∝ dE/dt = 0. (c) In region c of the graph,

iA dt

20. From Eq. 28-1, we have i = (ε / R ) e − t/τ since we are ignoring the self-inductance of the capacitor. Eq. 32-16 gives

B = µoid r

 2

2πR Furthermore, Eq. 25-9 yields the capacitance

C = εoπ(0.05 m)

0.003 m = 2.318 ×10 −1 F,

At t = 250 × 10 −6 s, the current is i = 12.0 V

20.0 x 106 Ω e

− t/τ = 3.50 × 10 −7 A .

Since i = id (see Eq. 32-15) and r = 0.0300 m, then (with plate radius R = 0.0500 m) we find

B = µoid r

21. (a) At any instant the displacement current id in the gap between the plates equals the conduction current i in the wires. Thus id = i = 2.0 A.

(b) The rate of change of the electric field is

dt A ddt i A

8Fm m V msc hb g

(c) The displacement current through the indicated path is d i

(d) The integral of the field around the indicated path is

2. (a) Fig. 32-34 indicates that i = 4.0 A when t = 20 ms. Thus, Bi = µoi/2πr = 0.89 mT. (b) Fig. 32-34 indicates that i = 8.0 A when t = 40 ms. Thus, Bi ≈ 0.18 mT.

(c) Fig. 32-34 indicates that i = 10 A when t > 50 ms. Thus, Bi ≈ 0.220 mT.

(d) Eq. 32-4 gives the displacement current in terms of the time-derivative of the electric field: id = εoA(dE/dt), but using Eq. 26-5 and Eq. 26-10 we have E = ρi/A (in terms of the real current); therefore, id = εoρ(di/dt). For 0 < t < 50 ms, Fig. 32-34 indicates that di/dt =

200 A/s. Thus, Bid = µoid /2πr = 6.4 × 10

−2 T.

(e) As in (d), Bid = µoid /2πr = 6.4 × 10

−2 T.

(f) Here di/dt = 0, so (by the reasoning in the previous step) B = 0.

(g) By the right-hand rule, the direction of i

BG at t = 20 s is out of page.

(h) By the right-hand rule, the direction of id

BG at t = 20 s is out of page.

23. (a) Eq. 32-16 (with Eq. 26-5) gives

B = µoid r µo Jd A r where we set A = πR2 (which led to several cancellations).

(b) Similarly, Eq. 32-17 gives B = µoid

2πr = µo Jd πR

2πr = 67.9 nT.

24. (a) Eq. 32-16 gives B = µoid r

2πR 2 = 2.2 µT.

(b) Eq. 32-17 gives B = µoid

2πr = 2.0 µT.

25. (a) Eq. 32-1 applies (though the last term is zero) but we must be careful with id,enc . It is the enclosed portion of the displacement current, and if we related this to the displacement current density Jd , then id,enc = ³ r o Jd 2πr dr = (4.0 A/m.s)(2π) ³ r o ()1 – r2 –

with SI units understood. Now, we apply Eq. 32-17 (with id replaced with id,enc) or start from scratch with Eq. 32-1, to get B= µo id enc

2πr = 27.9 nT.

(b) The integral shown above will no longer (since now r > R) have r as the upper limit; the upper limit is now R. Thus,

R2 – πR2 .

Now Eq. 32-17 gives B = µoid

2πr = 15.1 nT.

26. (a) Eq. 32-1 applies (though the last term is zero) but we must be careful with id,enc . It is the enclosed portion of the displacement current. Thus Eq. 32-17 (which derives from Eq. 32-1) becomes, with id replaced with id,enc,

B = µo id enc

= µo (3.0 A) r which yields (after canceling r, and setting R = 0.0300 m) B = 20.0 µ T.

(b) Here id = 3.0 A, and we get B = µoid

=12.0 µ T.

27. The horizontal component of the Earth’s magnetic field is given by Bhi =Bcosφ, where B is the magnitude of the field andφi is the inclination angle. Thus

B h

55T T.

28. (a) The flux through Arizona is

BAr inward. By Gauss’ law this is equal to the negative value of the flux Φ' through the rest of the surface of the Earth. So Φ' = 1.3 × 107 Wb.

(b) The direction is outward.

29. We use Eq. 32-31: µ orb, z = –mAµB.

(a) For mA = 1, µorb,z = –(1) (9.3 × 10–24 J/T) = –9.3 × 10–24 J/T.

(b) For mA = –2, µorb,z = –(–2) (9.3 × 10–24 J/T) = 1.9 × 10–23 J/T.

30. We use Eq. 32-27 to obtain ∆U = –∆(µs,zB) = –B∆µs,z, where µ

(see Eqs. 32-24 and 32-25). Thus,

31. (a) Since mA = 0, Lorb,z = mAh/2π = 0.

(b) Since mA = 0, µorb,z = –mAµB = 0.

(c) Since mA = 0, then from Eq. 32-32, U = –µorb,zBext = –mAµBBext = 0.

(d) Regardless of the value of mA, we find for the spin part

 =− =± =± × = ± ×−−µ, 9 271 0 35 321 024 25JT mT Jc hb g

sz B

(e) Now mA = –3, so

(f) and

(g) The potential energy associated with the electron’s orbital magnetic moment is now

(h) On the other hand, the potential energy associated with the electron spin, being independent of mA, remains the same: ±3.2 × 10–25 J.

32. Combining Eq. 32-27 with Eqs. 32-2 and 32-23, we see that the energy difference is

∆U = 2 µB B where µB is the Bohr magneton (given in Eq. 32-25). With ∆U = 6.0 × 10

−25 J, we obtain B = 32.3 mT.

3. (a) The potential energy of the atom in association with the presence of an external magnetic field G Bext is given by Eqs. 32-31 and 32-32:

orb ext orb, ext ext .zB UB B m Bµµ µ=− ⋅ = − = − AG

For level E1 there is no change in energy as a result of the introduction of G Bext, so U ∝mA

= 0, meaning that mA= 0 for this level.

(b) For level E2 the single level splits into a triplet (i.e., three separate ones) in the presence of G Bext, meaning that there are three different values of mA. The middle one in the triplet is unshifted from the original value of E2 so its mA must be equal to 0. The other two in the triplet then correspond to mA = –1 and mA = +1, respectively.

(c) For any pair of adjacent levels in the triplet |∆mA| = 1. Thus, the spacing is given by

34. (a) A sketch of the field lines (due to the presence of the bar magnet) in the vicinity of the loop is shown below:

(b) The primary conclusion of §32-9 is two-fold: G u is opposite to G

B, and the effect of G F is to move the material towards regions of smaller G B values. The direction of the magnetic moment vector (of our loop) is toward the right in our sketch, or in the +x direction.

(c) The direction of the current is clockwise (from the perspective of the bar magnet.)

(d) Since the size of G

B relates to the “crowdedness” of the field lines, we see that G F is towards the right in our sketch, or in the +x direction.

35. An electric field with circular field lines is induced as the magnetic field is turned on. Suppose the magnetic field increases linearly from zero to B in time t. According to Eq. 31-27, the magnitude of the electric field at the orbit is given by rd Bdt rB t = FHG IKJ = FHG IKJ22 where r is the radius of the orbit. The induced electric field is tangent to the orbit and changes the speed of the electron, the change in speed being given by

∆va t eE m t em rB t t erB m e e

HG I KJ FHG IKJ FHG IKJ =2

The average current associated with the circulating electron is i = ev/2πr and the dipole moment is µ == FHG IKJ =Ai r ev ch.

The change in the dipole moment is

er v er erBm er B m e

36. Reviewing Sample Problem 32-3 before doing this exercise is helpful. Let

2 G G G G µµµdi which leads to

K c hb g

c h

37. The magnetization is the dipole moment per unit volume, so the dipole moment is given by µ = M , where M is the magnetization and is the volume of the cylinder

( =πrL2, where r is the radius of the cylinder and L is its length). Thus,

38. (a) From Fig. 32-14 we estimate a slope of B/T = 0.50 T/K when M/Mmax = 50%. So

B = 0.50 T = (0.50 T/K)(300 K) = 1.5×102 T.

(b) Similarly, now B/T ≈ 2 so B = (2)(300) = 6.0×102 T.

(c) Except for very short times and in very small volumes, these values are not attainable in the lab.

39. For the measurements carried out, the largest ratio of the magnetic field to the temperature is (0.50 T)/(10 K) = 0.050 T/K. Look at Fig. 32-14 to see if this is in the region where the magnetization is a linear function of the ratio. It is quite close to the origin, so we conclude that the magnetization obeys Curie’s law.

40. Section 32-10 explains the terms used in this problem and the connection between M and µ. The graph in Fig. 32-37 gives a slope of

M/Mmax

Bext /T =

0.20 = 3/4 in Kelvins per Tesla. Thus we can write µmax = ¾ (0.80 T)/(2.0 K) = 0.30 .

41. (a) A charge e traveling with uniform speed v around a circular path of radius r takes time T = 2πr/v to complete one orbit, so the average current is eT ev r== 2π

The magnitude of the dipole moment is this multiplied by the area of the orbit:

Since the magnetic force with magnitude evB is centripetal, Newton’s law yields evB = rmveB= Thus, µ = FHG IKJ = FHG IKJ FHG IKJ = 12 1 2

2ev mv eB B mv K

B eeebg.

The magnetic force −×evB G G must point toward the center of the circular path. If the magnetic field is directed out of the page (defined to be +z direction), the electron will travel counterclockwise around the circle. Since the electron is negative, the current is in the opposite direction, clockwise and, by the right-hand rule for dipole moments, the dipole moment is into the page, or in the –z direction. That is, the dipole moment is directed opposite to the magnetic field vector.

(Parte 1 de 2)