Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch32

Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch32

(Parte 2 de 2)

(b) We note that the charge canceled in the derivation of µ = Ke/B. Thus, the relation µ = Ki/B holds for a positive ion.

(c) The direction of the dipole moment is –z, opposite to the magnetic field.

(d) The magnetization is given by M = µene + µini, where µe is the dipole moment of an electron, ne is the electron concentration, µi is the dipole moment of an ion, and ni is the ion concentration. Since ne = ni, we may write n for both concentrations. We substitute µe = Ke/B and µi = Ki/B to obtain n MK K

42. The Curie temperature for iron is 770°C. If x is the depth at which the temperature has this value, then 10°C + (30°C/km)x = 770°C. Therefore,

30 C km km.

43. (a) The field of a dipole along its axis is given by Eq. 30-29:

B z where µ is the dipole moment and z is the distance from the dipole. Thus,

T. c hc h

c h

(b) The energy of a magnetic dipole Gµ in a magnetic field G B is given by

UB B=− ⋅ = − Gµφcos, where φ is the angle between the dipole moment and the field.

The energy required to turn it end-for-end (from φ = 0° to φ = 180°) is

∆UB== × × = × ×−− − −2 2 15 10 30 10 9 0 1023 6 29 10µJT T J = 5.6 10 eV.c hc h

The mean kinetic energy of translation at room temperature is about 0.04 eV. Thus, if dipole-dipole interactions were responsible for aligning dipoles, collisions would easily randomize the directions of the moments and they would not remain aligned.

4. (a) The number of iron atoms in the iron bar is

.. .. gc m cm cm gm ol mol 32c hb gc h

b g c h

µ =× × = ⋅−21 10 4 3 10 8923 23JT A m2c hc h

Thus the dipole moment of the iron bar is

(b) τ = µB sin 90° = (8.9 A · m2 )(1.57 T) = 13 N · m.

45. The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by Msat = µn, where n is the number of atoms per unit volume and µ is the magnetic dipole moment of an atom. The number of nickel atoms per unit volume is n = ρ/m, where ρ is the density of nickel. The mass of a single nickel atom is calculated using m = M/NA, where M is the atomic mass of nickel and NA is Avogadro’s constant. Thus,

9.126 10 atoms cm 58.71g mol

The dipole moment of a single atom of nickel is

Mn sat 3


46. (a) Eq. 29-36 gives τ = µrod B sinθ = (2700 A/m)(0.06 m)π(0.003 m)2(0.035 T)sin(68°)= 1.49 × 10

We have used the fact that the volume of a cylinder is its length times its (circular) cross sectional area.

(b) Using Eq. 29-38, we have

= –(2700 A/m)(0.06 m)π(0.003m)2

∆U = – µrod B(cos θf – cos θi) (0.035T)[cos(34°) – cos(68°)]

= –72.9 µJ.

47. (a) The magnitude of the toroidal field is given by B0 = µ0nip, where n is the number of turns per unit length of toroid and ip is the current required to produce the field (in the absence of the ferromagnetic material). We use the average radius (ravg = 5.5 cm) to calculate n:

400 turns 1.16 10 turns/m .

N n

Thus, i B

(b) If Φ is the magnetic flux through the secondary coil, then the magnitude of the emf induced in that coil is ε = N(dΦ/dt) and the current in the secondary is is = ε/R, where R is the resistance of the coil. Thus,

NR d dt s = FHG IKJ Φ

The charge that passes through the secondary when the primary current is turned on is

Nd N N q i dt dt d

R dt R R

The magnetic field through the secondary coil has magnitude B = B0 + BM = 801B0, where BM is the field of the magnetic dipoles in the magnetic material. The total field is perpendicular to the plane of the secondary coil, so the magnetic flux is Φ = AB, where A is the area of the Rowland ring (the field is inside the ring, not in the region between the ring and coil). If r is the radius of the ring’s cross section, then A = πr2 . Thus,

Φ=80120πrB. The radius r is (6.0 cm – 5.0 cm)/2 = 0.50 cm and


48. From Eq. 29-37 (see also Eq. 29-36) we write the torque as τ = −µBh sinθ where the minus indicates that the torque opposes the angular displacement θ (which we will assume is small and in radians). The small angle approximation leads to h Bτµθ≈−, which is an indicator for simple harmonic motion (see section 16-5, especially Eq. 16-2). Comparing with Eq. 16-23, we then find the period of oscillation is

T = 2π I µBh where I is the rotational inertial that we asked to solve for. Since the frequency is given as 0.312 Hz, then the period is T = 1/f = 1/0.312 in SI units. Similarly, Bh = 18.0 × 10 −6

. The above relation then yields I = 3.19 × 10 −9 kg.m2 .

49. (a) If the magnetization of the sphere is saturated, the total dipole moment is µtotal = Nµ, where N is the number of iron atoms in the sphere and µ is the dipole moment of an iron atom. We wish to find the radius of an iron sphere with N iron atoms. The mass of such a sphere is Nm, where m is the mass of an iron atom. It is also given by 4πρR3 /3, where ρ is the density of iron and R is the radius of the sphere. Thus Nm = 4πρR3 /3 and

3πρ .

We substitute this into µtotal = Nµ to obtain µ ρµ total =

m .

We solve for R and obtain

R m= FHG IKJ3 4 13µρµ totalπ

kg m J T m. 3 c hc h

(b) The volume of the sphere is VRs == × = ×44

..mm3ch and the

volume of the Earth is

so the fraction of the Earth’s volume that is occupied by the sphere is

(Parte 2 de 2)