Física

Páginas: 6 (1258 palavras) Publicado: 29 de maio de 2012
CHAPTER 32

SOLUTION FOR PROBLEM 19

If the electric field is perpendicular to a region of a plane and has uniform magnitude over the region then the displacement current through the region is related to the rate of change of the electric field in the region by dE , id = 0 A dt where A is the area of the region. The rate of change of the electric field is the slope of the graph. For segment adE 6.0 × 105 N/C − 4.0 × 105 N/C = = 5.0 × 1010 N/C · s dt 4.0 × 10−6 s and id = (8.85 × 10−12 F/m)(1.6 m2 )(5.0 × 1010 N/C · s = 0.71 A. For segment b dE/dt = 0 and id = 0. For segment c dE 4.0 × 105 N/C − 0 = = 2.0 × 1011 N/C · s dt 2.0 × 10−6 s

and id = (8.85 × 10−12 F/m)(1.6 m2 )(2.0 × 1011 N/C · s = 2.8 A.

CHAPTER 32

SOLUTION FOR PROBLEM 31

(a) The z component of the orbitalangular momentum is given by L z = m h/2π, where h is orb, the Planck constant. Since m = 0, Lorb, z = 0. (b) The z component of the orbital contribution to the magnetic dipole moment is given by µorb, z = −m µB , where µB is the Bohr magneton. Since m = 0, µorb, z = 0.

(c) The potential energy associated with the orbital contribution to the magnetic dipole moment is given by U = −µorb, z Bext,where Bext is the z component of the external magnetic field. Since µorb, z = 0, U = 0. (d) The z component of the spin magnetic dipole moment is either +µB or −µB , so the potential energy is either U = −µB Bext = −(9.27 × 10−24 J/T)(35 × 10−3 T) = −3.2 × 10−25 J . or U = +3.2 × 10−25 J. (e) Substitute m into the equations given above. The z component of the orbital angular momentum is Lorb, z = mh (−3)(6.626 × 10−34 J · s) = = −3.2 × 10−34 J · s . 2π 2π

(f) The z component of the orbital contribution to the magnetic dipole moment is µorb, z = −m µB = −(−3)(9.27 × 10−24 J/T) = 2.8 × 10−23 J/T . (g) The potential energy associated with the orbital contribution to the magnetic dipole moment is U = −µorb, z Bext = −(2.78 × 10−23 J/T)(35 × 10−3 T) = −9.7 × 10−25 J .

(h) The potentialenergy associated with spin does not depend on m . It is ±3.2 × 10−25 J.

CHAPTER 32

SOLUTION FOR PROBLEM 49

(a) If the magnetization of the sphere is saturated, the total dipole moment is µ = N µ, where total 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 N m = 4πρR3/3 and N= Substitute this into µtotal = N µ to obtain µtotal = Solve for R and obtain 3mµtotal R= 4πρµ The mass of an iron atom is m = 56 u = (56 u)(1.66 × 10−27 kg/u) = 9.30 × 10−26 kg . So R= 4π(14 × 3(9.30 × 10−26 kg)(8.0 × 1022 J/T)
3 103 kg/m)(2.1 1/3

4πρR3 . 3m

4πρR3µ . 3m
1/3

.

(b) The volume of the sphere is Vs = and the volume of Earth is Ve =

×

10−23 J/T)

= 1.8 × 105 m .

4π 3 4π R = (1.82 × 105 m)3 = 2.53 × 1016 m3 3 3 4π (6.37 × 106 m)3 = 1.08 × 1021 m3 , 3

so the fraction of Earth’s volume that is occupied by the sphere is 2.53 × 1016 m3 = 2.3 × 10−5 . 1.08 × 1021 m3

The radius of Earth wasobtained from Appendix C.

CHAPTER 32

HINT FOR PROBLEM 9

Use the Maxwell law of induction. The magnetic field lines are circles concentric with the boundary of the region and the magnitude of the field is uniform on a line. Thus B · ds = 2πrB. Set this equal to µ0 0 dΦEenc/dt and solve for B. The region of electric flux extends only to r = R. ans: (a) 3.54 × 10−17 T; (b) 2.13 × 10−17 T CHAPTER 32

HINT FOR PROBLEM 23

The magnetic field a distance r from the central axis of a uniform distribution of displacement current is given by B = µo id /2πr, where id is the displacement current though a cross-section. For part (a) the point is inside the displacement current distribution so you should take id to be πr 2 Jd . For part (b) the point is outside so you should take it to be...
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