Física

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 a dE 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 orbital angular 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 = m h (−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 potential