Home Work Solutions 12

Home Work Solutions 12

12-1 In Fig.1, an electric field is directed out of the page within a circular region of radius R = 3.00 cm. The field magnitude is E = (0.500 V/m · s)(1 - r/R)t, where t is in seconds and r is the radial distance (r ≤ R). What is the magnitude of the induced magnetic field at radial distances (a) 2.00 cm and (b) 5.00 cm?

Sol

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

3 2

0 0

2 (0.500 V/m s)(2 ) 1 1

2 3

r r

E

r r

E rdr t rdr t r

R R

  ^{ } ^{ }

 





with SI units understood. Then (after taking the derivative with respect to time) Eq.

32-3 leads to

3 2 0 0

(2 ) 1

2 3

B r r r

   ^  R^

 .

For r = 0.0100 m and R = 0.0300 m, this gives B = 1.36  10^21 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,



3

2 2

1 1

2 3 6

E

t R R t R

^ R^ 

    

  .

Consequently, Eq. 32-3 becomes

2 0 0

(2 ) 1

B r 6  R which for r = 0.0500 m, yields

2 12 7 2 0 0 (8.85 10 )(4 10 )(0.030) 20

1.67 10 T .

12 12(0.0500)

B R

r

   ^  ^ _

   

12-2 In Fig. 2, a parallel-plate capacitor has square plates of edge length L = 1.0 m. A current of 2.0 A charges the capacitor, producing a uniform electric field between the plates, with perpendicular to the plates. (a) What is the displacement current i_d through the region between the plates? (b) What is dE/dt in this region? (c) What is the displacement current encircled by the square dashed path of edge length d = 0.50 m?

(d) What is around this square dashed path?

Sol

The electric field between the plates in a parallel-plate capacitor is changing, so there is a nonzero displacement current i_d ₀(d_E/dt) between the plates.

Let A be the area of a plate and E be 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. The current into the positive plate of the capacitor is

 

( ) _E

A d

dq d dV d Ed dE

i CV C A

dt dt dt d dt dt dt

   ^

      ,

which is the same as the displacement current.

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

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

dE

dt  1











  i_d



   

(c) The displacement current through the indicated path is

 

2

2

0.50m

2.0A 0.50A.

d d 1.0m i i d

L

   

      

 

 

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

r B  d r

s  

i

   1.26  10

H m   ^{0.50 A}  ^{ 6.3  10}

^{T  m.}

— 

12-3 Assume that an electron of mass m and charge magnitude e moves in a circular orbit of radius r about a nucleus. A uniform magnetic field is then established perpendicular to the plane of the orbit. Assuming also that the radius of the orbit does not change and that the change in the speed of the electron due to field is small, find an expression for the change in the orbital magnetic dipole moment of the electron due to the field.

Sol

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

E  r 2









dB dt  r

2









B t ,

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

v at  eE

m_e t  e m_e









r 2









B t







t  erB 2m_e .

The average current associated with the circulating electron is i = ev/2r and the dipole moment is

 

2 2

Ai r ev evr

  r



 

   

 

The change in the dipole moment is





2erv  1

2er erB 2m_e







  e²r²B 4m_e .

12-4 The magnetic field of Earth can be approximated as the magnetic field of a dipole. The horizontal and vertical components of this field at any distance r from Earth’s center are given by

where λ_m is the magnetic latitude (this type of latitude is measured from the geomagnetic equator toward the north or south geomagnetic pole). Assume that Earth’s magnetic dipole moment has magnitude μ = 8.00 × 10²² A‧m². (a) Show that the magnitude of Earth’s field at latitudeλm is given by

(b) Show that the inclination φ_i of the magnetic field is related to the magnetic (HRW 32-55)

latitudeλ_m by .

Sol: (a) The Pythagorean theorem leads to

2 2

2 2 0 0 0 2 2

3 3 3

0 2 3

cos sin cos 4sin

4 2 4

1 3sin , 4

h v m m m m

m

B B B

r r r

r

     

 

   

               

  



where cos² m + sin² m = 1 was used.

(b) We use Eq. 3-6:

 

 

3 0

3 0

2 sin

tan 2 tan .

4 cos

v m

i m

h m

B r

B r

  

 

 

   

 

12-5 A charge q is distributed uniformly around a thin ring of radius r. The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speed ω. (a) Show that the magnetic moment due to the rotating charge has magnitude

(b) What is the direction of this magnetic moment if the charge is positive?

(HRW 32-60)

Sol: (a) The period of rotation is T = 2/ and in this time all the charge passes any fixed point near the ring. The average current is i = q/T = q/2 and the magnitude of the magnetic dipole moment is

iA q  

r q r

2

1 2

2 2

  .

(b) We curl the fingers of our right hand in the direction of rotation. Since the charge is positive, the thumb points in the direction of the dipole moment. It is the same as the direction of the angular momentum vector of the ring.

12-6 Consider a solid containing N atoms per unit volume, each atom having a magnetic dipole momentμ. Suppose the direction ofμcan be only parallel or

antiparallel to an externally applied magnetic field B (this will be the case ifμis due to the spin of a single electron). According to statistical mechanics, the probability of an atom being in a state with energy U is proportional to e^-U/kT, where T is the temperature and k is Boltzmann’s constant. Thus, because energy U is -μ‧B, the fraction of atoms whose dipole moment is parallel to B is proportional to e^μB/kT and the fraction of atoms whose dipole moment is antiparallel to is proportional to e^-μB/kT. (a) Show that the magnitude of the magnetization of this solid is M = Nμtanh(μB/kT). Here tanh is the hyperbolic tangent function: tanh(x) = (e^x – e^-x)/( e^x + e^-x). (b) Show that the result given in (a) reduces to M = Nμ²B/kT forμB<<kT. (c) Show that the result of (a) reduces to M = Nμ forμB>>kT. (d) Show that both (b) and (c) agree qualitatively with Fig. 3. (HRW32-45)

Fig. 3

Sol: (a) We use the notation P() for the probability of a dipole being parallel to  B ,

and P(–) for the probability of a dipole being antiparallel to the field. The magnetization may be thought of as a “weighted average” in terms of these probabilities:

   

     

tanh .

B KT B KT

B KT B KT

N e e

N P N P B

M N

P P e e kT

 

 

      

 





    

       

(b) For B kT (that is, B kT/  1) we have e^±B/kT  1 ± B/kT, so

M N B

kT

N B kT B kT

B kT B kT

N B



F

H G IK J

    

 

tanh 1 1  .

1 1

b gb g

b gb g

(c) For B kT we have tanh (B/kT)  1, so M N B kT N



F

H G IK J

tanh  . (d) One can easily plot the tanh function using, for instance, a graphical calculator.

One can then note the resemblance between such a plot and Fig. 32-14. By adjusting the parameters used in one’s plot, the curve in Fig. 32-14 can reliably be fit with a tanh function.