CHAPTER 9
9.1. In
Fig. 9.4, let B let B = 0. 0 .2cos120π t T, and assume that the conductor joining the two ends of the resistor resistor is perfect. It may be assumed assumed that the magnetic magnetic field produced produced by I (t) is negligible. Find: a) V ab Sincee B is constant over the loop area, the flux is Φ = π(0. (0.15)2 B = 1.41 ab (t): Sinc 10 2 cos120π t Wb. Wb. No Now, w, emf = V ba dΦ/dt = (120π )(1. )(1.41 10 2 )sin120π t. ba (t) = Then V Then V ab ab (t) = V ba ba (t) = 5.33sin120 π t V. −
−
−
−
×
×
−
b) I (t) = V = V ba /R = 5.33 sin(120 sin(120π t)/250 = 21. 21.3 sin(120 sin(120π t) mA ba (t)/R = 9.2. In
the example described by Fig. 9.1, replace the constant magnetic flux density by the timevarying quantity B = B = B0 sin ωt az . Assume that v is constant and that the displacement y displacement y of the bar is zero at t at t = = 0. Find the emf at any time, t. t . The magnetic flux through the loop area is Φm =
vt
B · dS =
s
d
0
B0 sin ωt ( az · az ) dxdy = dxdy = B B 0 v t d sin ωt
0
Then the emf is
emf =
E · dL =
− dΦdtm = −B d v [sin ωt + ωt cos ωt] V 0
300 az cos(3 108 t y ) A/m in free space, find the emf developed in the general aφ direction about the closed path having corners at a) (0,0,0), (0,0,0), (1,0,0), (1,0,0), (1,1,0), (1,1,0), and (0,1,0): (0,1,0): The magnetic magnetic flux will be:
9.3. Given H =
×
1
Φ
=
0
−
1
0
= 300µ 300µ0
8
300µ 300 µ0 cos(3
8
× 10 t − y) dxdy = dxdy = 300µ 300µ sin(3 × 10 t − y )| sin(3 × 10 t − 1) − sin(3 × 10 t) Wb
0
8
8
1 0
Then
− ddtΦ = −300(3 × 10 )(4π × 10 ) cos(3 × 10 t − 1) − cos(3 × 10 t) = −1.13 × 10 cos(3 × 10 t − 1) − cos(3 × 10 t) V 8
emf =
5
7
−
8
8
8
8
b) corners at (0,0,0), (2 π,0,0), (2π ,2π,0), (0,2π ,0): In this case, the flux is Φ
= 2π
8
2π 0
× 300 300µ µ sin(3 × 10 t − y)| 0
The emf is therefore 0.
164
=0
9.4. A
rectangu rectangular lar loop of wire containin containingg a high-resi high-resistan stance ce voltmet voltmeter er has corners initiall initially y at (a/2 a/2, b/2 b/2, 0), ( a/2 a/2, b/2 b/2, 0), ( a/2 a/2, b/2 b/2, 0), and (a/ (a/22, b/2 b/2, 0). 0). The The loo loop p begin beginss to rotat rotatee about the x axis at constant angular velocity ω, with the first-named corner moving in the az direction at t = 0. Assume Assume a uniform uniform magnetic magnetic flux densit density y B = B0 az . Determ Determin inee the induced emf in the rotating loop and specify the direction of the current.
−
−
−
−
The magnetic flux though the loop is found (as usual) through Φm =
B · dS, where S = n da
s
Because the loop is rotating, the direction of the normal, n, changing, and is in this case given by n = cos ω t az sin ωt ay
−
Therefore,
b/2
Φm =
b/2
−
a/2
B0 az · (cos · (cos ωt az a/2
−
− sin ωt ay ) dxdy = dxdy = abB abB cos ωt 0
The integral is taken over the entire loop area (regardless of its immediate orientation). The important result is that the component of B that is normal to the loop area is varying sinusoidally, and so it is fine to think of the B field itself rotating about the x axis x axis in the opposite direction while the loop is stationary. Now the emf is emf =
E · dL =
− dΦdtm = ab ωB sin ωt V 0
The direction of the current is the same as the direction of E in the emf expression. It is easiest to picture this by considering the B field field rotati rotating ng and the loo loop p fixed. fixed. By convention, d convention, d L will be counter-clockwise when looking down on the loop from the upper half-space (in the opposite direction of the normal vector to the plane). The current will be counter-clockwise whenever the emf is positive, and will be clockwise whenever the emf is negative. location of the sliding bar in Fig. 9.5 is given by x = x = 5t + 2t 2t3 , and the separation of the two rails is 20 cm. Let B = 0.8x2 az T. Find the voltmeter reading at: a) t = 0.4 s: The flux through the loop will be
9.5. The
0.2
Φ
=
0
x
0.8(x 8(x )2 dx dy = dy =
0
0.16 3 0.16 x = (5t (5t + 2t 2t3 )3 Wb 3 3
Then emf =
− ddtΦ = 0.316 (3)(5t (3)(5t + 2t ) (5+6t (5+6t ) = −(0. (0.16)[5(. 16)[5(.4)+2(. 4)+2(.4) ] [5+6(. [5+6(.4) ] = −4.32 V 3 2
2
3 2
2
b) x = 0.6 m: Have 0. 0.6 = 5t 5 t + 2t 2t3 , from which we find t find t = 0.1193. Thus emf =
3 2
2
−(0. (0.16)[5(. 16)[5(.1193) + 2(. 2(.1193) ] [5 + 6(. 6(.1193) ] = −.293 V 165
9.6. Let
the wire loop of Problem 9.4 be stationary in its t = 0 position and find the induced emf that results from a magnetic flux density given by B(y, t) = B0 cos(ω t β y) az , where ω and β are are constants. We begin by finding the net magnetic flux through the loop:
−
Φm
=
b/2
B · dS =
s
=
b/2
−
a/2
B0 cos(ωt a/2
−
B0 a [sin(ωt + β b/2) b/2) β
− β y) az · az dxdy
− sin(ωt − β b/2)] b/2)]
Now the emf is
− dΦdtm = − B β aω [cos(ωt + β b/2) b/2) − cos(ωt − β b/2)] b/2)] Using the trig identity, cos(a cos(a ± b) = cos a cos b ∓ sin a sin b, we may write the above result emf =
0
E · dL =
as
ω β
emf = +2B +2 B0 a sin(ωt) sin( sin(β b/2) b/2) V
9.7. The
rails in Fig. 9.7 each each have have a resist resistanc ancee of 2.2 Ω/m. The The bar bar moves moves to the the righ rightt at a constant speed of 9 m/s in a uniform magnetic field of 0.8 T. Find I ( I (t), 0 < t < 1 < 1 s, if the bar is at x = 2 m at t = 0 and a) a 0.3 Ω resistor is present across the left end with the right end open-circuited: The flux in the left-hand closed loop is = B Φl = B
× area = (0. (0.8)(0. 8)(0.2)(2 + 9t 9t) = −dΦl /dt = −(0. (0.16)(9) = − 1.44 V. With With the bar bar in motio motion, n, the loop loop
Then, emf l resistance is increasing with time, and is given by R by Rl (t) = 0.3+2[2. 3+2[2.2(2+9t 2(2+9t)]. The current is now emf l 1.44 I l (t) = = A Rl (t) 9.1 + 39 39..6t
−
Note that the sign of the current indicates that it is flowing in the direction opposite that shown in the figure. b) Repeat part a, but with a resistor of 0.3 Ω across across each end: end: In this case, there will be a contributi contribution on to the current current from the right right loop, which which is now now closed. closed. The flux in the right loop, whose area decreases with time, is Φr
= (0. (0 .8)(0. 8)(0.2)[(16
− 2) − 9t]
and emf r = dΦr /dt = (0. (0.16)(9) = 1. 1.44 V. The resistanc resistancee of the right loop is Rr (t) = 0.3 + 2[2. 2[2.2(14 9t)], and so the contribution to the current from the right loop will be
− −
I r (t) =
−1.44 A 61. 61.9 − 39. 39.6t
The minus sign has been inserted because again the current must flow in the opposite direction direction as that indicated indicated in the figure, figure, with the flux decreasing decreasing with time. time. The total current is found by adding the part a part a result, or I T T (t) =
−1.44
1 61. 61.9 166
−
1 + 39. 39.6t 9.1 + 39. 39.6t
A
9.8. A
perfectly-conducting filament is formed into a circular ring of radius a. At one one poin point a resistance R is inserted into the circuit, and at another a battery of voltage V 0 is inserted. Assume that the loop current itself produces negligible magnetic field. a) Apply Apply Farada Faraday’s y’s law, Eq. (4), evaluat evaluating ing each side of the equation equation carefully carefully and indeindependent pendently ly to show show the equality equality:: With no B field present, and no time variation, the right-hand side of Faraday’s law is zero, and so therefore
E · dL =
0
This is just a statement of Kirchoff ’s ’s voltage law around the loop, stating that the battery voltage is equal and opposite to the resistor voltage. b) Repeat part a, assumi assuming ng the battery battery remov removed, ed, the ring ring closed closed aga again, in, and and a linear linearlylyincreasing B field applied applied in a direction direction normal normal to the loop surface: The situation situation now becomes becomes the same as that that shown shown in Fig. Fig. 9.4 9.4,, except except the loop radius radius is now now a, and the resist resistor or value value is not not specifie specified. d. Consid Consider er the loo loop p as in the x-y plane with the positive z axis axis direct directed ed out of the page. The aφ direction is thus counter-clockwise around the loop. The B field (out of the page as shown) can be written as B(t) = B 0 t az . With the normal to the loop specified as az , the direction of dL is, by the right hand convention, aφ . Since Since the wire is perfectly-c perfectly-condu onductin cting, g, the only voltage voltage appears across the resistor, resistor, and is given as V as V R . Faraday’s law becomes
E · dL = V = V R =
−
dΦm = dt
−
d dt
B0 t az · az da = da =
s
2
−πa B
0
This indicates that the resistor voltage, V voltage, V R = π a2 B0 , has polarity such that the positive terminal is at point a in the figure, while the negative terminal is at point b. Curre Curren nt 2 flows in the clockwise direction, and is given in magnitude by I = π a B0 /R. /R. 9.9. A
square filamentary loop of wire is 25 cm on a side and has a resistance of 125 Ω per meter length. The loop lies in the z the z = 0 plane with its corners at (0, (0, 0, 0), (0. (0.25, 25, 0, 0), (0. (0.25, 25, 0.25, 25, 0), and (0, (0, 0.25, 25, 0) at t = 0. The loop is moving moving with veloci velocity ty vy = 50 m/s in the field Bz = 8 8cos(1. 8cos(1.5 10 t 0.5x) µT. µT. Develop a function of time which expresses the ohmic power being deliver delivered ed to the loop: First, First, since since the field does not vary vary with y, the loop motion in the y direction direction does not produce produce any any time-v time-varying arying flux, and so this motion is imma immateria terial. l. We can evaluate the flux at the original loop position to obtain:
×
−
.25
Φ(t)
=
0
=
.25
8
0
6
−
−(4 × 10
6
−
× 10
8
× 10 t − 0.5x) dxdy sin(1. sin(1.5 × 10 t − 0.13) − sin(1. sin(1.5 × 10 t)
)
cos(1. cos(1.5 8
8
Wb
Now, emf = V (t (t) = dΦ/dt = /dt = 6.0 102 cos(1. cos(1.5 108 t 0.13) cos(1. cos(1.5 108 t) , The total loop resistance is R is R = = 125(0. 125(0.25 + 0. 0.25 + 0. 0.25 + 0. 0.25) = 125 Ω. Then the ohmic power is
−
V 2 (t) P ( P (t) = = 2.9 R
×
3
× 10
×
cos(1. cos(1.5
8
−
−
×
8
× 10 t − 0.13) − cos(1. cos(1.5 × 10 t)
167
2
Watts
9.10
a) Show that the ratio of the amplitudes of the conduction current density and the displacement current density is σ/ω for the applied field E = E m cos ωt. Assu Assume me µ = µ0 . Firs First, t, D = E = E m cos ω t. Then the displacemen displacementt current current density density is ∂ D/ D/∂ t = ωE m sin ωt. Second, Second, J c = σ E = σ E m cos ωt. Using these results we find | find |J J c |/|J d | = σ /ω.
−
b) What is the amplitude amplitude ratio if if the applied field field is E is E = E m e t/τ , where τ is is real? As before, before, τ τ t/ t/ find D = E = E m e , and so J d = ∂ D/∂ t = (/τ )E m e . Also, J c = σ E m e t/τ . σ τ /. Finally, |J c |/|J d | = στ −
−
−
−
−
9.11. Let
the internal dimension of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The homogeneous material inside the capacitor has the parameters = 10 11 F/m, µ F/m, µ = = 10 5 H/m, and σ = 10 5 S/m. If the electric field intensity is E = (106 /ρ) cos(10 cos(105 t)aρ V/m, find: a) J: Use 10 J = σ E = cos(105 t)aρ A/m2 −
−
−
ρ
b) the total conduction conduction current, current, I I c , through the capacitor: Have I c =
J · dS =
2πρlJ = 20 π l cos(105 t) = 8π cos(105 t) A
c) the total displacement displacement current, I current, I d , through the capacitor: First find ∂ D ∂ Jd = = (E) = ∂ t ∂ t
(105 )(10
11
−
−
ρ
)(106 ) sin(105 t)aρ =
− ρ1 sin(10 t) A/m 5
Now I d = 2πρlJ d =
5
5
−2πl sin(10 t) = −0.8π sin(10 t) A
d) the ratio ratio of the amplitude amplitude of I I d to that of I I c , the quality factor of the capacitor: This will be |I d | 0.8 = = 0. 0 .1 |I c | 8
168
9.12. Find
the displacement current density associated with the magnetic field (assume zero conduction current): current): H = A = A1 sin(4x sin(4x)cos(ωt
− β z) ax + A + A cos(4x cos(4x) sin( sin(ωt − β z ) az 2
The displacement current density is given by ∂ D = ∂ t
2
∇ × H = (4A (4A + β A ) sin(4 sin(4x x)cos(ωt − β z ) ay A/m 2
1
9.13. Consider
the region defined by |x|, |y |, and |z | < 1. Let r = 5, µr = 4, and σ = 0. If Jd = 20 cos(1 cos(1..5 108 t bx) bx)ay µA/m2 ;
×
a) find
D
−
and E: Since
Jd = ∂ D/∂ t,
D =
we write
20 10 6 sin(1. sin(1.5 108 bx) bx)ay 8 1.5 10 10 13 sin(1. sin(1.5 108 t bx) bx)ay C/m2
× × ×
Jd dt + dt + C C =
= 1. 1 .33
−
×
−
×
−
−
where the integration constant is set to zero (assuming no dc fields are present). Then 1.33 10 13 E = = sin(1. sin(1.5 (5 8.85 10 12 ) D
= 3. 3 .0
×
3
−
× 10
×
−
−
×
8
× 10 t − bx) bx)ay
8
sin(1. sin(1.5
× 10 t − bx) bx)ay V/m
b) use the point form of Faraday’s araday’s law and an integration with respect to time to find H: In this case, Ey ∇ × E = ∂ ∂ E az = −b(3. (3.0 × 10 x
3
−
Solve for B =
Now
B by
b(3. (3.0 1.5
∂ B × 10 t − bx) bx)az = − ∂ t 8
integrating over time:
× 10 × 10
3
−
8
) sin(1. sin(1.5
8
B
11
−
× 10 t − bx) bx)az = (2. (2.0)b 0)b × 10
(2. (2.0)b 0)b 10 11 sin(1. sin(1.5 µ 4 4π 10 7 = (4. (4 .0 10 6 )b sin(1. sin(1.5 108 t
H =
−
× × ×
=
−
−
×
c) use case
)cos(1. )cos(1.5
B and
∇ × H = Jd + J to find Jd:
×
sin(1. sin(1.5
8
× 10 t − bx) bx)az − bx) bx)az A/m
Since Since σ = 0, there is no conduction current, so in this
Hz ∇ × H = − ∂ ∂ H ay = 4.0 × 10 x
6 2
−
b cos(1. cos(1.5
8
2
× 10 t − bx) bx)ay A/m
d) What is the numerical numerical value value of b? We set the given expression for of part c part c to obtain: 20
6
−
× 10
8
× 10 t − bx) bx)az T
= 4.0
6 2
−
× 10 169
b
⇒
b =
√
Jd equal
1
−
5.0 m
= J d to the result
9.14. A
voltage source, V 0 sin ωt, is connected between two concentric conducting spheres, r = a and r and r = = b b,, b > a, where the region between them is a material for which = r 0 , µ = µ = µ µ0 , and σ = 0. Find the total displacemen displacementt current current through through the dielectri dielectricc and compare compare it with the source source current current as determine determined d from the capacitance capacitance (Sec. 6.3) and circuit analysis analysis methods: methods: First, First, solving solving Laplace’s Laplace’s equation equation,, we find the voltage voltage between between spheres (see Eq. 39, Chapter Chapter 6): (1/r (1/r)) (1/b (1/b)) V (t (t) = V 0 sin ω t (1/a (1/a)) (1/b (1/b))
− −
Then E =
Now
V sin ωt −∇V = r (1/a ar ⇒ (1/a − 1/b) /b) 0
2
Jd =
r 0 V 0 sin ω t
D =
r2 (1/a (1/a
ar − 1/b) /b)
∂ D r 0 ω V 0 cos ω t ar = 2 ∂ t r (1/a (1/a 1/b) /b)
−
The displacement current is then I d = 4π r2 J d =
4π r 0 ωV 0 cos ωt dV = C (1/a (1/a 1/b) /b) dt
−
where, from Eq. 6, Chapter 6, C = 9.15.
4π r 0 (1/a (1/a 1/b) /b)
−
Let µ Let µ = = 3 10 5 H/m, = 1.2 10 10 F/m, and σ = 0 everywhere. If H cos(1010 t β x)az H = 2 cos(10 A/m, use Maxwell’s equations to obtain expressions for B, D, E, and β : First, First, B = µH = 5 10 6 10 cos(10 t β x)az T. Next we use −
×
−
×
−
×
−
−
∂ D ∇ × H = − ∂ ∂ Hx a y = 2 β sin(10 sin(10 t − β x)ay = ∂ t 10
from which D =
2β sin(10 sin(1010 t
2β − β x) dt + dt + C C = − cos(10 t − β x)ay C/m 10 10
2
10
where the integration constant is set to zero, since no dc fields are presumed to exist. Next, E =
D
=
2β − (1. (1.2 × 10 )(10 10
10
−
Now
)
cos(1010 t
10
− β x)ay = −1.67β cos(10 cos(10 t − β x)ay V/m
∂ B Ey ∇ × E = ∂ ∂ E az = 1. 1 .67β sin(10 t − β x)az = − ∂ t x 2
So
−
10
)β 2 cos(1010 t
We require this result to be consistent with the expression for
B originally
B =
1.67β 2 sin(1010 t
10
(1. (1.67
10
−
× 10
−
− β x)az dt = dt = (1. (1.67 × 10
)β 2 = 6
5
−
× 10 170
⇒
β = ± = ±600 600 rad/ rad/m
− β x)az found. So
9.16. Derive
the continuity continuity equation from Maxwell’s equations: First, take the divergence of both sides of Ampere’s circuital law:
∇ · ∇ × H = ∇ · J + ∂ ∂ t ∇ · D = ∇ · J + ∂ρ∂ tv = 0
0
where we have used
∇ · D = ρv , another Maxwell equation.
9.17. The
electric field intensity in the region 0 < x < 5, 0 < y < π/12, 0 < z < 0.06 m in free E space is given by E = C sin(12y sin(12y ) sin( sin(az az)) cos(2 cos(2 1010 t) ax V/m. Beginnin Beginning g with the relationship, use Maxwell’s equations to find a numerical value for a, if it is known that a is greater than zero: In this case we find
×
∇ ×
∂ E Ex Ez ∇ × E = ∂ ∂ E ay − az ∂ y z
= C [a [a sin(12y sin(12y)cos(az )cos(az))ay
10
− 12 cos(12 cos(12y ) sin( sin(az az))az ] cos(2 cos(2 × 10
t) =
− ∂ ∂ Bt
Then H =
=
−
1 µ0
∇ ×
E dt + dt + C C 1
− µ (2 C × 10
10
0
[a sin(12y sin(12y)cos(az )cos(az))ay
10
− 12 cos(12 cos(12y ) sin( sin(az az))az ] sin(2 sin(2 × 10
t) A/m
where the integration constant, C constant, C 1 = 0, since there are no initial conditions. Using this result, we now find
∇ × H =
∂ H Hz ∂ y
=
∂ H Hy ∂ z
−
2
ax =
(144 + a + a ) − µC (144 sin(12y sin(12y) sin( sin(az az)) sin(2 sin(2 × 10 (2 × 10 )
10
10
0
t) ax =
∂ D ∂ t
Now E =
D
0
1 0
∇×
C (144 (144 + a + a2 ) dt + C C 2 = sin(12y sin(12y ) sin( sin(az az)) cos(2 cos(2 H dt + µ0 0 (2 1010 )2
×
10
× 10
t) ax
where C 2 = 0. This field must must be the same as the original original field field as stated, stated, and so we require require that C (144 (144 + a + a2 ) =1 µ0 0 (2 1010 )2
×
Using µ0 0 = (3
8
× 10 )
2
−
, we find
(2 1010 )2 a = (3 108 )2
× ×
− 144
171
1/2 1
−
= 66 m
9.18. The
parallel plate transmission line shown in Fig. 9.7 has dimensions b dimensions b = = 4 cm and d and d = = 8 mm, while the medium between plates is characterized by µ by µr = 1, r = 20, and σ = 0. Neglect fields outside outside the dielectri dielectric. c. Given Given the field H = 5 cos(10 cos(109 t β z )ay A/m, use Maxwell’s equations to help find: β > 0: Take a) β , if β ∂ H ∂ E Hy H = ax = 5β sin(10 sin(109 t β z )ax = 20 0 ∂ z ∂ t So β 5β E = sin(109 t β z )ax dt = dt = cos(109 t β z )ax 9 200 (4 10 )0 Then ∂ E β 2 ∂ H Ex 9 E = ay = β ay = µ0 sin(10 t z ) ∂ z ∂ t (4 109 )0 So that
−
∇×
−
−
−
−
H =
−
×
2
−β
9
sin(10 t
9
−
×
∇×
−
−
β 2
cos(109 t
− β z)ax dt = dt = (4 × 10
18
(4 10 )µ0 0 )µ0 0 = 5 cos(10 cos(109 t β z )ay where the last equality is required to maintain consistency. Therefore
×
−
β 2
(4
18
× 10
)µ0 0
=5
⇒
− β z)
1
−
β = = 14. 14.9 m
b) the displacement displacement current density density at z at z = 0: Since σ = 0, we have 9
9
∇ × H = J d = −5β sin(10 sin(10 t − β z ) = −74. 74.5 sin(10 sin(10 t − 14. 14.9z )ax = −74. 74.5 sin(10 sin(10 t)ax A/m at z = 0 9
c) the total displacement displacement current crossing the surface x = 0.5d, 0 < y < b, b, and 0 < 0 < z < 0. 0 .1 m in the ax direction. We evaluate the flux integral of Jd over the given cross section:
0.1
I d = 9.19. In
−
1 2
−74. 74.5b
sin(109 t
0
− 14. 14.9z ) ax · ax dz = dz = 0.20
cos(109 t
9
− 1.49) − cos(10 t)
the first sectio section n of this this chap chapter ter,, Farada araday’s y’s law law was was used used to show show that that the field kt kt kB0 ρe aφ results from the changing magnetic field B = B = B 0 e az .
A E
=
a) Show that that these fields do not not satisfy Maxwell’s Maxwell’s other curl equation: equation: Note that B as stated is constan constantt with position, position, and so will have have zero curl. The electric field, howeve however, r, varies ∂ D H = ∂ t would have with time, and so have a zero left-hand side and a non-zero right-hand side. The equation is thus not valid with these fields.
∇×
b) If we let B0 = 1 T and k = 106 s 1 , we are establishing a fairly large magnetic flux H equation to show that the rate at which B density in 1 µ 1 µs. s. Use the which B z should (but 6 does not) change with ρ is only about 5 10 T/m in free space at t = 0: Assumi Assuming ng that B varies with ρ , we write ∂ H ∂ E Hz 1 dB0 kt 1 2 H = aφ = 0 k B0 ρekt e = 0 = ∂ρ ∂ t µ0 dρ 2 Thus dB0 1 1012 (1)ρ = µ0 0 k2 ρB0 = = 5.6 10 6 ρ dρ 2 2(3 108 )2 which is near the stated value if ρ is on the order of 1m. −
∇ ×
∇×
−
×
−
−
−
×
172
×
−
Maxwell’s equations in point form, assume that all fields vary as est and write the equation equationss without without explicitly explicitly involvin involvingg time: Write all fields fields in the general form A(r, t) = st A0 (r)e , where r is a position vector in any coordinate system. Maxwell’s equations become:
9.20. Given
∇ × E (r) est = − ∂ ∂ t 0
∇ × H (r) est = J (r)est + ∂ ∂ t 0
0
st B0 (r) e
st D0 (r) e
=
−sB (r) est 0
= J 0 (r)est + sD0 (r) est
∇ · D (r) est = ρ (r) est 0
0
∇ · B (r) est = 0 0
In all cases, the est terms divide out, leaving:
∇ × E (r) = −sB (r) ∇ × H (r) = J (r) + s + sD (r) ∇ · D (r) = ρ (r) ∇ · B (r) = 0 0
0
0
0
0
0
0
0
9.21.
a) Show that under under static field condition conditions, s, Eq. (55) reduces reduces to Ampere’s Ampere’s circuital circuital law. First use the definition of the vector Laplacian: 2
∇ A = −∇×∇× A + ∇(∇ · A) = −µJ which is Eq. (55) with the time derivative set to zero. We also note that A, (55) becomes state (from Eq. (54)). Now, since B =
∇× −∇ × B = −µJ ⇒ ∇ × H = J
∇ · A = 0 in steady
b) Show Show that Eq. (51) becomes Farad Faraday ay’s ’s law when taking the curl: Doing Doing this gives
∇ × E = −∇×∇V − − ∂ ∂ t ∇ × A The curl of the gradient is identially zero, and
∇ × A = B. We are left with ∇ × E = −∂ B/∂ t
173
9.22. In
a sourceless medium, in which J = 0 and ρ v = 0, assume a rectangular coordinate system in which E and H are functions only of z and z and t t.. The medium has permittivity and permeability permeability µ. a) If E = E x ax and H = H y ay , begin with Maxwell’s equations and determine the second order partial diff erential erential equation that E x must satisfy. First use ∂ B ∂ E ∂ H Ex Hy E = ay = µ ay ∂ t ∂ z ∂ t in which case, the curl has dictated the direction that H must must lie in. Similarly Similarly,, use the other Maxwell curl equation to find
∇×
−
⇒
−
∂ E Hy Ex ∇ × H = ∂ ∂ Dt ⇒ − ∂ ∂ H ax = ax ∂ t z
Now, Now, diff erentiate erentiate the first equation with respect to z , and the second equation with respect to t to t:: ∂ 2 E x ∂ 2 H y ∂ 2 H y ∂ 2 E x = µ and = ∂ z 2 ∂ t∂ z ∂ z ∂ t ∂ t2 Combining these two, we find ∂ 2 E x ∂ 2 E x = µ 2 ∂ z 2 ∂ t
−
b) Show that E that E x = E = E 0 cos(ωt Substituting, we find ∂ 2 E x = ∂ z 2
2
−
− β z) is a solution of that equation for a particular value of β :
−β E cos(ωt − β z) 0
and µ
∂ 2 E x = ∂ t2
2
−ω µE cos(ωt − β z) 0
These two will be equal provided the constant multipliers of cos( ωt
− β z) are equal.
c) Find β as a function of given parameters. Equating the two constants in part b, b , we find β = = ω µ.
√
174
region 1, z < 0, 1 = 2 10 11 F/m, µ1 = 2 10 6 H/m, and σ1 = 4 10 3 S/m; in region 2, z > 0, 2 = 1 /2, µ2 = 2µ1 , and σ2 = σ 1 /4. It is known known that E1 = (30ax + 20 ay + 9 10az ) cos(10 cos(10 t) V/m at P at P 1 (0, (0, 0, 0 ).
9.23. In
−
×
×
−
×
−
−
a) Find
EN 1 , Et1 , DN 1 ,
and
DN 1 = 1 EN 1 =
b) Find
JN 1
(2
c) Find
× 10
11
−
× 10
11
−
(2
Et1 =
(30ax + 20 ay ) cos(10 cos(109 t) V/m
)(10)cos(1 )(10) cos(1009 t)az C/m2 = 200cos( 200 cos(10 109 t)az pC/ pC/m2
)(30ax + 20 ay )cos(109 t) = (600ax + 400ay ) cos(10 cos(109 t) pC/ pC/m2
and Jt1 at P 1 : JN 1 = σ 1 EN 1
Jt1
These will be
10 cos(10 cos(109 t)az V/m
EN 1 =
Dt1 = 1 Et1 =
Dt1 :
= (4
× 10
Et2 , Dt2 ,
Jt2
and
3
−
= σ 1 Et1 = (4
3
−
× 10
)(10cos(1 )(10 cos(1009 t))az = 40 cos(10 cos(109 t)az mA mA//m2
)(30ax + 20ay ) cos(10 cos(109 t) = (120ax + 80 ay ) cos(10 cos(109 t) mA mA//m2
at P at P 1 : By continuity of tangential Et2 = E t1 =
E,
(30ax + 20 ay ) cos(10 cos(109 t) V/m
Then Dt2
= 2 Et2 = (10
Jt2 = σ 2 Et2 =
11
−
(10
)(30ax + 20 ay ) cos(10 cos(109 t) = (300ax + 200ay ) cos(10 cos(109 t) pC/ pC/m2
3
−
)(30ax + 20ay ) cos(10 cos(109 t) = (30ax + 20ay ) cos(10 cos(109 t) mA mA//m2
d) (Harder) Use the continuity continuity equation equation to help show that J that J N N 1 J N N 2 = ∂ DN 2 /∂ t ∂ DN 1 /∂ t and then determine EN 2 , DN 2 , and JN 2 : We assume the existence of a surface surface charge 2 layer at the boundary having density ρs C/m . If we draw a cylindrical “pillbox” whose top and bottom surfaces (each of area ∆a) are on either side of the interface, we may use the continuity condition to write
−
− J N N )∆a = − ∂ρ∂ ts ∆a
(J N N 2 where ρs = D = DN 2
−
1
− DN . Therefore, 1
J N N 1
− J N N = ∂ ∂ t (DN − DN ) 2
2
1
In terms of the normal electric field components, this becomes σ1 E N N 1
− σ E N N 2
2
=
∂ (2 E N N 2 ∂ t
− E N N ) 1
1
Now let E N = A cos(109 t) + B + B sin(10 sin(109 t), while from before, E before, E N cos(109 t). N 2 = A N 1 = 10 cos(10
175
9.23d
(continued) These, along with the permittivities and conductivities, are substituted to obtain (4
3
−
× 10
=
∂ 10 ∂ t
=
−(10
)(10) )(10) cos(10 cos(109 t) 11
−
2
−
3
−
− 10
[A cos(109 t) + B + B sin(10 sin(109 t)]
[A cos(109 t) + B + B sin(10 sin(109 t)]
A sin(109 t) + 10
2
−
11
−
− (2 × 10 B cos(10 t) + (2 × 10 9
)(10) )(10) cos(10 cos(109 t)
1
−
) sin(10 sin(109 t)
We now equate coefficients of the sin and cos terms to obtain two equations: 2
−
4
3
A = 10
2
A + 2
−
× 10 − 10 −10 B = −10 3
−
−
2
−
B 1
−
× 10
These are solved together to find A find A = = 20. 20.2 and B = 2.0. Thus EN 2 =
20. 20.2 cos(10 cos(109 t) + 2. 2.0 sin(10 sin(109 t)
az
= 20. 20.3 cos(10 cos(109 t + 5. 5.6 )az V/m ◦
Then DN 2 = 2 EN 2 =
203 cos(10 cos(109 t + 5. 5.6 )az pC/ pC/m2 ◦
and JN 2 = σ 2 EN 2
= 20. 20 .3 cos(10 cos(109 t + 5. 5.6 )az mA mA//m2 ◦
176
9.24. A
vector potential is given as A = A 0 cos(ω t possible are zero, find H, E, and V and V ; ; With
A
− kz) kz ) ay .
a) Assuming Assuming as many many components components as
y-directed only, and varying spatially only with z, z , we find H =
1 µ
∇ × A = − µ1 ∂ ∂ Azy ax = − kAµ sin(ωt − kz) kz ) ax A/m 0
Now, in a lossle Now, lossless ss mediu medium m we will have have zero zero condu conducti ctivit vity y, so that that the point point form form of Ampere’s circuital law involves only the displacement current term:
∇ × H = ∂ ∂ Dt = ∂ ∂ Et Using the magnetic field as found above, we find
∇×
∂ H Hx H = ∂ z
ay
k2 A0 = cos(ωt µ
− kz) kz ) ay
∂ E = ∂ t
⇒
Now, E =
−∇V − −
or
∂ A ∂ t
∇V = A ω 1 − 0
⇒ ∇V = −
k2 sin(ωt ω 2 µ
k2 A0 E = sin(ωt ω µ
− kz) kz ) ay V/m
∂ A +E ∂ t
∂ V V − kz) ay kz ) ay = ∂ y
Integrating over y we find
V = A0 ω y 1
−
k2 sin(ωt ω 2 µ
− kz) kz ) + C + C
where C where C ,, the integration constant, can be taken as zero. In part b, b , it will be shown that k = ω µ, which means that V that V = 0.
√
b) Specify k in terms of A0 , ω, and the constants of the lossless medium, and µ. Use the other Maxwell curl equation:
∇ × E = − ∂ ∂ Bt = −µ ∂ ∂ Ht so that
∂ H = ∂ t
3
Ey k A − µ1 ∇ × E = µ1 ∂ ∂ E ax = − cos(ωt − kz) kz ) ax ωµ z 0
2
Integrate over t (and set the integration constant to zero) and require the result to be consistant with part a: a : 3
H =
kA − ωk µA sin(ωt − kz) kz ) ax = − sin(ω t − kz) kz ) ax µ 2
0 2
0
We identify
√
k = ω µ 177
from part a
9.25. In
a region where µ where µ r = r = 1 and σ = 0, the retarded potentials are given by V = x( x (z V and A = x = x[( [(z/c z/c)) t]az Wb/m, where c where c = = 1/ µ0 0 .
√
−
− ct) ct)
a) Show Show that
∇ · A = −µ(∂ V /∂ t):
First,
∇ · A = ∂ ∂ Azz = xc = x√ µ
0 0
Second,
∂ V V = ∂ t
x −cx = cx = − √ µ
0 0
so we observe that · A = µ0 0 (∂ V /∂ t) in free space, implying that the given statement would hold true in general media.
∇
b) Find
B, H, E,
and
−
D:
Use B =
∇ × A = −
Then H =
Now, E =
B
µ0
=
− −
∂ Ax ay = t ∂ x
1 t µ0
z c
z c
ay
ay
T
A/m
−∇V − − ∂ ∂ At = −(z − ct) ct)ax − xaz + x + xaz = (ct ( ct − z )ax V/m
Then D = 0 E = 0 (ct
2
− z)ax C/m
c) Show Show that these results satisfy Maxwell’s equations if J and ρ v are zero: i. · D = · 0 (ct z )ax = 0
∇ ∇ − ii. ∇ · B = ∇ · (t (t − z/c) z/c)ay = 0
iii.
∇ × H = −
∂ H Hy 1 ax = ax = ∂ z µ0 c
which we require to equal ∂ D/∂ t: ∂ D = 0 cax = ∂ t
iv.
0
µ0
ax
Ex ∇ × E = ∂ ∂ E ay = −ay z which we require to equal
−∂ B/∂ t: ∂ B = a y ∂ t
So all four Maxwell equations are satisfied. 178
0
µ0
ax
9.26. Write
Maxwell’s equations in point form in terms of E and H as they apply to a sourceless medium, where J and ρv are both both zero. zero. Replac Replacee by µ, µ by , E by H, and H by E, and show show that the equations equations are unchang unchanged. ed. This is a more general general expressio expression n of the duality principle in in circuit theory.
−
Maxwell’s equations in sourceless media can be written as:
∇ × E = −µ ∂ ∂ Ht ∇ × H = ∂ ∂ Et ∇ · E = 0 ∇ · µH = 0
(1) (2) (3) (4)
In making the above substitutions, we find that (1) converts to (2), (2) converts to (1), and (3) and (4) convert to each other.
179