Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig....

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Transcript of Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig....

Page 1: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary
Page 2: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary
Page 3: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary
Page 4: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary
Page 5: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 4.48 With reference to Fig. 4-19, findE1 if E2 = x3− y2+ z2 (V/m),ε1 = 2ε0, ε2 = 18ε0, and the boundary has a surface charge densityρs = 3.54×10−11 (C/m2). What angle doesE2 make with thez-axis?

Solution: We know thatE1t = E2t for any 2 media. Hence,E1t = E2t = x3− y2.Also, (D1−D2) · n = ρs (from Table 4.3). Hence,ε1(E1 · n)− ε2(E2 · n) = ρs, whichgives

E1z =ρs+ ε2E2z

ε1=

3.54×10−11

2ε0+

18(2)

2=

3.54×10−11

2×8.85×10−12 +18= 20 (V/m).

Hence,E1 = x3− y2+ z20 (V/m). Finding the angleE2 makes with thez-axis:

E2 · z = |E2|cosθ , 2 =√

9+4+4cosθ , θ = cos−1(

2√17

)

= 61◦.

Page 6: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel,conducting plates separated by a distanced. The space between the plates containstwo adjacent dielectrics, one with permittivityε1 and surface areaA1 and anotherwith ε2 andA2. The objective of this problem is to show that the capacitanceC of theconfiguration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, asillustrated in Fig. P4.56(b), with

C = C1 +C2 (19)

where

C1 =ε1A1

d(20)

C2 =ε2A2

d(21)

To this end, proceed as follows:

(a) Find the electric fieldsE1 andE2 in the two dielectric layers.

(b) Calculate the energy stored in each section and use the result to calculateC1

andC2.

(c) Use the total energy stored in the capacitor to obtain an expression forC. Showthat (19) is indeed a valid result.

(a)

(b)

ε1

A1 A2

ε2d

+

−V

C1 C2V

+

Figure P4.56: (a) Capacitor with parallel dielectricsection, and (b) equivalent circuit.

Page 7: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Solution:

V+

-

(c)

E1 E2

ε2

d

ε1

Figure P4.56: (c) Electric field inside of capacitor.

(a) Within each dielectric section,E will point from the plate with positive voltageto the plate with negative voltage, as shown in Fig. P4-56(c). FromV = Ed,

E1 = E2 =Vd

.

(b)

We1 =12

ε1E21 · V =

12

ε1V2

d2 ·A1d =12

ε1V2A1

d.

But, from Eq. (4.121),

We1 =12

C1V2.

HenceC1 = ε1A1

d. Similarly,C2 = ε2

A2

d.

(c) Total energy is

We = We1 +We2 =12

V2

d(ε1A1 + ε2A2) =

12

CV2.

Hence,

C =ε1A1

d+

ε2A2

d= C1 +C2.

Page 8: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.12 Two infinitely long, parallel wires are carrying 6-A currents inopposite directions. Determine the magnetic flux density at pointP in Fig. P5.12.

I2 = 6 AI1 = 6 A

0.5 m

2 m

P

Figure P5.12: Arrangement for Problem 5.12.

Solution:

B = φφφµ0I1

2π(0.5)+φφφ

µ0I22π(1.5)

= φφφµ0

π(6+2) = φφφ

8µ0

π(T).

Page 9: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.14 Two parallel, circular loops carrying a current of 40 A each arearranged as shown in Fig. P5.14. The first loop is situated in thex–y plane with itscenter at the origin, and the second loop’s center is atz = 2 m. If the two loops havethe same radiusa = 3 m, determine the magnetic field at:

(a) z = 0

(b) z = 1 m

(c) z = 2 m

z = 2 m

0

z

y

x

a

a

I

I

Figure P5.14: Parallel circular loops of Problem 5.14.

Solution: The magnetic field due to a circular loop is given by (5.34) for a loop inthe x–y plane carrying a currentI in the+φφφ-direction. Considering that the bottomloop in Fig. is in thex–y plane, but the current direction is along−φφφ,

H1 = −zIa2

2(a2 + z2)3/2,

wherez is the observation point along thez-axis. For the second loop, which is at aheight of 2 m, we can use the same expression butz should be replaced with(z−2).Hence,

H2 = −zIa2

2[a2 +(z−2)2]3/2.

The total field is

H = H1 +H2 = −zIa2

2

[

1

(a2 + z2)3/2+

1

[a2 +(z−2)2]3/2

]

A/m.

Page 10: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

(a) At z = 0, and witha = 3 m andI = 40 A,

H = −z40×9

2

[

133 +

1

(9+4)3/2

]

= −z10.5 A/m.

(b) At z = 1 m (midway between the loops):

H = −z40×9

2

[

1

(9+1)3/2+

1

(9+1)3/2

]

= −z11.38 A/m.

(c) At z = 2 m,H should be the same as atz = 0. Thus,

H = −z10.5 A/m.

Page 11: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 4.49 An infinitely long conducting cylinder of radiusa has a surfacecharge densityρs. The cylinder is surrounded by a dielectric medium withεr = 4and contains no free charges. The tangential component of the electric field in theregionr ≥ a is given byEt = −φφφcosφ/r2. Since a static conductor cannot have anytangential field, this must be cancelled by an externally applied electric field. Findthe surface charge density on the conductor.

Solution: Let the conducting cylinder be medium 1 and the surrounding dielectricmedium be medium 2. In medium 2,

E2 = rEr − φφφ1r2 cosφ ,

with Er, the normal component ofE2, unknown. The surface charge density is relatedto Er. To findEr, we invoke Gauss’s law in medium 2:

∇ ·D2 = 0,

or1r

∂∂ r

(rEr)+1r

∂∂φ

(

− 1r2 cosφ

)

= 0,

which leads to∂∂ r

(rEr) =∂

∂φ

(

1r2 cosφ

)

= − 1r2 sinφ .

Integrating both sides with respect tor,

∫ ∂∂ r

(rEr) dr = −sinφ∫

1r2 dr

rEr =1r

sinφ ,

or

Er =1r2 sinφ .

Hence,

E2 = r1r2 sinφ .

According to Eq. (4.93),n2 · (D1−D2) = ρs,

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wheren2 is the normal to the boundary and points away from medium 1. Hence,n2 = r . Also,D1 = 0 because the cylinder is a conductor. Consequently,

ρs = −r ·D2|r=a

= −r · ε2E2|r=a

= −r · εrε0

[

r1r2 sinφ

]∣

r=a

= −4ε0

a2 sinφ (C/m2).

Page 13: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 4.50 If E = R150 (V/m) at the surface of a 5-cm conducting spherecentered at the origin, what is the total chargeQ on the sphere’s surface?

Solution: From Table 4-3,n · (D1−D2) = ρs. E2 inside the sphere is zero, since weassume it is a perfect conductor. Hence, for a sphere with surface areaS= 4πa2,

D1R = ρs, E1R =ρs

ε0=

QSε0

,

Q = ERSε0 = (150)4π(0.05)2ε0 =3πε0

2(C).

Page 14: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectriclayers. Use energy considerations to show that the equivalent capacitance of theoverall capacitor,C, is equal to the series combination of the capacitances of theindividual layers,C1 andC2, namely

C =C1C2

C1 +C2(22)

where

C1 = ε1Ad1

, C2 = ε2Ad2

(a) Let V1 andV2 be the electric potentials across the upper and lower dielectrics,respectively. What are the corresponding electric fieldsE1 and E2? Byapplying the appropriate boundary condition at the interface between the twodielectrics, obtain explicit expressions forE1 andE2 in terms ofε1, ε2, V, andthe indicated dimensions of the capacitor.

(b) Calculate the energy stored in each of the dielectric layers and then use the sumto obtain an expression forC.

(c) Show thatC is given by Eq. (22).

(a)

(b)

V+

C1

C2

+

d1

d2 V

A

ε1

ε2

Figure P4.58: (a) Capacitor with parallel dielectriclayers, and (b) equivalent circuit (Problem 4.58).

Page 15: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Solution:

+

- VE1

E2

ε1

ε1

V1

+

-

V1

+

-

d1

d2

Figure P4.58: (c) Electric fields inside of capacitor.

(a) If V1 is the voltage across the top layer andV2 across the bottom layer, then

V = V1 +V2,

and

E1 =V1

d1, E2 =

V2

d2.

According to boundary conditions, the normal component ofD is continuous acrossthe boundary (in the absence of surface charge). This means that at the interfacebetween the two dielectric layers,

D1n = D2n

orε1E1 = ε2E2.

Hence,

V = E1d1 +E2d2 = E1d1 +ε1E1

ε2d2,

which can be solved forE1:

E1 =V

d1 +ε1

ε2d2

.

Similarly,

E2 =V

d2 +ε2

ε1d1

.

Page 16: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

(b)

We1 =12

ε1E21 · V 1 =

12

ε1

V

d1 +ε1

ε2d2

2

·Ad1 =12

V2[

ε1ε22Ad1

(ε2d1 + ε1d2)2

]

,

We2 =12

ε2E22 · V 2 =

12

ε2

V

d2 +ε2

ε1d1

2

·Ad2 =12

V2[

ε21ε2Ad2

(ε1d2 + ε2d1)2

]

,

We = We1 +We2 =12

V2[

ε1ε22Ad1 + ε2

1ε2Ad2

(ε1d2 + ε2d1)2

]

.

But We = 12 CV2, hence,

C =ε1ε2

2Ad1 + ε21ε2Ad2

(ε2d1 + ε1d2)2 = ε1ε2A(ε2d1 + ε1d2)

(ε2d1 + ε1d2)2 =ε1ε2A

ε2d1 + ε1d2.

(c) Multiplying numerator and denominator of the expression forC by A/d1d2, wehave

C =

ε1Ad1

· ε2Ad2

ε1Ad1

+ε2Ad2

=C1C2

C1 +C2,

where

C1 =ε1Ad1

, C2 =ε2Ad2

.

Page 17: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.9 The loop shown in Fig. P5.9 consists of radial lines and segments ofcircles whose centers are at pointP. Determine the magnetic fieldH at P in terms ofa, b, θ, andI.

θ

b

a P

I

Figure P5.9: Configuration of Problem 5.9.

Solution: From the solution to Example 5-3, if we denote thez-axis as passing outof the page through pointP, the magnetic field pointing out of the page atP due tothe current flowing in the outer arc isHouter= −zIθ/4πb and the field pointing outof the page atP due to the current flowing in the inner arc isH inner = zIθ/4πa. Theother wire segments do not contribute to the magnetic field atP. Therefore, the totalfield flowing directly out of the page atP is

H = Houter+H inner = zIθ4π

(

1a− 1

b

)

= zIθ(b−a)

4πab.

Page 18: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.10 An infinitely long, thin conducting sheet defined over the space0≤ x≤w and−∞≤ y≤∞ is carrying a current with a uniform surface current densityJs = y5 (A/m). Obtain an expression for the magnetic field at pointP = (0,0,z) inCartesian coordinates.

Solution:

z

x

x

R

P = (0, 0, z)

0

|R| = x2 + z2

w

Figure P5.10: Conducting sheet of widthw in x–y plane.

The sheet can be considered to be a large number of infinitely long but narrow wireseachdx wide lying next to each other, with each carrying a currentIx = Js dx. Thewire at a distancex from the origin is at a distance vectorR from pointP, with

R = −xx+ zz.

Equation (5.30) provides an expression for the magnetic field due to an infinitely longwire carrying a currentI as

H =Bµ0

=φφφI2πr

.

We now need to adapt this expression to the present situation by replacingI withIx = Jsdx, replacingr with R = (x2+z2)1/2, as shown in Fig. P5.10, and by assigningthe proper direction for the magnetic field. From the Biot–Savart law, the directionof H is governed byl×××R, wherel is the direction of current flow. In the present case,l is in they direction. Hence, the direction of the field is

l×××R|l×××R| =

y××× (−xx+ zz)|y××× (−xx+ zz)| =

xz+ zx

(x2 + z2)1/2.

Page 19: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Therefore, the fielddH due to the currentIx is

dH =xz+ zx

(x2 + z2)1/2

Ix

2πR=

(xz+ zx)Jsdx2π(x2 + z2)

,

and the total field is

H(0,0,z) =∫ w

x=0(xz+ zx)

Jsdx2π(x2 + z2)

=Js

∫ w

x=0(xz+ zx)

dxx2 + z2

=Js

(

xz∫ w

x=0

dxx2 + z2 + z

∫ w

x=0

x dxx2 + z2

)

=Js

(

xz

(

1z

tan−1(

xz

))∣

w

x=0+ z

(

12 ln(x2 + z2)

)∣

w

x=0

)

=5

[

x2πtan−1(

wz

)

+ z12(ln(w2 + z2)− ln(0+ z2))

]

for z 6= 0,

=5

[

x2πtan−1(

wz

)

+ z12 ln

(

w2 + z2

z2

)]

(A/m) for z 6= 0.

An alternative approach is to employ Eq. (5.24a) directly.

Page 20: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.11 An infinitely long wire carrying a 25-A current in the positivex-direction is placed along thex-axis in the vicinity of a 20-turn circular loop locatedin thex–y plane (Fig. P5.11). If the magnetic field at the center of the loop is zero,what is the direction and magnitude of the current flowing in the loop?

1 m

d = 2 m

I1

x

Figure P5.11: Circular loop next to a linear current(Problem 5.11).

Solution: From Eq. (5.30), the magnetic flux density at the center of the loop due to

I2

Figure P5.11: (b) Direction ofI2.

the wire isB1 = z

µ0

2πdI1

wherez is out of the page. Since the net field is zero at the center of the loop,I2 mustbe clockwise, as seen from above, in order to opposeI1. The field due toI2 is, fromEq. (5.35),

B = µ0H = −zµ0NI2

2a.

Equating the magnitudes of the two fields, we obtain the result

NI22a

=I1

2πd,

or

I2 =2aI1

2πNd=

1×25π×20×2

= 0.2 A.

Page 21: Untitled OmniPage Documentece381/SECURE/HW_Solutions/... · Problem 4.48 With reference to Fig. 4-19, find E1 if E2 =xˆ3−yˆ2+ˆz2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary

Problem 5.13 A long, East-West–oriented power cable carrying an unknowncurrentI is at a height of 8 m above the Earth’s surface. If the magnetic flux densityrecorded by a magnetic-field meter placed at the surface is 15µT when the current isflowing through the cable and 20µT when the current is zero, what is the magnitudeof I?

Solution: The power cable is producing a magnetic flux density that opposes Earth’s,own magnetic field. An East–West cable would produce a field whose direction atthe surface is along North–South. The flux density due to the cable is

B = (20−15) µT = 5µT.

As a magnet, the Earth’s field lines are directed from the South Pole to the NorthPole inside the Earth and the opposite on the surface. Thus the lines at the surface arefrom North to South, which means that the field created by the cable is from Southto North. Hence, by the right-hand rule, the current direction is toward theEast. Itsmagnitude is obtained from

5 µT = 5×10−6 =µ0I2πd

=4π×10−7I

2π×8,

which givesI = 200 A.