Untitled OmniPage Documentece381/SECURE/HW...x θ x F31' F32' I3 I1 I2 x Figure P5.19: (b) Forces acting on I3. equidistant from I3, our intuitive answer might be that the net force

Problem 5.19 Three long, parallel wires are arranged as shown in Fig. P5.19.Determine the force per unit length acting on the wire carryingI3.

I1 = 10 A

I2 = 10 A

I3 = 10 A

2 m

2 m

2 m

Figure P5.19: Three parallel wires of Problem 5.19.

Solution: Since I1 and I2 are equal in magnitude and opposite in direction, and

x

x

2 m

2 m

R1

B2 R2

R1B1

R2 = R1

x

z

into the page (y)

out of the page (-y)

I1

I2

I3

Figure P5.19: (a)B fields due toI1 andI2 at location ofI3.

x

xθ

F31'

F32'

I3

I1

I2

x

Figure P5.19: (b) Forces acting onI3.

equidistant fromI3, our intuitive answer might be that the net force onI3 is zero.As we will see, that’s not the correct answer. The field due toI1 (which is alongy) atlocation ofI3 is

B1 = b1µ0I12πR1

where b1 is the unit vector in the direction ofB1 shown in the figure, which isperpendicular toR1. The force per unit length exerted onI3 is

F′31 =

µ0I1I32πR1

(y××× b1) = −R1µ0I1I32πR1

.

Similarly, the force per unit length excited onI3 by the field due toI2 (which isalong−y) is

F′32 = R2

µ0I2I32πR2

.

The two forces have opposite components alongx and equal components alongz.Hence, withR1 = R2 =

√8 m andθ = sin−1(2/

√8) = sin−1(1/

√2) = 45◦,

F′3 = F′

31+F′32 = z

(

µ0I1I32πR1

+µ0I2I32πR2

)

sinθ

= z2

(

4π×10−7×10×20

2π×√

8

)

× 1√2

= z2×10−5 N/m.

Problem 5.21 CurrentI flows along the positivez-direction in the inner conductorof a long coaxial cable and returns through the outer conductor. The inner conductorhas radiusa, and the inner and outer radii of the outer conductor areb and c,respectively.

(a) Determine the magnetic field in each of the following regions: 0≤ r ≤ a,a ≤ r ≤ b, b ≤ r ≤ c, andr ≥ c.

(b) Plot the magnitude ofH as a function ofr over the range fromr = 0 tor = 10 cm, given thatI = 10 A, a = 2 cm,b = 4 cm, andc = 5 cm.

Solution:(a) Following the solution to Example 5-5, the magnetic field in the regionr < a,

H = φφφrI

2πa2 ,

and in the regiona < r < b,

H = φφφI

2πr.

The total area of the outer conductor isA = π(c2−b2) and the fraction of the areaof the outer conductor enclosed by a circular contour centered atr = 0 in the regionb < r < c is

π(r2−b2)

π(c2−b2)=

r2−b2

c2−b2 .

The total current enclosed by a contour of radiusr is therefore

Ienclosed= I

(

1− r2−b2

c2−b2

)

= Ic2− r2

c2−b2 ,

and the resulting magnetic field is

H = φφφIenclosed

2πr= φφφ

I2πr

(

c2− r2

c2−b2

)

.

For r > c, the total enclosed current is zero: the total current flowing on the innerconductor is equal to the total current flowing on the outer conductor, but they areflowing in opposite directions. Therefore,H = 0.

(b) See Fig. P5.21.

0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Mag

netic

field

mag

nitu

deH

(A/c

m)

Radial distance r (cm)

Figure P5.21:Problem 5.21.

Problem 5.28 A uniform current density given by

J = zJ0 (A/m2)

gives rise to a vector magnetic potential

A = −zµ0J0

4(x2 + y2) (Wb/m)

(a) Apply the vector Poisson’s equation to confirm the above statement.

(b) Use the expression forA to findH.

(c) Use the expression forJ in conjunction with Ampere’s law to findH. Compareyour result with that obtained in part (b).

Solution:(a)

∇ 2A = x ∇ 2Ax + y ∇ 2Ay + z∇ 2Az = z(

∂ 2

∂x2 +∂ 2

∂y2 +∂ 2

∂ z2

)[

−µ0J0

4(x2 + y2)

]

= −zµ0J0

4(2+2) = −zµ0J0.

Hence,∇ 2A = −µ0J is verified.(b)

H =1µ0

∇ ×××A =1µ0

[

x(

∂Az

∂y− ∂Ay

∂ z

)

+ y(

∂Ax

∂ z− ∂Az

∂x

)

+ z(

∂Ay

∂x− ∂Ax

∂y

)]

=1µ0

(

x∂Az

∂y− y

∂Az

∂x

)

=1µ0

[

x∂∂y

(

−µ0J0

4(x2 + y2)

)

− y∂∂x

(

−µ0J0

4(x2 + y2)

)]

= −xJ0y2

+ yJ0x2

(A/m).

(c)

n

∫

CH ·dl = I =

∫

SJ ·ds,

φφφHφ ·φφφ2πr = J0 ·πr2,

H = φφφHφ = φφφJ0r2

.

r

J0

z

Figure P5.28:Current cylinder of Problem 5.28.

We need to convert the expression from cylindrical to Cartesian coordinates. FromTable 3-2,

φφφ = −xsinφ+ ycosφ = −xy

√

x2 + y2+ y

x√

x2 + y2,

r =√

x2 + y2 .

Hence

H =

(

−xy

√

x2 + y2+ y

x√

x2 + y2

)

· J0

2

√

x2 + y2 = −xyJ0

2+ y

xJ0

2,

which is identical with the result of part (b).

Problem 5.33 Given that a current sheet with surface current densityJs = x8 (A/m)exists aty = 0, the interface between two magnetic media, andH1 = z11 (A/m) inmedium 1(y > 0), determineH2 in medium 2(y < 0).

Solution:

y

H1

Js

H2

n2

x

Figure P5.33:Adjacent magnetic media withJs on boundary.

Js = x8 A/m,

H1 = z11 A/m.

H1 is tangential to the boundary, and thereforeH2 is also. Withn2 = y, from Eq.(5.84), we have

n2××× (H1−H2) = Js,

y××× (z11−H2) = x8,

x11− y×××H2 = x8,

ory×××H2 = x3,

which implies thatH2 does not have anx-component. Also, sinceµ1H1y = µ2H2y

andH1 does not have ay-component, it follows thatH2 does not have ay-componenteither. Consequently, we conclude that

H2 = z3.

Problem 5.20 A square loop placed as shown in Fig. P5.20 has 2-m sides andcarries a currentI1 = 5 A. If a straight, long conductor carrying a currentI2 = 10 A isintroduced and placed just above the midpoints of two of the loop’s sides, determinethe net force acting on the loop.

z

x

ya

a

1 3

4

2

I1

I2

Figure P5.20: Long wire carrying currentI2, just abovea square loop carryingI1 (Problem 5.20).

Solution: SinceI2 is just barely above the loop, we can treat it as if it’s in the sameplane as the loop. For side 1,I1 andI2 are in the same direction, hence the force onside 1 is attractive. That is,

F1 = yµ0I1I2a2π(a/2)

= y4π×10−7×5×10×2

2π×1= y2×10−5 N.

I1 andI2 are in opposite directions for side 3. Hence, the force on side 3 is repulsive,which means it is also alongy. That is,F3 = F1.

The net forces on sides 2 and 4 are zero. Total net force on the loop is

F = 2F1 = y4×10−5 N.

Problem 5.22 A long cylindrical conductor whose axis is coincident with thez-axishas a radiusa and carries a current characterized by a current densityJ = zJ0/r,whereJ0 is a constant andr is the radial distance from the cylinder’s axis. Obtain anexpression for the magnetic fieldH for

(a) 0≤ r ≤ a

(b) r > a

Solution: This problem is very similar to Example 5-5.(a) For 0≤ r1 ≤ a, the total current flowing within the contourC1 is

I1 =∫∫

J ·ds=∫ 2π

φ=0

∫ r1

r=0

(

zJ0

r

)

· (zr dr dφ) = 2π∫ r1

r=0J0 dr = 2πr1J0.

Therefore, sinceI1 = 2πr1H1, H1 = J0 within the wire andH1 = φφφJ0.(b) Forr ≥ a, the total current flowing within the contour is the total current flowing

within the wire:

I =∫∫

J ·ds=∫ 2π

φ=0

∫ a

r=0

(

zJ0

r

)

· (zr dr dφ) = 2π∫ a

r=0J0 dr = 2πaJ0.

Therefore, sinceI = 2πrH2, H2 = J0a/r within the wire andH2 = φφφJ0(a/r).

Problem 5.26 With reference to Fig. 5-10:

(a) Derive an expression for the vector magnetic potentialA at a pointP locatedat a distancer from the wire in thex–y plane.

(b) DeriveB from A. Show that your result is identical with the expression givenby Eq. (5.29), which was derived by applying the Biot–Savart law.

Solution:(a) From the text immediately following Eq. (5.65), that equation may take the

form

A =µ4π

∫

ℓ′

IR′ dl′ =

µ0

4π

∫ ℓ/2

z′=−ℓ/2

I√

z′2 + r2zdz′

=µ0

4π

(

zI ln(z′ +√

z′2 + r2))∣

∣

∣

ℓ/2

z′=−ℓ/2

= zµ0I4π

ln

ℓ/2+

√

(ℓ/2)2 + r2

−ℓ/2+

√

(−ℓ/2)2 + r2

= zµ0I4π

ln

(

ℓ+√

ℓ2 +4r2

−ℓ+√

ℓ2 +4r2

)

.

(b) From Eq. (5.53),

B = ∇ ×××A

= ∇ ×××(

zµ0I4π

ln

(

ℓ+√

ℓ2 +4r2

−ℓ+√

ℓ2 +4r2

))

= −φφφµ0I4π

∂∂ r

ln

(

ℓ+√

ℓ2 +4r2

−ℓ+√

ℓ2 +4r2

)

= −φφφµ0I4π

(

−ℓ+√

ℓ2 +4r2

ℓ+√

ℓ2 +4r2

)

∂∂ r

(

ℓ+√

ℓ2 +4r2

−ℓ+√

ℓ2 +4r2

)

= −φφφµ0I4π

(

−ℓ+√

ℓ2 +4r2

ℓ+√

ℓ2 +4r2

)

×(

(−ℓ+√

ℓ2 +4r2) ∂∂ r (ℓ+

√ℓ2 +4r2)− (ℓ+

√ℓ2 +4r2) ∂

∂ r (−ℓ+√

ℓ2 +4r2)

(−ℓ+√

ℓ2 +4r2)2

)

= −φφφµ0I4π

(

(−ℓ+√

ℓ2 +4r2)− (ℓ+√

ℓ2 +4r2)

(−ℓ+√

ℓ2 +4r2)(ℓ+√

ℓ2 +4r2)

)

4r√ℓ2 +4r2

= −φφφµ0I4π

(−2ℓ

4r2

)

4r√ℓ2 +4r2

= φφφµ0Iℓ

2πr√

ℓ2 +4r2(T).

which is the same as Eq. (5.29).

Problem 5.29 A thin current element extending betweenz = −L/2 andz = L/2carries a currentI along+z through a circular cross-section of radiusa.

(a) FindA at a pointP located very far from the origin (assumeR is so much largerthanL that pointP may be considered to be at approximately the same distancefrom every point along the current element).

(b) Determine the correspondingH.

Solution:

z

L/2

-L/2

θ

P

Cross-section

R

πa2

I

Figure P5.29:Current element of lengthL observed at distanceR ≫ L.

(a) Since R ≫ L, we can assume thatP is approximately equidistant from allsegments of the current element. Hence, withR treated as constant, (5.65) gives

A =µ0

4π

∫

V′

JR′ dV

′ =µ0

4πR

∫

V′z

I(πa2)

πa2 dz = zµ0I4πR

∫ L/2

−L/2dz = z

µ0IL4πR

.

(b)

H =1µ0

∇ ×××A

=1µ0

[

x∂Az

∂y− y

∂Az

∂x

]

=1µ0

{

x∂∂y

[

µ0IL4π

(

1√

x2 + y2 + z2

)]

− y∂∂x

[

µ0IL4π

(

1√

x2 + y2 + z2

)]}

=IL4π

[ −xy+ yx

(x2 + y2 + z2)3/2

]

.

Problem 5.30 In the model of the hydrogen atom proposed by Bohr in 1913, theelectron moves around the nucleus at a speed of 2× 106 m/s in a circular orbit ofradius 5×10−11 m. What is the magnitude of the magnetic moment generated by theelectron’s motion?

Solution: From Eq. (5.69), the magnitude of the orbital magnetic moment of anelectron is

|m0| =∣

∣−12eur

∣

∣ = 12 ×1.6×10−19×2×106×5×10−11 = 8×10−24 (A·m2).

Problem 5.32 The x–y plane separates two magnetic media with magneticpermeabilitiesµ1 andµ2 (Fig. P5.32). If there is no surface current at the interfaceand the magnetic field in medium 1 is

H1 = xH1x + yH1y + zH1z

find:

(a) H2

(b) θ1 andθ2

(c) EvaluateH2, θ1, andθ2 for H1x = 2 (A/m), H1y = 0, H1z = 4 (A/m), µ1 = µ0,andµ2 = 4µ0

θ1μ1

μ2

H1

z

x-y plane

Figure P5.32: Adjacent magnetic media(Problem 5.32).

Solution:(a) From (5.80),

µ1H1n = µ2H2n,

and in the absence of surface currents at the interface, (5.85) states

H1t = H2t.

In this case,H1z = H1n, andH1x andH1y are tangential fields. Hence,

µ1H1z = µ2H2z,

H1x = H2x,

H1y = H2y,

andH2 = xH1x + yH1y + z

µ1

µ2H1z.

(b)

H1t =√

H21x +H2

1y ,

tanθ1 =H1t

H1z=

√

H21x +H2

1y

H1z,

tanθ2 =H2t

H2z=

√

H21x +H2

1y

µ1

µ2H1z

=µ2

µ1tanθ1.

(c)

H2 = x2+ z14·4 = x2+ z (A/m),

θ1 = tan−1(

24

)

= 26.56◦,

θ2 = tan−1(

21

)

= 63.44◦.

Problem 5.34 In Fig. P5.34, the plane defined byx− y = 1 separates medium 1 ofpermeabilityµ1 from medium 2 of permeabilityµ2. If no surface current exists onthe boundary and

B1 = x2+ y3 (T)

find B2 and then evaluate your result forµ1 = 5µ2.Hint: Start by deriving the equation for the unit vector normal to the given plane.

y

x(1, 0)

(0, −1)

μ2

Medium 2

Plane x − y = 1

μ1

Medium 1

Figure P5.34: Magnetic media separated by the planex− y = 1 (Problem 5.34).

Solution: We need to findn2. To do so, we start by finding any two vectors in theplanex− y = 1, and to do that, we need three non-collinear points in that plane. Wechoose(0,−1,0), (1,0,0), and(1,0,1).

VectorA1 is from (0,−1,0) to (1,0,0):

A1 = x1+ y1.

VectorA2 is from (1,0,0) to (1,0,1):

A2 = z1.

Hence, if we take the cross productA2×××A1, we end up in a direction normal to thegiven plane, from medium 2 to medium 1,

n2 =A2×××A1

|A2×××A1|=

z1××× (x1+ y1)

|A2×××A1|=

y1− x1√1+1

=y√2− x√

2.

In medium 1, normal component is

B1n = n2 ·B1 =

(

y√2− x√

2

)

· (x2+ y3) =3√2− 2√

2=

1√2

,

B1n = n2B1n =

(

y√2− x√

2

)

· 1√2

=y2− x

2.

Tangential component is

B1t = B1−B1n = (x2+ y3)−(

y2− x

2

)

= x2.5+ y2.5.

Boundary conditions:

B1n = B2n, or B2n =y2− x

2,

H1t = H2t, orB2t

µ2=

B1t

µ1.

Hence,

B2t =µ2

µ1B1t =

µ2

µ1(x2.5+ y2.5).

Finally,

B2 = B2n+B2t =

(

y2− x

2

)

+µ2

µ1(x2.5+ y2.5).

For µ1 = 5µ2,B2 = y (T).