205

y

20 turns

/Frol = IIlIlIIBI,

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l x B is in the downward direction, and I = V IR = 12 V/ 4 Q = 3 A. Therefore,

IBI = IFrol = 0.123 =410 ( T)IIlIll 3 x 0.1 m .

Figure P5A: Hinged rectangular loop of Problem 504.

x

Problem 5.4 The rectangular loop shown in Fig. 5-33 (P5A) consists of 20 closelywrap turns and is hinged along the z-axis. The plane of the loop makes anangle of 30° with the y-axis, and the current in the windings is 0.5 A. Whatis the magnitude of the tC!rque exerted on the loop in the p~sence of a unifonnfield B =Y1.2 T? When viewed from above, is the expected direction of rotationclockwise or counterclockwise?

Solution: The magnetic torque on a loop is given by T = m x B (Eq. (5.20», wherem = iiNIA (Eq. (5.19». For this problem, it is given that! = 0.5 A, N = 20 turns, andA = 0.2 m xOA m =0.08 m2• From the figure, n= -xcos 30° +Ysin30°. Therefor¢,m = nO.8 (A· m2) and T = fiO.S(A· m2) x Yl.2 T = -zO.83 (N·m) .. As the torqJeis negative, the direction of rotation is clockwise, looking from above.

Problem 5.s In a cylindrical coordinate system, a 2-m-long straight wire canyinga current of 5 A in the positive z-direction is located at r = 4 em, cp= Te12, and-1 m < z < 1m.

(a) IfB = rO.2coscp (T). what is the magnetic force acting on the wire?

x

207CHAPTER 5

m = fiN/A = fi20 x 10x (15 x 5) x 10-4 = fi 1.5 (A·m2),

Figure P5.6: Rectangular loop of Problem 5.6.

when -n/2< 4> ~ n/2 and negative over the second half of the circle. Thus, workis provided by the force between cp = nl2 and cp = -n!2 (when rotated in the-+-direction), and work is supplied for the second half of the rotation, resulting in anet work of zero.

(c) The force F is maximum when coscp = 1, or cP= O.

z

(a) If the coil, which carnes a current / = 10 A, is in the presence of a magneticflux density

(f)robJem s.§.) A 20-turn rectangular coiJ with side I = 15 cm and w = 5 cm is placedin the y-z plane as shown in Fig. 5-34 (P5.6).

B = 2 x 10-2 (x+y2) (T),

determine the torque acting on the coil.

(b) At what angle cP is the torque zero?

(c) At what angle cP is the torque maximum? Determine its value.

Solution:

(a) The magnetic field is in direction (x + y2), which makes an angleCPo= tan-I f = 63.43°.

The magnetic moment of the loop is

Thus, when fi is parallel to B, T = O.(c) The torque is a maximum when fi is perpendicular to D, which occurs at

CHAPTER 5

y2

z

cp = 63.43° or - 116.57°.

Figure P5.6: (a) Direction of B.

/x

tancp = 2,

cP = 63.43 ±90° = -26.57° or + 153.43°.

2cos</>- sincp = 0,

Ii = icos$+ysin$,

T=mxD=n1.5x2x 1O-2(i+y2)

= (icos$+ysincp) 1.5 x 2 x 1O-2(i+y2}

= i3 x W-2[2coscp-sin<PJ (N·m).

(b) The torque is zero when

208

where it is the surface normal in accordance with the right-hand rule. Wben the loopis in the negative-y of the y-z plane, II is equal to i, but when the plane of the loop ismoved to an angle $, fi becomes

or

Mathematically, we can obtain the same result by taking the derivative of T andequating it to zero to find the values of cpat which ITI is a maximum. Thus,

aT a ., .acp = acp(3 x 1O--(2cosCP - sm$» = 0

x

209

y

-2sinlj!+cos<p =0,

cp = -26.57° or 153.43°.

CHAP1ER 5

or

which gives tancjl= -~. or

z

Figure P5.7: Problem 5.7.

B} = -z I'J/21try' 4,-2 -+ 12

= -z 41t X 10-7 X 25 x 0.0821t x 0.06 x v4 X 0.062 +0.082 = -z4.62 x 10-5 (T).

For aU segments shown in Fig. P5.7. the combination of the direction of the currentand the right-hand rule gives the direction of the magnetic field as -z direction at theorigin. With r = 6 em and 1 = 8 em,

Section 5-2: Biot-Savart Law

C Problem si) An 8 cm x 12 em rectangular loop of wire is situated in the x-yplane with the center of the loop at the origin and its long sides parallel to the x-axis.The loop has a current of 25 A flowing with clockwise direction (when viewed fromabove). Determine the magnetic field at the center of the loop.

Solution: The total magnetic field is the vector sum of the individuai fieids or eachof the four wire segments: B =B1 + B2 +B3 +B4. An expression for the magneticfield from a wire segment is given by Eg. (5.29).

CHAPTER 5

B4 =-z 10040 X 10-5 (T).

I

92P2(Z2)t" J"",""" "," """ ""

--- "_:::.~ ..•- - - - r - - Per, <P. z)

B3= -z4.62 X 10-5 (T),

Figure P5.8: Current-carrying linear conductor of Problem 5.8.

z

82 = -z f11121trN+-]2

= -z 41t X 10-7 X 25 x 0.1221t x 0.04 x v4 x 0.042+0.122 = -zlOAO x 10-5 (T).

210

For segment 2, r = 4 em and I = 12 em,

Similarly.

The total field is then B = BI + B2+B3 +84 = -ZO.30 (roT).

Problem 5.8 . Use the approach outlined in Example 5-2 to develop an expressionfor the magnetic field H at an arbitrary point P due to the linear conductor defined bythe geometry shown in Fig. 5-35 (P5.8). If the conductor extends between z 1 = 3 mand Zz = 7 m and carries a current I = 5 A. find H at P(2, $, 0).

Solution: The solution follows Example 5-2 up through Eq. (5.27), but theexpressions for the cosines of the angles should be generalized to read as

21JCHAFTER 5

instead of the expressions in Eq. (5.28), which are specialized to a wire centered atthe origin. Plugging these expressions back into Eq. (5.27), the magnetic field isgiven as

Figure P5.9: Configuration of Problem 5.9.

H=Houter+Hinner=ZI9 (!_!) =/9(b-a)41t a b --'

• 1 (z-z, Z-z,)H =+ 47tr Jr2+ (Z-Zl)2 - Jr2+(Z-Z2)2 .

For the specific geometry of Fig. P5.8,

Solution: From the solution to Example 5-3, if we denote the z-axis as passing outof the page through point P, the magnetic field pointing out of the page at P due tothe current flowing in the outer arc is Houler = -ZI9/41tb and the field pointing outof the page at P due to the current flowing in the inner arc is HinDer = ZI9/41ta. Theother wire segments do not contribute to the magnetic field at P. Therefore, the totaJfield flowing directly out of the page at P is

CProbJem 5.?) The loop shown in Fig. 5-36 (p5.9) consists of radiaJ lines andsegments of circles whose centers are at point P. Determine the magnetic field Hat P in tenus of a, b, e, and J.

Problem 5.10 An infinitely long, thin conducting sheet defined over the spaceo ~ x ~ w and -00 < Y =::; 00 is carrying a current with a uniform surface current

which gives I = 320 A.

1a2

H2= -z 2[a2 + (1._ 2)2]3/2 .

215CHAPTER 5

Problem 5.13 A long, East-West oriented power cable carrying an unknowncurrent I 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 12 f.lT when the current isflowing through the cable and 20 p.T when the current is zero, what is the magnitudeof J?

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 = (20-12),uT =8f.lT.

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 South

to North. Hence, by the right-hand rule, the current direction is toward the East. Itsmagnitude is obtained from

8 T=8 X 10-6= J.lQI = 41t x 10-1/f-l 21td 21t x 8 '

~blem s--:iU Two parallel, circular loops carrying a current of 20 A each arearranged as shown in Fig. 5-39 (p5.14). The first loop is situated in the x-y planewith its center at the origin and the second loop's center is at z = 2 m. If the twoloops have the same radius a = 3 m, determine the magnetic field at:

(a) z = 0,(b) z = 1m.

(c) z = 2 m.

Solution: The magnetic field due to a circular loop is given by (5.34) for a loop inthe x-y plane canying a current / in the +~irection. Considering that the bottomloop in Fig. P5.14 is in the x-y plane, but the current direction is along -+,

H A It?I = -z 2(a2 + 1.2)3/2 '

where z is the observation point along the z-axis. For the second loop, which is at aheight of 2 m, we can use the same expression but z should be replaced with (1. - 2).Hence,

H = -z5.25 Afm.

CHAPTER 5

y

~[a2 [I 1]H=HI + H2 = -zT (02 + z2)3/2 + [a2 + (z _ 2)2]3/2 Afm.

(a) At z = 0, and with a = 3 m and 1= 20 A,

A20X9[1 1]_H=-Z-2- 33 + (9+4)3/2 =-z5.25AIm.(b) At z = 1 m (midway between the loops):

H A 20 x 9 [1 1] A 5 69 Af= -z-2- (9+ 1)3/2 + (9+ 1)3/2 = -z . m.(c) At z = 2 m, H should be the same as at z =O. Thus,

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

z

x

216

The total field is

Section 5-3: Forces between Currents

~oblem s.lD The long, straight conductor shown in Fig. 5-40 (p5.15) lies in theplane of the rectangular loop at a distance d = 0.1 ffi. The loop has dimensions

a = 0.2 m and b = 0.5 m. and the currents are II = 10 A and /2 = 15 A. Determinethe net magnetic force acting on the loop.

Solution: The net magnetic force on the loop is due to the magnetic field surroundingthe wire canying current l). The magnetic forces on the loop as a whole due to thecurrent in the loop itself are canceled out by symmetry. Consider the wire carryingII to coincide with the z-axis, and the loop to lie in the +x side of the x-z plane.Assuming the wire and the loop are surrounded by free space or other nonmagneticmaterial, Eg. (5.30) gives

217

Ib=05m

a = O.2md=O.lm

Figure P5.15: Current loop next to a conducting wire (Problem 5.15).

CHAPTER 5

B =• .uoh .21tr

In the plane of the loop, this magnetic field is

B=Y~~.

Then, from Eq. (5.12), the force on the side of the loop nearest the wire is

The force on the side of the loop farthest from the wire is

z

CHAPTER 5

z

toO

· " ~1F'·

mx~

:'2

----.......-......d

x y(a)(b)(c)

218

Problem 5.16 In the ammgement shown in Fig. 5-41 (p5.16), each of the two long,

parallel conductors carries a current I, is supported by 8-cm-long strings, and has amass per unit length of 0.3 glcm. Due to the repulsive force acting on the conductors,the angle a between the supporting strings is 10°. Determine the magnitude ofl andthe relative directions of the currents in the two conductors.

Figure P5.16: Parallel conductors supported by strings (Problem 5.16).

F = Fml +Fm2

,.,/lOh lzb ,., p.o1ihb=-x---+x---21td 21t(a +d)

,., /lOll lzab=-x----21td(a+d)

_ ,.,41tx 10-7 X lOx 15xO.2xO.5 _ ,.,0-4 (N)- => 1 (mN)_-x----------- - -xl - -AU ••21t X 0.1 x 0.3

The force is pulling the loop toward the wire.

The other two sides do not contribute any net forces to the loop because they areequal in magnitude and opposite in direction. Therefore, the total force on the loop is

Solution: While the vertical component of the tension in the strings is counteractingthe force of gravity on the wires, the horizontal component of the tension in the stringsis counteracting the magnetic force, which is pushing the wires apart. Accordingto Section 5-3, the magnetic force is repulsive when the currents are in oppositedirections.

Figure P5.16(b) shows forces on wire 1 of part (a). The quantity F' is the tension

force per unit length of wire due to the mass per unit length m' = 0.3 g/cm. The

ClLA..PTER 5

side 1 is attractive. That is,

F = ~JlQltha = A 4n x 10-7 X 5 x 10 x 2 = A 2 x 10-5 NI Y 2n(aj2) y 21t X 1 Y .

h and /z are in opposite directions for side 3. Hence, the force on side 3 is repulsive,which means it is also along y. That is, F3 = F] .

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

F=2F, =y4x 10-5 N.

Section 5-4: Gauss's Law for Magnetism and Ampere's Law

CProblem 5.2V Current I flows along the positive z-direction in the inner conductorof a long coaxial cable and returns through the outer conductor: The inner conductorhas radius a, and the inner and outer radii of the outer conductor are b and c.respectively.

(a) Determine the magnetic field in each of the following regions: 0:5 r :5 a,a ~ r ~ b, b ~ r ~ c. and r ~ c.

(b) Plot the magnitude of H as a function of r over the range from r = 0 tor = 10 em, given that I = lOA, a = 2 em, b = 4 em, and c = 5 em.

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

A rI

H =If>2mz2 '

and in the region a < r < b,

A IH=If>-.2nr

The total area of the outer conductor is A = n( c2 - b2) and the fraction of the areaof the outer conductor enclosed by a circular contour centered at r = 0 in the regionb < r < c is

n(? _b2) _ ?_b2rc( c2 - b2) - c2 - b2 .

The total current enclosed by a contour of radius r is therefore

225

0.8

5'0.7

~ '-'0.6

::I:G)

0.5"0 ::J-'c 0.4~

E"0

03(jj I;::(,) 0.2'';: 4)c~ 0.1~0.0

O.1.

Figure P5.20: Problem 5.20(b).

if i21[iT!(Vo) ir'h= I J·ds= - .(zrdrdcp)=21t Jodr=21tr]Jo.L $:0 r:0 r r:0

Radial distance r (em)

CHAPTER 5

and the resulting magnetic field is

H = • _len_cl_osed_ = +_1 (_t?_-_,2_)21tr 21tr \.c2 - JJ2 / •

For r > c, the total enclosed current is zero: the total current flowing on the inner

conductor is equal to the total current flowing on the outer conductor, but they areflowing in opposite directions. Therefore, H =O.

(b) See Fig. P5.20.

~blem 5~ A long cylindrical conductor whose axis is coincident with the z-axishas a radius a and carries a current characterized by a current density J = Vo/ r,where Jo is a constant and r is the radial distance from the cylinder's axis. Obtain anexpression for the magnetic field Hfor (a) 0 5r 5a and (b) r> a.

Solution: This problem is very similar to Example 5-5.(a) For ()5 rI =:; a, the total current flowing within the contour CI is

CHAPTER 5

Therefore, since /1 = 2nr, HI , HI = Jo within the wire and HI =~Jo.(b) For r 2: a, the total current flowing within the contour is the total current flowing

within the wire:

Therefore, since 1=21trH2. H2 = loa/r within the wire and H2 =~lo(a/r).

Problem 5.22 Repeat Problem 5.21 for a current density J = Voe-r•

z

s

Figure P5.22: Cylindrical current.

Solution: