.~•• IImy.fit.edu/~blail/ECE3442/ECE3443F05HW7.pdfcrroblem ~ A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centered at the origin

Figure P6.1: Loops of Problem 6.1.

249CHAPTER 6

Chapter 6

-A a -ABoro1=/fatBosinrot= R cosrot (A),

Solution: The magnetic coupling will be strongest at the point where the wires ofthe two loops come closest. When the switch is closed the cUlTent in the bottom loopwill start to flow clockwise, which is from left to right in the top portion of the bottomloop. To oppose this change, a eUlTent will momentarily flow in the bottom of thetop loop from right to left. Thus the current in the top loop is momentarily clockwisewhen the switch is closed. Similarly, when the switch is opened, the current in thetop loop is momentarily counterclockwise.

.~ II •• I

tl: I-1 ~~./ TT

RI

Sections 6-1 to 6-6: Faraday's Law and its Applications

Problem 6.1 The switch in the bottom loop of Fig. 6-17 (p6.1) is closed at t = 0and then opened at a later time tl. What is the direction of the current I in the toploop (clockwise or counterclockwise) at each of these two times?

blem. The loop in Fig. 6-18 (P6.2) is in the x-y plane and B = ZBo sin rot

with 0 oSltlVe. What is the direction of / (+ or -+) at (a) t =0, (b) rot =1&/4, and(c) rot =1t/21

Solution: / = Vemr/ R. Since the single-turn loop is not moving or changing shapewith time, V~f = 0 V and Vemf = Ve~f' Therefore, from Eq. (6.8),

IT' -1 ls,dB1=Y. c/R=- -·dsem R S dt .

If we take the surface normal to be +2, then the right hand rule gives positiveflowing current to be in the ++ direction.

. d 1 d 10.1251°.125vemf= -N- B·ds=-N- B· (zdxdy),dt s dt -0.125 -0.125

CHAPTER 6

y

z

x

Figure P6.2: Loop of Problem 6.2.

250

where N = 100 and the surface nonnal was chosen to be in the +z direction.(a) For B = iIOe-2t (T),

Vemf = -100; (lOe-2t(0.25)1) = 125e-2t . (V).

where A is the area of the loop.(a) A, co and R are positive quantities. At t = 0, cos rot = 1 so I < 0 and the

current is flowing in the -~ direction (so as to produce an induced magnetic fieldthat opposes B).

(b) At rot = 1t/4, cosCOt = ../2/2 so I < 0 and the current is still flowing in the -+direction.

(c) At cot = 1t/2, cos cot =0 so I= O. There is no current flowing in either direction.

crroblem ~ A coil consists of 100 turns of wire wrapped around a square frameof sides 0.25 m. The coil is centered at the origin with each of its sides parallel tothe x- or y-axis. Find the induced emf across the open-circuited ends of the coil if themagnetic field is given by

(a) B = z IOe-2t (T),

(b) B = zlOcosx cos 103t (T),

(c) B =i lOcosx sin2y cos 103f (T).

Solution: Since the coil is not moving or changing shape; Ve'::tf = 0 V andVemf= Ve~f' From Eq. (6.6),

~mf = 2.5 A x 0.5 .Q = 1.25 V.

1=Vemc/{20+0.5 a) = 1.25V/2.5 0=0.5 (A).

251

(kV).

d ( 10.125 10.125 )Vemf = -l00d lOcos lift cosxsin2ydxdy = O.t .r-:-O.l25 y=-0.125

(c) For B = ilOcosxsin 2ycos !03t (T),

(b) For B = ilOcosxcos l<Pt (T),

d ( LOol25 100125 )Vemf=-l00;T lOcos103t cosxdxdy = 62.3sin103tt =:-0.125 y=-O.I25

CHAPTER 6

Problem 6.4 A stationary conducting loop with internal resistance of 0.5 .0 is

placed in a time-varying magnetic field. When the loop is closed, a current of 2.5 Aflows through it What will the current be if the loop is opened to create a small gapand a 2-.0 resistor is connected across its open ends?

Solution: Vernf is independent of the resistance which is in the loop. Therefore, whenthe loop is intact and the internal resistance is only 0.5 a,

Problem 6.5 A circular-loop TV antenna with 0.01 m2 area is in the presence of auniform-amplitude 300-MHz signal. When oriented for maximum response, the loopdevelops an emf with a peak value of 20(m V). What is the peak magnitude of B ofthe incident wave? .

When the small gap is created, the total resistance in the loop is infinite and thecurrent flow is zero. With a 2-.Q resistor in the gap,

Solution: TV loop antennas have one turn. At maximum orientation, Eq. (6.5)evaluates to ct> = J B . ds = ±BA for a loop of area A and a unifonn magnetic fieldwith magnitude B = IBI. Since we know the frequency of the field is f = 300 MHz,we can express B as B =Bocos (rot +~) with co= 21t X 300 x 106 cadis and ao anarbitrary reference phase. From Eq. (6.6),

Vemr= -N~~ = -A ~[Bocos(rot+ao)J =ABorosin(rot+ao).

Vernf is maximum when sine rot +ao) = 1. Hence,

20x 10-3 =ABoro= ]0-2 xBox61tx 108,

CHAPTER 6

y

TIOcm

1

I-IOcm-f

5cm

I(t)

x

z

i(t) = 2.5cos2n x 104t (A).

Figure P6.6: Loop coplanar with long wire (Problem 6.6).

(a) Determine the emf induced across a small gap created in the loop.

(b) Determine the direction and magnitude of the current that would flow througha 4-Q resistor connected across the gap. The loop has an internal resistance of1Q.

which yields Bo = 1.06 (nAJm).

252

~Iem 69The square loop shown in Fig. 6-19 (P6.6) is coplanar with a long,straight WIre carrying a current

Solution:

(a) The magnetic field due to the wire is

where in the plane of the loop, ~ = -x and r =y. The flux passing through the loop

Problem 6.7 The rectangular cOIJducting loop shown in Fig. 6-20 (P6.7) rotates at6,000 revolutions per minute in a uniform magnetic flux density given by

At t = 0, B is a maximum, it points in -i-direction, and since it varies as

eos(21t X l04t), it is decreasing. Hence, the induced eUITent has to be CCW whenlooking down on the loop. as shown in the figure.

253

Is 115em ( #0/)<P= B·ds= -x- '[-x1O(cm)]dy. s 5~ 2~

PQ/ X lO-1 15=' In-5= 41t X lO-7 X 2.5cos(2x X 104t) X 10-1 xLI

2x .

=0.55 X lO-7 cos(21t X 1041) (Wb).

Vernf = - ~~ = 0.55 X 2x X 104 sin(2x X 104t) X lO-7

= 3.45 X 10-3 sin(2x X 1041) (V).

Vemf 3.45X 10-3 • ( 4 ) . ( 4 )Iind = 4+ I = ~ sm 2x X 10 1 =0.69sm 2x X 10 1 (mA).

(b)

B =y50 (mT).

<P= iB.dS = y50 X 10-3 ·Y(2 X 3 X 1O-4)cosCP(t) = 3 X 10-5 cos CP(t),

2xx6x 1&<P(t) = cot= n. t = 2001tt (rad/s),

<P= 3 X 10-5 cos(2007tt) (Wb),

Vemf =- ~~= 3 X 10-5 X 2007tsin(2007tt) = 18.85 x 10-3 sin (2001tt) (V),

Vemf '( )l;.nd = 0.5 = 37.7 sm 2001tt (mA) ..

CHAPTER 6

is

Detennine the CUITentinduced in the loop if its internal resistance is 0.5 .0.

Solution:

The magnetic field B is created by the wire carrying IJ. Choosing z to coincide withthe direction of It, Eq. (5.30) gives the external magnetic field of a long wire to be

255

y

o

0'-, 0\

x

" __ 17m _- -1. (n v It). AI

remr - "'emf - "r' \ .... ".aI}. CA-••JC .

From Eq. (6.24),

V12 = Ve-::n= fI (u x B).dJ = fO (~61tr X i3 X 10-4). idr12 1r=O.5

= 181t X 10-4 fO rdrJr=.O.5

= 91tx 1O-4?r0.5

= -91t x 10-4 X 0.25= -707 (uV).

Figure P6.9: Rotating rod of Problem 6.9.

B = .JlOh21tr •

z

.' CHAPTER 6

<&;;bIem 6.UO The loop shown in Fig. 6-22 (P6.1O) moves away from a wirecanyinga current II = 10 (A) at a constant velocity u = y5 (mls). If R = 10.Q andthe direction of h is as defined in the figure, find lz as a function of Yo, the distancebetween the wire and the loop. Ignore the internal resistance of the loop.

Solution: Assume that the wire carrying current /J is in the same plane as the loop.The two identical resistors are in series. so h=Vemr/2R. where the induced voltageis due to motion of the loop and is given by Eg. (6.26):

B=r6 (T).

CHAPTER 6

(nA).

+-10cm--!If

II = lOA.

T,I----- u

lOr

hI----- u

~~f (1 1)lz = -2R- = 100 Yo- y-o-+-O-.l

Ve~f = rO.2 (i,uo1Iu) I . (idz) + rO (z,uo1Iu) I .(idz)10 21tr r=)'O 10.2 21tr r=yo+O.1

= 41t X 10-7 X 10 x 5 x 0.2 ( ..!.. _ 1 )21t Yo Yo + 0.1

= 2 x 10-6 ( ..!.. _ 1 ) (V),YO Yo +0.1

For positive values of Yo in the y-z plane, y = T, so

'" '" '"POll ",Poll UU x B =ylulx B = rlulx.- = z--.21tr 21tr

256

Yo

R

Figure P6.1O: Moving loop of Problem 6.10.

Integrating around the four sides of the loop with dl = z dz and the limits ofintegration chosen in accordance with the assumed direction of h, and recognizingthat only the two sides without the resistors contribute to Ve'::tf' we have

and therefore

Problem 6.11 The conducting cylinder shown in Fig. 6-23 (P6.11) rotates about itsaxis at 1,200 revolutions per minute in a radial field given by

Problem 6.13 The circular disk shown in Fig. 6-24 (p6.13) lies in the x-y planeand rotates with uniform angular velocity ro about the z-aXis. The disk is of radius aand is present in a uniform magnetic flux density B =Wo. Obtain an expression forthe emf induced at the rim relative to the center of the disk.

257

z

ISliding contact

TIOcm

1

Figure P6.11: Rotating cylinder in a magnetic field (Probl~m 6.11):

CHAPTER 6

The eylinde~ whose radius is 5-em and height 10 em, has sliding contacts at its topand bottom connected to a voltmeter. Detennine the induced voltage.

Solution: The surface of the cylinder has velocity u given by

~ A 1200 2 A

u=,ror='21tX~X5xl0- =cft21t (mls),

rL rO.1V12= Jo {uxB).dI= Jo (~21txr6).zdz=-3.77 (V).

(Problem 6.12;)The electromagnetic generator shown in Fig. 6-12 is connected to anelectric bulb with a resistance of loo!1. If the loop area is 0.1 m2 and it rotatesat 3,600 revolutions per minute in a uniform magnetic flux density Bo = 0.2 T,detennine the amplitude of the current generated in the light bulb.

Solution: From Eg. (6.38). the sinusoidal voltage generated by the a-c generator is

Vemf =Ac.oBosin(rot + Co) .:..-Vosin(c.or+Co): Hence.

21t x 3,600Vo=AroBo=O.1 x 60 xO.2=7.54 (V),

Vo 7.54I = Ii = 100 =75.4 (mA).