I.~ I ./ TT. Figure P6.1: Loops of Problem 6.1.

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Aubrey Thornton
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1 CHAPTER Chapter 6 Sections 6-1 to 6-6: Faraday's Law and its Applications Problem 6.1 The switch in the bottom loop of Fig (p6.1) is closed at t = 0 and then opened at a later time tl. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? tl: I.~ I -1 ~~./ TT RI Figure P6.1: Loops of Problem 6.1. I Solution: The magnetic coupling will be strongest at the point where the wires of the two loops come closest. When the switch is closed the cultent in the bottom loop will start to flow clockwise, which is from left to right in the top portion of the bottom loop. To oppose this change, a eultent will momentarily flow in the bottom of the top loop from right to left. Thus the current in the top loop is momentarily clockwise when the switch is closed. Similarly, when the switch is opened, the current in the top loop is momentarily counterclockwise. blem. The loop in Fig (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 shape with time, V~f = 0 V and Vemf = Ve~f' Therefore, from Eq. (6.8), 1=Y. c/r=- -1ls,dB - ds em IT' R S dt. If we take the surface normal to be +2, then the right hand rule gives positive flowing current to be in the ++ direction. -A a -ABoro 1=/fatBosinrot= R cosrot (A),

2 250 CHAPTER 6 z y x Figure P6.2: Loop of Problem 6.2. 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 field that 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 soi = O. There is no current flowing in either direction. crroblem ~ 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 with each of its sides parallel to the x- or y-axis. Find the induced emf across the open-circuited ends of the coil if the magnetic 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 and Vemf = Ve~f' From Eq. (6.6), 1 vemf=. -N-d B ds=-nd B (zdxdy), dt s dt 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).

3 CHAPTER (b) For B = ilocosxcos l<pt (T), Vemf=-l00;Td t ( locos103tlool25 =: y=-o.i25 cosxdxdy ) = 62.3sin103t (kv). (c) For B = ilocosxsin 2ycos!03t (T), Vemf = -l00d d t ( locos lift r-:-O.l y= cosxsin2ydxdy ) = O. Problem 6.4 A stationary conducting loop with internal resistance of is placed in a time-varying magnetic field. When the loop is closed, a current of 2.5 A flows through it What will the current be if the loop is opened to create a small gap and a 2-.0 resistor is connected across its open ends? Solution: Vernf is independent of the resistance which is in the loop. Therefore, when the loop is intact and the internal resistance is only 0.5 a, ~mf = 2.5 A x 0.5.Q = 1.25 V. When the small gap is created, the total resistance in the loop is infinite and the current flow is zero. With a 2-.Q resistor in the gap, 1= Vemc/{ a) = 1.25V/2.5 0=0.5 (A). Problem 6.5 A circular-loop TV antenna with 0.01 m2 area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop develops an emf with a peak value of 20(m V). What is the peak magnitude of B of the incident wave?. 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 field with magnitude B = IBI. Since we know the frequency of the field is f = 300 MHz, we can express B as B = Bocos (rot+~) arbitrary reference phase. From Eq. (6.6), with co= 21t X 300 x 106 cadis and ao an 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,

4 252 CHAPTER 6 which yields Bo = 1.06 (najm). ~Iem 69The square loop shown in Fig (P6.6) is coplanar with a long, straight WIre carrying a current i(t) = 2.5cos2n x 104t (A). (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 through a 4-Q resistor connected across the gap. The loop has an internal resistance of 1Q. z I(t) 5cm I-IOcm-f T IOcm 1 y x Figure P6.6: Loop coplanar with long wire (Problem 6.6). 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

5 CHAPTER is <P= B ds= -x- '[-x1o(cm)]dy. Iss 115em 5~ ( 2~ #0/) PQ/ X =' lo-1 In = 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). (b) Vernf = - ~~ = 0.55 X 2x X 104sin(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). 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 when looking down on the loop. as shown in the figure. Problem 6.7 The rectangular coijducting loop shown in Fig (P6.7) rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B = y50 (mt). Detennine the CUITentinduced in the loop if its internal resistance is Solution: <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) = x 10-3 sin (2001tt) (V), Vemf '( ) l;.nd = 0.5 = 37.7 sm 2001tt (ma)..

6 .' CHAPTER z 0'-, 0 \ o y x Figure P6.9: Rotating rod of Problem 6.9. From Eq. (6.24), V12 = Ve-::n= 12 fi (u x B).dJ = 1r=O.5 fo (~61tr X i3 X 10-4). idr = 181t X 10-4 Jr=.O.5 fo rdr = 91tx 1O-4?r0.5 = -91t x 10-4 X 0.25= -707 (uv). <&;;biem 6.UO The loop shown in Fig (P6.1O) moves away from a wire canyinga current II = 10 (A) at a constant velocity u = y5 (mls). If R = 10.Q and the direction of h is as defined in the figure, find lz as a function of Yo, the distance between 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 voltage is due to motion of the loop and is given by Eg. (6.26): " 17m _- -1. (n v It). AI remr - "'emf - JC "r' \... ".ai}.. CA- The magnetic field B is created by the wire carrying IJ. Choosing z to coincide with the direction of It, Eq. (5.30) gives the external magnetic field of a long wire to be B =.JlOh 21tr

7 256 CHAPTER 6 II = loa. lor I----- If u h T, I----- u + -10cm--! R Yo Figure P6.1O: Moving loop of Problem For positive values of Yo in the y-z plane, y = T, so '" '" '"POll ",Poll U 21tr 21tr U x B = ylulx B = rlulx.- = z--. Integrating around the four sides of the loop with dl = z dz and the limits of integration chosen in accordance with the assumed direction of h, and recognizing that only the two sides without the resistors contribute to Ve'::tf' we have and therefore Ve~f = ro.2 (i,uo1iu) I. (idz) + ro (z,uo1iu) I.(idz) 10 21tr r=)'o tr r=yo+o.1 = 41t X 10-7 X21t10 x 5 x 0.2 ( Yo..!.. _ Yo ) = 2 x 10-6 ( YO..!.. _ Yo ) (V), lz = -2R- ~~f = 100 (1Yo- y-o-+-o-.l 1) (na). Problem 6.11 The conducting cylinder shown in Fig (P6.11) rotates about its axis at 1,200 revolutions per minute in a radial field given by B=r6 (T).

8 CHAPTER z T IOcm 1 I Sliding contact Figure P6.11: Rotating cylinder in a magnetic field (Probl~m 6.11): The eylinde~ whose radius is 5- em and height 10 em, has sliding contacts at its top and bottom connected to a voltmeter. Detennine the induced voltage. Solution: The surface of the cylinder has velocity u given by ~ A A u=,ror='21tx~x5xl0- =cft21t (mls), V12= Jo rl {uxb).di= Jo ro.1(~21txr6).zdz=-3.77 (V). (Problem 6.12;)The electromagnetic generator shown in Fig is connected to an electric bulb with a resistance of loo!1. If the loop area is 0.1 m2 and it rotates at 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 3,600 Vo=AroBo=O.1 x 60 xo.2=7.54 (V), Vo 7.54 I =Ii = 100 = 75.4 (ma). Problem 6.13 The circular disk shown in Fig (p6.13) lies in the x-y plane and rotates with uniform angular velocity ro about the z-axis. The disk is of radius a and is present in a uniform magnetic flux density B = Wo. Obtain an expression for the emf induced at the rim relative to the center of the disk.

/R = If we take the surface normal to be +ẑ, then the right hand rule gives positive flowing current to be in the +ˆφ direction.

/R = If we take the surface normal to be +ẑ, then the right hand rule gives positive flowing current to be in the +ˆφ direction. Problem 6.2 The loop in Fig. P6.2 is in the x y plane and B = ẑb 0 sinωt with B 0 positive. What is the direction of I (ˆφ or ˆφ) at: (a) t = 0 (b) ωt = π/4 (c) ωt = π/2 V emf y x I Figure P6.2: Loop of