hw7_solutions

# hw7_solutions - 164 (a) At t TIME- VARYING FIELDS AND...

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164 TIME-VARYING FIELDS AND MAXWELL'S EQUATIONS (a) At t = 0, the induced current is 1=0. (b) At t = 10 ms, the induced current is I ~ - 7 .83 mAo (Note that at t = 10 ms, since dBz(t) > 0 dt Based on Lenz's law, the induced current I through the IO.Q resistor must flow in the clockwise direction. (c) At t = 100 IDS, the induced current is I = O. (d) At t = 1 s, the induced current is I = O. 7-3. Two circular coils. (a) Using the result of Example 6-6, the magnetic flux density produced on the axis of coil 1 with radius al = 5 cm, current II = l00cos(20007rt) A, and number of turns NI = 10, at a distance of z = 1 m away from its center can be calculated as p.oN I l l aI 2 BIz -2( 2 2)3/2 al +z 2 7 (411" x 1O- )(IO)[100cos(20007rt)](0.05) ~ 1.56cos(20001l"t) fLT - '11') - 2[(0.05)2 + 12] The voltage induced across the terminals of coil 2 (a2 = 5 cm and N2 = 100) due to the time- varying flux '1'12 produced by coil 1 linking coil 2 can be evaluated as d'l'12 2 dBb 'Vind2 = - dt = -N2(1I" a 2 )--;u;- ~ -(100)[11"(0.05)2][-(1.56 x 10- 6 )(200011") sin(20007rt)] ~7.72 x 10- 3 sin(20001l"t) V Therefore, the amplitude of the induced voltage'Vind2 is "-'7.72 mY. (b) Repeating the same calculations for the same current II oscillating at 10 kHz [i.e., II(t) = l00cos(200007rt) A] results in an induced voltage'Vind2 of amplitude 1'V77.2 mY. 7-4. Two concentric coils. (a) From Example 7-4, we have p.oNII -5 B ctr = 2-- = 22.5 x 10 IT 2a Substituting NI = 30, we find the radius of the larger coil as a ~ 0.754 m. (b) From Example 7-4, since a ~ 75.4 cm» 1 cm=b, the total flux linking the smaller coil can be approximated as , '1'12 = N2 p.o~tl (1I"b 2 ) = (75)(2.5 X 10- 5 1)[11"(10- 2 )2] ~ 5.89 X 10- 7 1 (c) For I(t) = 10 cos(l 2011"t) A, the induced voltage across the terminals of the smaller loop can be found as 'Vind = - d'l'12 = -(5.89 x 10- 7 )[-(10)(12011") sin(l201l"t)] V ~ 2.22 sin(1201l"t) mV dt

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168 TIME-VARYING FIELDS AND MAXWELL'S EQUATIONS opposing the current driven by the source, consistent with Lenz's law. This process continues until the velocity of the bar reaches a value Vfinal such that the induced emf is equal to the source voltage making the net current in the loop zero. Beyond that point, there is no current, hence no force, hence no acceleration, and the bar simply continues to move at velocity Vfinal. The terminal velocity vo can be found by simply equating the motional emf to vo. We have Vo 'Vemf = f (Vfinal' X Bo) . dI = vfinalBOl = Vo ~ Vfinal
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## This note was uploaded on 11/16/2010 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.

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hw7_solutions - 164 (a) At t TIME- VARYING FIELDS AND...

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