Electromagnetics I Exam No. 3 December 1, 2003 Solution

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1 Electroagnetics Ea No. 3 Deceber 1, 2003 Solution Please read the ea carefull. Solve the folloing 4 probles. Each proble is 1/4 of the grade. To receive full credit, ou ust sho all ork. f cannot understand hat ou are doing, ou ill receive no credit. f ou need to assue anthing, state our assuptions clearl. Reasonable assuptions that are necessar to solve the proble ill be accepted. n all probles assue properties of free space ( 0 = F/, µ 0 =410 7 H/) unless otherise specified. You can rite on both sides of the page. f ou need additional space SK for additional paper and ake sure ou rite our nae on it. Level of difficult: 2 (ost difficult), 3, 4, 1 (easiest) e n Chapter9.probles.etra 5. a. Which of the folloing fields are possible agnetic fields in free space? 1. B 1 = 2 + (2z + 3) + z(5 2z) 2. B 2 = ( ) + 3 zz 3. B 3 = (9 3 ) + (z 3 3 ) + z( 3 3 ) b. For the possible agnetic fields in question, deterine the current densit J. Note: guess, even if correct, ill not be accepted. Solution: To sho that a field is a agnetic field e need to sho that its divergence is zero and its curl is non-zero. That is, e appl the postulates: H = J,.B = 0 Hoever, J can be zero so it is iportant to calculate the divergence first. a. We first calculate the divergence. B definition e ust have: For the four fields:.b 1 = (2).B 2 = (2 + 2 ).B 3 = (9 3 ).B = B + B + B z z = 0 (2z + 3) + + ( 3) + + (z3 3 ) (5 2z) z (z) z + (3 3 ) z = 2 0 = 2 2 = 0 = Thus, onl B 1 can be a agnetic field. (b). The curl of B is: H = J B = µ 0 J

2 Thus: = B 1 = µ 0 J = B z B + B z z B z (5 2z) (2z + 3) + (2) (5 2z) z z = 2z + 2 z2 + z B B + z (2z + 3) (2) Therefore, the current densit is: J = 2z + 2 µ 0 z 2 µ 0 2 e8.91. n Chapter8.probles.etra. 2. To circular, conducting clinders lie ith their aes at =+c and =c (c<b/2) and each of radius a (2a<b) and each carries a uniforl distributed current /2 in the positive z-direction (out of the page). Surrounding these is a thin, circular clindrical shell ith its ais at =0 and radius b, carring a uniforl distributed current in the reverse direction (into the page). Figure shos the configuration in cross-section. the solid clinders and the shells a be assued to be infinitel long. Calculate the agnetic field intensit (direction and agnitude) at an point inside the outer clinder but outside the saller clinders. c /2 c /2 Figure. Solution: Use pere's la since the clinders are infinitel long. a. First, e note that the outer shell produces no field inside itself. The field is therefore onl due to the inner clinders. Taking a general point (,) outside the inner clinders e have the configuration in Figure B for the left clinder. This clinder produces a agnetic field intensit as follos:

3 H1 H 1 H 1 (,) R /2 a c c /2 Figure B. The and coponents are: H 1 = /2 2R = 4R n vector for e have: H 1 = 4R sin = 4R H 1 = 4R cos = ( + c) = 4R R R = 4R = 2 ( + c) 4R 2 = 4 ( + c) ( + c) 4 ( + c) H 1 = 4 ( + c) ( + c) + 4 ( + c) Repeating the process for the right hand side clinder (see Figure C): H 2 = /2 2R = 4R n vector for: H 2 = 4R sin = 4R R = 4R = 2 H 2 = 4R cos = ( c) = 4R R 4 ( c) ( c) 4R = ( c) 2 4 ( c) H 2 = 4 ( c) ( c) + 4 ( c) The total field is: = 4 H = H 1 + H ( c) ( + c) ( c) ( c) ( + c) ( + c) 2 + 2

4 H2 H 2 H 2 (,) R /2 a c /2 c Figure C. 3. e9.55. (Fro probl.tosolve). Calculate the inductance per unit length of the parallel plate conductors shon. ssue that the current in the conductors is equal to aperes, uniforl distributed and that the plates are ver thin. ssue b<<. b b<< Figure. Solution: Since the distance beteen the plates is ver sall, e a assue that the field beteen the plates is the sae as for infinitel ide plates (see eaple 8.8). Thus, using pere's la for, sa, the upper plate, e get (see Figure B): H = 2 The agnetic field intensit due to the loer plate points in the sae direction as in Figure B and is also equal to /2. Thus, the agnetic field intensit beteen the plates is: H = The agnetic flu, hich crosses the area parallel to the -z plane is (for a length in the z direction equal to 1:

5 = µ 0 HS = µ 0 1b = µ 0b The inductance per unit length is therefore: L = = µ 0b H H Figure B. e9.104 Fro Bastos.scan.additional n infinite ire and a square loop are given as shon in Figure. The loop and ire are in the sae plane (see figure) but there is no electrical contact beteen the to (that is, the ire sits on top of the loop ithout actuall touching it). Calculate the utual inductance beteen the ire and loop for the configuration shon. 1 Figure Solution: Using eaple 9.11 as a guide, e solve for the agnetic flu for the general configuration in Figure. ssuing an arbitrar current in the ire, the agnetic flu densit at a radial distance r fro the ire is: B = µ 0 T 2r We note (see Figure B), that the flu to the right of the ire points up hile the flu to the left points don. The total net flu through the loop is the difference beteen these to flues. Taking an eleent of area as shon, e get for the flu to the right of the ire: The flu to the left of the ire is: 1 = r=0 r= µ 0 2r dr = µ 0 2 ln r 0

6 Calculating the difference: 2 = r=0 r= µ 0 2r dr = µ 0 2 ln r 0 = 1 2 = µ 0 2 = µ 0 2 ln r 0 µ 0 2 ln r 0 = µ 0 2 ln ( ) ln (0) ln () + ln (0) = µ 0 2 ln ( ) ln (0) ln () + ln (0) ln Note: norall ou ould not rite ln(0) but anted to sho that this ter cancels out due to the subtraction and so the fact that it is not defined does not ean that a solution does not eist. Since this is the total flu that links the ire and loop e can calculate the utual inductance b dividing the total flu b the current in the ire. Thus: L 12 = µ 0 ln 2 H Note that this akes sense since, for =/2, L 12 =0 as required. r dr Figure B.