Question

Question-3 [25 Points] An infinitely long filamentary wire carries a current of 3 A in the +y-direction. a) [10 pts] Formulate and calculate the magnetic field flux density, B at P(6,8,12) b) [15 pts] Formulate and calculate the magnetic flux, Ø, through the loop described by where 1<p <5,-2<z< 2,9 = TT

Answer 1

Sal! given that An infinitely long filamentary wire current of 3A the ty-direction. carries a in the magnetic field clue to infinitely Hong wire : Ampere's law: According to Amperels creuit law the line integral of magnetic field H around a closed path net current I enclosed by the path. is the same as the 2 fu.de - I Let an infinitely long wire carrying current of magnitude I A in +2- direction Then by Ampere's circuit daw J = I magnetic field fit4 ap) updf ap) Sedo Berlin de=pdpad = I x Ho Hogne EI I H = ITP ap so that magnet freed of a long current carrying conductor circular path around the wire is in H 1

know that I B = Mot M = Mo for air B = MOI 2 Т Р 96 where ap=9ex ap and point in given question I = 3A in ty-direction of observation is at p(6,8,12 12) perpendicular distance of point from wire- 20 91 (6,0,12) I-3A 6 x

Take a point A on -axis as . (ه, 8 ره ) A perpendicular distance reetor ول زه ره ره )P ( 6 / 5 ) A بوا 6 ax +2 4z مه و 12 x 6 62+22 مو و 12 x 6 180 | = م 22) مJ62 م J/8o

And direction of By а) = 44 хар /jao 1 * (46, + 12 92-) (3x+) +12 (°11 ч. ) 4. + 2 Jiao јао G ау Jilo //go ар – 12 at — Ч. Гјво so that field flux density , B at P (6,8,12) is, ра Ког 9 ТР аф Mo33 #т 100 Х 12 ах -6 ч. јдо с 3 Чо 9x180 (11 4 – 6 ч. ) put Mo что 10-7 3 xчто-? 20 x 100 (12 а, - 6ч. ) b = B % х? (iz q -(а.) в – чах — 242) 30 * tuhhht )

(Ь. 2 P-5 2=2! ds = dydz az A P=1 y I=34 e, Z=-2 1$ Na ✓ -2 x Ø = 17/2 is -2 2 2 2 2 12PL 5, , The described loop by in figure shown

magnetic flux clensity in this situation is given as- ap B = MOI 212 we To avoid con fusion take P=2. because e is perpendreular distance of point of observation from wire. AT point A ap = az, au= ay so that, as = aetap ayxa2 ap= ax so that aa B,= Mol 2112 ae=ay AT point B: ap=-92 / so that ap=aetap =(taglx (292) - Cay+92) -ax Hence i B. Mo I 212 are magnedic filuz & is defreel as- &= fbids

flux above f Birds g-anus through half loop where ds dy dz ax, K425, 02222 - को St Mo I ax dy dzae 2152 opy MOI . 1 2725) olz R2 2172 dydz flux below y-aris through half cloop- d2 = SBid's y where asa clydzak ,129 25, -22220 Mo Itax) s dy dz dzane g1T2 are we can &2 = - f Mot dy dz ; 12925 , • ~2 2220 see that di and Q2 equal in magnitude but opposite sign that totul flux flux through cloop will be do +02 क -0,- 6=0 Hence, (9) В Cu ax – 292.) x1050 (wbl me) $=0 b