Question

(5%) Problem 19: In a region of space the magnetic field is given by B = (a + Api + (x2), where a = 543.7T m. A = 945 nt, and y = 382 nT m. Y Determine the magnitude of the current passing through a box in thew-plane bounded by x1 10.0 cm and x = 70.0 cm in the x-direction and by = 20.0 cm and 50,0 cm in the direction Grade Summary 343 mA Deductions 34. I ma Potential 34.7 mA Submissions 25 mA Attempts tenanang: per attemap detailed view

Answer 1

Solution Y B y = $0.0 X-700 XE100 y = 20.01 from formula - Bodi = 4o I enclosed, % B = (dy+ B) { + (re) de = dait d=543.7 nT/m B = 94 Snt 382 nT / m² dy ) slay+B)i + (ra)]]. ( dxi + dy}) scay+B) dx + fredy y a 20.0 so.o from AMB 22=100cm z: 0.1m or y = 0:2 + 0.5 dx=0 OS 0.5 gir r 10.1²) dy 0.018 Ly] D 2 0.2 B₂ = 0.0037 Oism 3- dy=0 from Btc yo so.oem De = 0.1 0.7 X=10.0-700 0.7 B2 = for co-s)+]dz [260.5)+6] (72017 B2 = 0.6 [d10.5)+B] Oil (2

from cud x = 0.7m da=0 Я 0.5 0.2 0:2 B₂ r(0.73 dy si 110.49)(y)... 0.5 B₂ = r(0.49) 10.3) 3 from DHA y = = 20 cm y = 0.2m x = 0.74 0.Im B4 = [0(0-2) + B] da 0.7 ca 10.2) +B) [e.001 0.7 :- ((0:2)+p](0.6) 4 on addition BJ + B2 + B3 + B4 DJ lo I enclosed, 0.003r + Z 210.30) +0.6B - 0.1477 - [160.12) + 0.65) D.1801 0.1447 0,18 ( 543.7) -0.144 (382) CL 97.886- 55.008 c = 42.878 OT ri Jenclosed z 1 x 42.878 Mo d x 42.878x109 417xio7 I 34:12 X1 x 103 Jenclosed 34.12 mA therefore option © és correct 34.1 mA therefore option