Question

Problem:

Problem #2 A current of 1 A flows in a square conducting loop situated on the z = 0 plane in free space, as shown. Determine H at (a) (x, y,2) (0.5, 0.5, 0) m (b) (x, y, ) (0, 0, 1) m (c) (x, y, 2) (0.5, 0.5, 1) m y(m) (1, 1) (0,1) x(m) (1,0) (0,0)

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Answer 1

(a) at point (0.5, 0.5,olm due to Symmetory all element have same magnetic field so if i element field is known Then total magnetic field is a time so Ħ = u Ĥ and Hi can be calculate as (0.5,0.5) - - - - - - I OGåx 110.50) (10) ad, = - ax and ap- ay so ap,= -az xây=-å. PS Jesmo - 0.5 0. = tant (ais) = 450 2, - 1800 -0= 135° t2 = 45° so H. I ( cos 45º – Cos 1350) (-áz) S . 4 axo.5 Hi I (1/2 + 1/2 ) az = - un - 4 az 42 √2 4T X0.5

| ۸۱۲ ه 0 . 3- = 4 = (5ر0 . 5 0 . 5 ) 1 (5 , ۱ ره ت ) (b) (۰٫۱ ,۵) و (ه) - ريه) : (أ) 5 2 = 135° د (هره ) ) ه ه ود * ه (دره ره) °90 = ه و. = ره ها = (اره ره) = (مهره) ۔ (5 , ۱ ره ) - | k so Cups = = &xx azza ap, = å P, = 1, ن = (+5) = | 5x1ہا زهر ارا (در ارا) ) » لم (ن) ا (اره ر(-) = (هره را ) - ((زهره) = م وهو لا ورور او در دوره اهر - ر له = | * * ( 2 )( * | = = - (4 + 4 ) 2583

Similarly (110) (0, 1,0) (id) F₂ = -(ay+az) 82 x 3 P(0,0,1) (iv) and - Hy = 4722 So ---- Poon) (0,0,0) Ft (0,0,1)= o Conselhos they cay tane), ora] 23 F1 (0,0,1) auta (] ax + (the Islay # oz] # (0,0,1)= 0.033 åx + 0,033ây - gougaz Alm (c) at point (0.5, 0.5)) due to symmetry Ĥ = uh where ti is

Pco.s.0.5, 1) 02 (0.5, o,o) (1,0,0) Co, o.s, 1) P = (0.5, 0.5,1 ) - (0.5,0,0) = P = 1 4 - 5 ap, = 2(o.say + a2) a., -ax 25 24 = - * [ 2605 t'] ap, - - & Lay + o. Sáz] only z - component exist so api = - 1az H . unx is - 92 x 2 4TX SX To - 4 az 4 RX 5x 56 F (0.5, 0.5)) = -4a2 = -6.104 az Alm SARTO

75 (1.1) (o, +0.5, 0.5) (0,0) (10) x The total magnatic field at any point is rector Sum due to each current element of square H = H, + Flz + Fiz + Hu and magnetic field due to a finite current element is given as # I (cast-casa, say 56 = xã 1d2