Question

4. In this problem and in the next, you will calculate the torque on a current loop. In lecture we analysed the torque on a rectangle, while in the recitation you analysed a circle. Here you will consider a current loop in the shape of an equilateral triangle, where each side has a length a. In this problem you will evaluate the torque by directly finding the torque on each side due to a magnetic field B = Bý. Please use the general formula for evaluating the torque about the origin, = 7, Ti = 1 =1/7x(dē, xĒ). To make this calculation easier for you, I shall set up the parametrisation of each side for you, 71 = x'$-ay - a sx's a ose'sa 73 = fa 8 - zt a 9 +e+(+£#+239) Osl' sa 0

Answer 1

mos paroa metrisation of the sides, a non ha -> Q The magnetic field B = B mo x' ; - - asa's a faa chat 9 + (+ os d'<a a (c) 23 mg =

1/3 a 9 - d' (b + para ) 213 of l'isa How, the torque on the ith arom is, I frotx (als x B) 30) 7 = I l m x ( L x B) < How Mi de' att Mix B 1) Z (all 2 x 187) / = 6 cul ñ 1so - I x Now X = da SO) Mix 3 = = B dal SO all 21 I -alz avg_h) x XZ B dal alz BI 51-19 - ) to - alz

→ B ale = BT X! qacha 213 I-ala [-2 BI ( ) B1992 So / TE BI az How n dlá > 2 60° Now, sol = dł" = d' (-27560° + sin soºſ) di (- 12 + 3 % M X de' (-23 + 3 ô x 8 - Boll T Srx (du x B) Z So BI xt 23 x ² di'

t BI 2. - 0 - S + ( del 나 3. 33 0 - 숫 t 3 2 3 BI 2 a BI 3 ( 8 - (-18 - ) + (+3 [ ( ) + (30g + 1 grg(-) S- E BI a BIO 3 BIa 2 ( +④) Howy EP ㅓ 숫 600 7 이 에 (c0860° 숫 - sin60°5 de

SK) > allax B Ball (1 2 + 3 9 ) x B de २ 92 - I S 'X ( x B) E-By In SI 899-1019 số 6) x cư - l' ( - 2 -IB + IB 2 + * ) } del 23 ll IB 2 3 x 213 - IB २ a l-kuġ + 25 11 IB २ { content {l Bagan + to az 13 3a 213 2 + az 4 y â

SO wel BIA² x + 9 3 V3 8 ig TZ Hence the total toroque on the comment loop M + + P BIG² ô tŷ stala + BIA xt 8 213 BIG 23 BIG² 413 BIA2 463 sol total toroque on the comment loop is BIA 473 &