Question

Assuming a current I in a semi-circular wire with radius a. What is the magnetic flux density on a point at the center of the semi-circle and above the semi-circle plane by a distance d. Also find the magnetic flux density if the circle is full.

4. Assuming a current I in a semi-circular wire with radius a. What is the magnetic flux density on a point at the center of the semi-circle and above the semi-circle plane by a distance d. Also find the magnetic flux density if the circle is full.

Answer 1

-- - Semy Now) dB = a no Idd 41 82 I doi= Mo Idesino we know that 4T 82 do - Mo Iditar i Horia sino = L 41 R12 !! 0 = 90 T ãą z onit veetor Current element is chave del Mo Ide sino 4X R² a part of ciruar wire. Therefore the angle o 1 Sino = angle between as gou which is tangent ! the cell. cevrent - to the bircle] I Ap element & the P... Lire joining p to the element. Integrate on both / direction is given by 1 Right Hand rule JAB = . lo i de pro I. KG X Fa ? " X a gts dibreetton 4a is outward and denoted by Cirilo with dot.ie. ☺ B = fo FOT side. 82 r= a . 40 C Scanned with CamScanner

DAHULU density (3) at the center of magnetic fleex full wrebe IdBl= Mo Ide sino 4T 82 | Thi sino = 1 0=90 Cuorent element is tangent to the lirular wire. So o is goo 248. Mo ide uit 1 dB J 4T 82 dB - 28 76 rra B = MOLT 2 a 7² 760,014) i diz (n=al here atzea eta ã or(0,0, he di magnetic flux density at a distance de 2B = Mo Idex AT IRS CScanned with CamScanner

where do = P doão R = 10,0,d) -(x,y,o) k= fapt blåz 1R1 = (p2 +62) de x R = ap ad azi To edo o ttaa 2Ã = pbbdo ar + P2 do âz Hente de los [P-d-dø af +pdoâz] . 4* (d2 +82) 32 ] dB = dBp âp 4 dbzâz LAB. a C A9I LÝ: ddd a1 a. map] ap = cose case - 42 (d2 +682 + sinpay | || Beat - HĐ1 [P, densº dº +sind dº 41 44 (424/ => post for [+0] =>0 dBp =0 =) only me lomponent be present Therefore, dB = dBzaz foL P² doâril 2 4 4² + p 27% CSScanned with CamScanner

2 24 Грох – dФа, 41 (4”- е охрана, 2 (440% В - Ког ф2 — и x 4°Ру here eza в юга“ или 244-+aj% CScanned with CamScanner