Question

Problem 3 (5000) Biot-Savart Law: The law states that the magnetic field at a point P due to a length element ds carrying a steady current / located at i away is given by dR - Holds xi _4.1 ds xi 40 p² 403 Question: Derive the Biot-Savart Law by using the vector magnetic potential A (2000). Then, find B at the center of the loop which is an 8(cm) x 6(cm) rectangular conducting loop. The loop lies in the xy-plane. A direct current of 5(A) flows in a clockwise direction viewing from the top (3000).

Answer 1

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form magnetic Leeter potential v%A(7) ---J47)4400 solution of above equation. Axlr) = no (Jxr av 45 Juli Ay(2) de A(7) tas ir de 40, A(7) = the s how for length element de AC) = 411 8 121 = nold d3 ALOQ, 00 0(7): V XALT) otr) - VX 1693 dus no Sox Jo) dr. Out +x:16) - 7()

classmate Date Page For ds elemet Br)= Mol b. dixr Small element I dB = No I 41 do xro 713 Now, magnetic field due to straight Conductor :- = R 2 - 1 I do not de ĝx (RE- 41 y 32 B = 100 deri TR2+) de Rsee odoo la R tamo 7 - Hot csinatsinor 45RL for o chor Bi tof sino 2TR

Dets y = 8 cm n = 60m ISA Que to AB o=45° at center. Ba N. I sinus (-2) TY same for DC so total.se B² 12 16 1 ² similashly. BTAD (21or (-2) Bi= 1200F ( + 1 + 1 th) - fax 98xion 5 | **+/xia24-18) 2 6 GF : 8.25 410 41-5)

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