Question

4

Problem 4: Each of the 3 wires carries 4 A in the directions shown in the figure. The value of B at point P is: 4.00 A out 2.00 cm 4.00 A out 2.00 cm 4.00 A out B) 4.42 x 10-6 T C) 1.77 x 10-6 T D) 1.77 x 10-5 T A) 5.66 x 10-5 T E) 1.26 x 10-4 T

Answer 1

(0,2,0) → Direchon of ☺ i Coming out of paper or board. Om, the choosen Co-ordinale System shown in dug 2, is +7 derechion tó P (1, 1,0) 1 6 =) qz derechos X (0,0,0) A (2,0,0) →X, Z Fig2 as Point p in the midpoint figi of AB and since AY OAB is right Angle Isosceles triangle point p is at the mid point of AB and given P11,10) → The magnehe field density 8 due to infinielly Jong which can canay current I is given by lo I QR k AN where a Distance ble wire and point AN Direchon of B Dinechon of B is detexmined by Cross booduct of direchon of Current and line joining the wire to the point le Scalmed vath dinechon of Current X direchon Current wire to point. wire con A of line joining CamScanner 2

Uo 1 A 4 Distance blw Jowo point (as, 40, 81) 8 (20.42,32) given as d = 62-732 +42-4,)? +(2-8,)2 and equahon of line is given as 941,3) (22, 92, 32) = 293-20) cia + (46-4,) ay +(32-3,798 Using Above equahon 2 =V12+12 Distance AP1 = 761-2) +(1-0)2 +0 (5 1BP! (1-0)2 + (1-2)2 +02 12.cm Equahon of line Art (1-2)(a + (1-0) ag Ā charged wit_o) a + (1-2) ag 12 cm CH = 1 A ay A an - ay

(3 ONI ay 1 Direchon of Magnehi field due to A with help of efn © ağ x l-ai taý ani ağ - ar as x a smidaly Direchon of B dure to B q8 xl ar - ağ 6 ggrot = ang - C 2 OM2 1 ANE ay ta Now B due to Current wine at A , with the help of and © л T am -Qy - QQ BA = Mo da Ti2+ 12 Mo T = da ra (-ay - ) -a-an 2 roo BA ga lo I [ Qâ + aģ) B due do Current at B Dion'larly Bg Setel T (09 tarp z lo la + ay) - Ja Combined me mag field at Point P due do Current wire at locahon A and B Mo BA+ BB = I BAB ch-aș + ane tau at 2 BAB =0 10 CE Scanned with CamScanner

Eqn 16 show that mag field B at point P is due to wire at origin only as as they both cancel ou each other . 1 bl at point P due to wire at point to I lo 1Bl= 11 lopi TOPI Now no need to worry aboul duechon, as we have to calcedate magnitude only. (1-0)2 + (1-0)2 v2 cm = r2x10-2m Now Mo= 4K X10 7 Pesme a rebility of free space. I 4A all values in eqh @ 4* X1007 Qa X 10 810-5 12 5.66 x10-5 Pulting 1812 X T2 X10-9 N Ans T 1B1 7 Socenophon Ä is comed ophon. CamScanner