Hello there. I’m having a very hard time understanding a concept and it’s driving me crazy.lol. I’ve had help with this problem already but I don’t understand the work that you guys completed. I was just hoping that someone could clarify why r^2 becomes just r in the attached problem. I just want to understand the work that you guys did! Thank you so much! CHAP 22 ( 35)A current path shaped as shown in Figure, produces a magnetic field at P, the centerof the arc. If the 'arc subtends an angle ot e = 30° and the radius of the arcis .bm, what are the magnitude and direction of the field produced at P if thecurvent is 3A ?truban OuSE BIOT - SAVART LAWdB = M, Idrs.lomSince Pis on the axis ef the wires,0:0FOR B, B3So sine = Oe= 30°r=.lomI = 3AMo= 4T x 10 T.mA→B=0B2(CUEVED PATH)M. I4n r(4m x 10T.m/A)( 3A)TT3.7.704.524

Arc radius = 0.6 m I = 3A θ = 30° = (π/6)rad To find = Magnitude of field produced at P

Answered: A current path of the arc. If the arc…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Hello there. I’m having a very hard time understanding a concept and it’s driving me crazy.lol. I’ve had help with this problem already but I don’t understand the work that you guys completed. I was just hoping that someone could clarify why r^2 becomes just r in the attached problem. I just want to understand the work that you guys did! Thank you so much!

CHAP 22 ( 35)
A current path shaped as shown in Figure, produces a magnetic field at P, the center
of the arc. If the 'arc subtends an angle ot e = 30° and the radius of the arc
is .bm, what are the magnitude and direction of the field produced at P if the
curvent is 3A ?
truban O
uSE BIOT - SAVART LAW
dB = M, Id
rs.lom
Since Pis on the axis ef the wires,0:0
FOR B, B3
So sine = O
e= 30°
r=.lom
I = 3A
Mo= 4T x 10 T.mA
→B=0
B2
(CUEVED PATH)
M. I
4n r
(4m x 10T.m/A)( 3A)
TT
3.7.70
4.524

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