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Find magnetic field Bp/Bc=?

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Asked by Vipulgautam050 27th October 2021, 11:47 PM
Answered by Expert
Answer:
Let us consider a small current element of length dl at angular position θ as shown in figure.
 
magnetic field dB at P is given as
 
begin mathsize 14px style d B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction cross times fraction numerator i space stack d l with rightwards arrow on top space cross times r with hat on top over denominator r squared end fraction end style...............................(1)
let us condier dl as arc length ( R dθ ), where dθ is angle subtended by dl at centre C
 
vector begin mathsize 14px style r with rightwards arrow on top end style = ( -R cosθ ) begin mathsize 14px style i with hat on top end style - ( R + R sinθ ) begin mathsize 14px style j with hat on top end style
unit vector begin mathsize 14px style r with hat on top end style = (1/r ) [ ( -R cosθ ) begin mathsize 14px style i with hat on top end style - ( R + R sinθ ) begin mathsize 14px style j with hat on top end style]
where r is magnitude of vector begin mathsize 14px style r with rightwards arrow on top end style
We get magnetic field dB from eqn.(1) as
 
begin mathsize 14px style d B space equals space fraction numerator mu subscript o space i over denominator 4 pi space r squared end fraction cross times space left parenthesis space space stack d l with rightwards arrow on top space cross times r with hat on top space right parenthesis end style.........................(2)
 
begin mathsize 14px style stack d l space with rightwards arrow on top cross times space r with hat on top space equals 1 over r space open vertical bar table row cell i with hat on top end cell cell j with hat on top end cell cell k with hat on top end cell row cell negative R cos theta d theta end cell cell R sin theta d theta end cell 0 row cell negative R cos theta end cell cell negative left parenthesis R sin theta plus R right parenthesis end cell 0 end table close vertical bar space equals space R squared over r left parenthesis space 2 space sin theta space cos theta space plus cos theta space right parenthesis space d theta end style
From above expression of  ( dl × begin mathsize 14px style r with hat on top end style )  , we see that if we integrate ( dl × begin mathsize 14px style r with hat on top end style ) from θ = 0 to θ = π
to get magnetic field due to full semicircular arc , we get zero
 
hence BP = 0
 
Magnetic field BC at centre is determined as
 
begin mathsize 14px style B subscript C space equals space fraction numerator mu subscript o i over denominator 4 pi end fraction integral subscript 0 superscript pi fraction numerator R space d theta over denominator R squared end fraction space equals space fraction numerator mu subscript o i over denominator 4 R end fraction end style
Answered by Expert 28th October 2021, 11:47 AM
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