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Find magnetic field Bp/Bc=?

Asked by Vipulgautam050 | 27 Oct, 2021, 11:47: PM

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 B_P = 0

Magnetic field B_C 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 Thiyagarajan K | 28 Oct, 2021, 11:47: AM

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