JEE Class main Answered
Find magnetic field Bp/Bc=?
![question image](http://images.topperlearning.com/topper/new-ate/102906IMG20211027WA0002.jpg)
Asked by Vipulgautam050 | 27 Oct, 2021, 23:47: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/7ea843daefb782c0e878e58027d3cf48617a3a19869766.27886842f5.png)
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](https://images.topperlearning.com/topper/tinymce/cache/8c889777ebe1c7b980f3db24ef954906.png)
let us condier dl as arc length ( R dθ ), where dθ is angle subtended by dl at centre C
vector
= ( -R cosθ )
- ( R + R sinθ )
![begin mathsize 14px style r with rightwards arrow on top end style](https://images.topperlearning.com/topper/tinymce/cache/18d9063f1aa24fc3e615246827a59066.png)
![begin mathsize 14px style i with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/a5a79159fc1a127cc872f245943c4834.png)
![begin mathsize 14px style j with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/ef647306fae7d3b52276fc8b581df5b6.png)
unit vector
= (1/r ) [ ( -R cosθ )
- ( R + R sinθ )
]
![begin mathsize 14px style r with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/57111c07fb27191d059d2c555091dada.png)
![begin mathsize 14px style i with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/a5a79159fc1a127cc872f245943c4834.png)
![begin mathsize 14px style j with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/ef647306fae7d3b52276fc8b581df5b6.png)
where r is magnitude of vector ![begin mathsize 14px style r with rightwards arrow on top end style](https://images.topperlearning.com/topper/tinymce/cache/18d9063f1aa24fc3e615246827a59066.png)
![begin mathsize 14px style r with rightwards arrow on top end style](https://images.topperlearning.com/topper/tinymce/cache/18d9063f1aa24fc3e615246827a59066.png)
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](https://images.topperlearning.com/topper/tinymce/cache/9fc9bca98545296974161c1c1026e575.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/db3c72fa53fd2515aa35a72c2ba1fe81.png)
From above expression of ( dl ×
) , we see that if we integrate ( dl ×
) from θ = 0 to θ = π
![begin mathsize 14px style r with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/57111c07fb27191d059d2c555091dada.png)
![begin mathsize 14px style r with hat on top end style](https://images.topperlearning.com/topper/tinymce/cache/57111c07fb27191d059d2c555091dada.png)
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](https://images.topperlearning.com/topper/tinymce/cache/c68844de88afeccabc14b4fa727cdb2c.png)
Answered by Thiyagarajan K | 28 Oct, 2021, 11:47: AM
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