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If the magnetic field at P can be written as `K tan (alpha/2)`,
image
A. `(mu_0I)/(4pid)`
B. `(mu_0I)/(2pid)`
C. `(mu_0I)/(pid)`
D. `(2mu_0I)/(pid)`

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Correct Answer - B
(b) image
Let us compute the magnetic field due to any one segment:
`B=(mu_0i)/(4pi(d sin alpha)) (cos0^@+cos(180-alpha))`
`=(mu_0I)/(4pi(d sin alpha))(1-cos alpha)=(mu_0I)/(4pid) tan (alpha)/2`
Resultant field will be
`B_(n et)=2B=(mu_0I)/(2pid)tan (alpha)/2 implies K=(mu_0I)/(2pid)`

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