Find the magnetic field at `P` due to the arrangement shown
.

Options

A. `(mu_0i)/(sqrt2pid)(1-(1)/sqrt2)ox`
B. `(2mu_(0)i)/(sqrt2pid)ox`
C. `(2mu_(0)i)/(sqrt2pid)ox`
D. `(mu_(0)1)/(sqrt2pi)(1+(1)/(sqrt2))ox`

Answer

A

Explanation

`(x)/(d) = sin45^(@) implies x = (d)/(sqrt2)`
we can extend wire and then subtract the field of extended part
`B_(1) = (mu_(0))/(4pi) (I)/(d//sqrt2) (sin 0^(@) + sin 90^(@))-`
`= (mu_(0))/(4pi) (I)/(d//sqrt2) (sin 0^(@) + sin 45^(@))`
`(mu_(0))/(4pi) (sqrt2I)/(d)(1 - (1)/(sqrt2)) = (mu_0I)/(2sqrt2pid)(1 -(1)/sqrt2)ox`
`B_(1)ox, B_(2)ox`
`B = (B_(1) + B_(2)) ox = 2B_(1) ox = (mu_(0)I)/(sqrt2 pid)(1-(1)/(sqrt2))ox`
.