Current i flows through a long conducting wire bent at right angle as shown in f

Question

Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at a distance r from O is

Options:

A . μ0iπr

B . 2μ0iπr

C . μ0i4πr(√2+1)

D . μ04π.2ir(√2+1)

Answer: Option D
:
D
By using

B = \frac{μ_{0}}{4 π} . \frac{i}{d} (s i n ϕ_{1} + s i n ϕ_{2})

, from figure

d = r s i n 45^{\circ} = \frac{r}{\sqrt{2}}

Magnetic field due to each wire at P

B = \frac{μ_{0}}{4 π} . \frac{i}{(r / \sqrt{2})} (s i n 45^{\circ} + s i n 90^{\circ})

= \frac{μ_{0}}{4 π} . \frac{i}{r} (\sqrt{2} + 1)

Hence net magnetic field at P

B_{n e t} = 2 \times \frac{μ_{0}}{4 π} . \frac{i}{r} (\sqrt{2} + 1)

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