A long straight wire, carrying I is bent at its mid-point to form an angle of `45^@)`. Induction of magnetic field ( in tesla) at point P, distant R from point of bending is equal to

Explanation

Since point P lies on the axis of straight part OA, magnetic field at point P is zero due to part OA of wire

For part OC From `DeltaOPN,d = Rcos45°`
Since both the ends O and C are on the same side of normal PN, `phi_(1)=-45^(@)` and `phi_(2)=+90^(@)`
So `B=(mu_(0)i)/(4pid)(sinphi_(1)+sinphi_(2))`
`=(mu_(0)i)/(4piRcos45^(@))[sin(-45^(@))+sin90^(@)]`
`=(mu_(0)ixxsqrt2)/(4piR)(-sin45^(@)+1)`
`=(sqrt2mu_(0)i)/(4piR)(1-1/(sqrt2))` (radially inward)
`=((sqrt2-1)mu_(0)i)/(4piR)` (radially inward)