According to Biot-Savarat's law, magentic field due to a straight current carrying wire at a point at a distance r form it is given by
`B=(mu_0I)/(4pir)(sinphi_1+sinphi_2)`
The direction of magnetic field being perpendicular to the plane containing the wire and the point.

Find magnetic field at point O in Fig.

Options

A. `-(mu_(0)I)/(4pi sqrt(2)a)(hati+hatk)`
B. `-(mu_(0)I)/(4pi sqrt(2)a)(hati+hatk)`
C. `(-mu_(0)I)/(2sqrt(2)pia)(hatj+hatk)`
D. None of these

Answer

B

Explanation

At B: Magnetic field will be zero due to CB and AB parts
`vecB_(AO)=(mu_(0))/(4pi)I/a[sin 0^(@)+sin 45^(@)](-hatj)=(-mu_(0)I)/(4pi sqrt(2)a)hatj`
`vecB_(OC)=(mu_(0))/(4pi)I/a[sin 0^(@)+sin 45^(@)](-hatk)=(-mu_(0)I)/(4pi sqrt(2)a)hatk`
Net magnetic field at B
`vecB_(AO)+vecB_(OC)=(-mu_(0)I)/(4pisqrt(2)a)(hatj+hatk)`