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)/(4pia)(hati+hatk)`
B. `-(mu_(0)I)/(4pia)(hatj+hatk)`
C. `(-mu_(0)I)/(2pia)(hatj+hatk)`
D. `(-mu_(0)I)/(2sqrt(2)pia)(hatj+hatk)`

Answer

C

Explanation

At O : Magnetic field due to AO and OC will be zero.
Due to `CB:r=a cos 45^(@)=a/(sqrt(2))`
`vecB_(CB)=(mu_(0))/(4pi)I/r[sin 45^(@)+sin45^(@)](-hatk)=(-mu_(0)I)/(2pia)hatk`
similarly, `vecB_(BA)=(-mu_(0))/(2pi)I/a hatj :. vecB_("net")=vecB_(CB)+vecB_(BA)=(-mu_(0))/(2pi) I/a(hatj+hatk)`