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.

Figure, shows a closed loop AOCBA in which current I is flowing as shown. Given `OA=OB=OC=a`. Find the magnetic field at point B due to this loop.

Options

A. `-(mu_0I)/(4pi(sqrt2a)) (hati+hatj)`
B. `-(mu_0I)/(4pi(sqrt2a)) (hatj+hati)`
C. `(-mu_0I)/(2(sqrt2pi)) (hatj+hati)`
D. none of these

Answer

B

Explanation

(b)
At B, magnetic field will be zero due to CB and AB parts.

`vecB_(AO)=(mu_0)/(4pi) I/a[sin0^@+sin45^@](-hatj)=(-mu_0I)/(4pisqrt2a)hatj`
`vecB_(OC)=(mu_0)/(4pi) I/a[sin0^@+sin45^@](-hatk)=(-mu_0I)/(4pisqrt2a)hatk`
Net magnetic field at B
`=vecB_(AO)+vecB_(OC)=-(mu_0I)/(4pisqrt2a)(hatj+hatk)`
At O, magnetic field due to AO and OC will be zero.
Due. to BC: `r=acos45^@=a/(sqrt2)`
`vecB_(CB)=(mu_0)/(4pi) I/r[sin45^@+sin45^@](-hatk)=(-mu_0I)/(2pia)hatk`
Similarly, due to BA: `vecB_(BA)=(-mu_0I)/(2pia)hatj`
Net magnetic field `=vecB_(CA)+vecB_(BA)=(-mu_0I)/(2pia)(hatj+hatk)`