A long straight wire along the Z-axis carries a current I in the negative z-direction. The magnetic vector field B at a point having coodinates (x, y) in the z = 0 plane is

Options

A. `(mu_(0)I(yhati-xhatj))/(2pi(x^(2)+y^(2)))`
B. `(mu_(0)I(xhati+yhatj))/(2pi(x^(2)+y^(2)))`
C. `(mu_(0)I(xhatj-yhati))/(2pi(x^(2)+y^(2)))`
D. `(mu_(0)I(xhati-yhatj))/(2pi(x^(2)+y^(2)))`

Answer

A

Explanation

Magnetic field at P is B. Perpendicular to OP in the direction shown in figure.
So `B=B sin theta hati-B cos theta hatj` Here `B=(mu_(0))/(2pir) sin theta=Y/r` and `cos theta =x/r`
`:.B=(mu_(0)I)/(2pi) . 1/(r^(2))(yhati-xhati)=(mu_(0)(yhati-xhatj))/(2pi(x^(2)+y^(2))` (as `r^(2)=x^(2)+y^(2))`