A long straight wire along the `z`- axis carries a current ` I` in the negative ` z- direction` . The magnetic vector field ` vec(B)` at a point having coordinates (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 `vecB`, perpendicular to `OP` in the direction shown in figure.

So, `vecB=Bsin thetahati-Bcos thetahatj`
Here, `B=(mu_(0))/(2pi) I/r`
`sin theta=y/r` and `cos theta=x/r`
`:. vecB=(mu_(0)I)/(2pi).1/(r^(2))(yhati-xhatj)=(mu_(0)I(yhati-xhatj))/(2pi(x^(2)+y^(2)))`
(as `r^(2)=x^(2)+y^(2)`)