Question

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

• A. $\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$
  
• B. $\frac{{\mu }_{0}I(x\hat{i}+y\hat{j})}{2\pi (x^2+y^2)}$
  
• C. $\frac{{\mu }_{0}I(x\hat{j}-y\hat{i})}{2\pi (x^2+y^2)}$
  
• D. $\frac{{\mu }_{0}I(x\hat{i}-y\hat{j})}{2\pi (x^2+y^2)}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$


Magnetic field at P is $\overrightarrow{B}$ , perpendicular to OP in the direction shown in figure.
So, $\overrightarrow{B}=Bsin\theta \hat{i}-Bcos\theta \hat{j}$
Here $B=\frac{{\mu }_0}{2\pi }\frac{I}{r}$
$sin\theta =\frac{y}{r}$ and $cos\theta =\frac{x}{r}$
$\therefore \overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi }.\frac{1}{r^2}(y\hat{i}-x\hat{j})=\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}(as{r}^2=x^2+y^2)$