Question

A long straight wire along the z - axis carries a current i in the negative z - direction. The magnetic field $\overrightarrow{B}$ at a point having co - ordinate (x, y) on 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 direction shown in figure.

So, $\overrightarrow{B}=Bsin\theta \hat{i}-Bcos\theta \hat{j}$
Here, $B=\frac{{\mu }_{0}I}{2\pi r}$
$sin\theta =\frac{y}{r}andcos\theta =\frac{x}{r}$
$\therefore \overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi }.\frac{1}{r^2}(y\hat{i}-x\hat{i})=\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$
(as $r^2=x^2+y^2$)