Question

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

• A. $\frac{{\mu }_{0}I}{2\pi }\left(\frac{x\hat{i}-y\hat{j}}{x^2+y^2}\right)$
• B. $\frac{{\mu }_{0}I}{2\pi }\left(\frac{x\hat{i}+y\hat{j}}{x^2+y^2}\right)$
• C. $\frac{{\mu }_{0}I}{4\pi }\left(\frac{x\hat{j}-y\hat{i}}{x^2+y^2}\right)$
• D. $\frac{{\mu }_{0}I}{2\pi }\left(\frac{y\hat{i}-x\hat{j}}{x^2+y^2}\right)$

Answer 1

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

The given situation in the question is shown in the below diagram.


From the figure, we get,

$r=\sqrt{x^2+y^2};\sin \theta =\frac{y}{r};\cos \theta =\frac{x}{r}$

We can see that, field at point $\text{P}(x,y)$ will have two components.

The horizontal component of the field $B_x$ is,

$B_x=B\sin \theta \hat{i}=\frac{{\mu }_{0}I}{2\pi \sqrt{x^2+y^2}}\left(\frac{y}{\sqrt{x^2+y^2}}\right)\hat{i}$

$B_x=\frac{\mu Iy}{2\pi (x^2+y^2)}\hat{i}$

The vertical component of the field $B_y$ is,

$B_y=B\cos \theta (-\hat{j})=\frac{{\mu }_{0}I}{2\pi \sqrt{x^2+y^2}}\left(\frac{x}{\sqrt{x^2+y^2}}\right)(-\hat{j})$

$B_y=\frac{\mu Ix}{2\pi (x^2+y^2)}(-\hat{j})$

The resultant field at point $\text{P}(x,y)$ is,

$\overrightarrow{B}=B_x\hat{i}+B_y(-\hat{j})=\frac{{\mu }_{0}I}{2\pi (x^2+y^2)}(y\hat{i}-x\hat{j})$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.