Question

A long wire is carrying current $I$ vertically inwards and placed at the origin. The value of magnetic field at point $\text{P}(x,y)$ due to the wire is

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

Answer 1

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




Magnetic field due to a long straight wire is,

$B=\frac{{\mu }_{0}I}{2\pi r}$

From the figure we get,

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

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

$\overrightarrow{B}=B_x\hat{i}-B_y\hat{j}$

$=B\sin \theta \hat{i}-B\cos \theta \hat{j}$

$=B[\frac{y}{r}\hat{i}-\frac{x}{r}\hat{j}]$

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r^2}[y\hat{i}-x\hat{j}]$

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

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