Question

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

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

Answer 1

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

$MagneticField$ = $B=\frac{{\mu }_{O}l}{2\pi \sqrt{x^2+y^2}}$

Unit vector perpendicular to the position vector is

$⟹\frac{(y\stackrel{\wedge }{i}-x\stackrel{\wedge }{j})}{\sqrt{x^2+y^2}}$

$\left|\overrightarrow{B}\right|=\frac{{\mu }_{O}l}{2\pi (x^2+y^2)}(y\stackrel{\wedge }{i}-x\stackrel{\wedge }{j})$