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

 

  • Option 1)

    \frac{\mu 0I\left (yi^{\wedge }-xj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

     

  • Option 2)

    \frac{\mu 0I\left (xi^{\wedge }-yj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

  • Option 3)

    \frac{\mu 0I\left (xj^{\wedge }-yi^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

  • Option 4)

    \frac{\mu 0I\left (xi^{\wedge }-yj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

     

 

Answers (1)

As we learnt in 

For Infinite Length -

\phi_{1}=\phi_{2}=90^{\circ}

B=\frac{\mu_{o}}{4\pi}\:\frac{2i}{r}\:

- wherein

 

Magnitude of magnetic field =\frac{\mu _oI}{2 \pi r}=\frac{\mu _oI}{2 \pi (x^{2}+y^{2})^{1/2}}

Direction of field along tangential direction=\frac{y\hat{i}{-x\hat{j}}}{\sqrt{x^{2}+y^{2}}}

\therefore \vec{B}=\frac{\mu_oI}{2\pi (x^{2}+y^{2})}.(y\hat{i}{ -x\hat{j}})


Option 1)

\frac{\mu 0I\left (yi^{\wedge }-xj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

 

correct

Option 2)

\frac{\mu 0I\left (xi^{\wedge }-yj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

incorrect

Option 3)

\frac{\mu 0I\left (xj^{\wedge }-yi^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

incorrect

Option 4)

\frac{\mu 0I\left (xi^{\wedge }-yj^{\wedge }\right )}{2\pi\left ( x^{2}+y^{2} \right )}

 

incorrect

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