Question

A long straight wire along the \(z\)-axis carries a current '\(I\)' in the negative z-direction. The magnetic induction '\(\vec{B}\)' at a point having co-ordinates \((x,y)\) 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. \(\frac{\mu _{0}I(x\hat{i}-y\hat{j})}{2\pi (x^{2}+y^{2})}\)

Hint:

Use the magnitude from the long wire formula and find the direction with the right-hand rule or the cross product.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The magnitude is \(B=\dfrac{\mu_{0}I}{2\pi r}\) with \(r^{2}=x^{2}+y^{2}\). The direction is along \(\hat{I}\times\hat{r}\).

Step 2: Direction
Current direction \(-\hat{k}\) and \(\hat{r}=\dfrac{x\hat{i}+y\hat{j}}{r}\).
\(-\hat{k}\times(x\hat{i}+y\hat{j})=-(x\hat{j}-y\hat{i})=y\hat{i}-x\hat{j}\).

Step 3: Result
\(\vec{B}=\dfrac{\mu_{0}I\,(y\hat{i}-x\hat{j})}{2\pi(x^{2}+y^{2})}\).

Final Answer:
The field is mu zero I times (y i minus x j) over 2 pi (x squared plus y squared). \[ \boxed{\text{(A) }\dfrac{\mu_{0}I(y\hat{i}-x\hat{j})}{2\pi(x^{2}+y^{2})}} \]