Question

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:

• A. \(\dfrac{\mu_0 I(\hat{i}-\hat{j})}{2\pi(x^2+y^2)}\)
• B. \(\dfrac{\mu_0 I(\hat{i}+\hat{j})}{2\pi(x^2+y^2)}\)
• C. \(\dfrac{\mu_0 I(y\hat{i}-x\hat{j})}{2\pi(x^2+y^2)}\)
• D. \(\dfrac{\mu_0 I(x\hat{i}-y\hat{j})}{2\pi(x^2+y^2)}\)

Hint:

For long straight wire: \[ \vec{B}\propto \hat{\phi} \] Use right-hand thumb rule for direction.

Answer 1

The Correct Option is C

Step 1: Magnetic field magnitude: \[ B=\frac{\mu_0 I}{2\pi r} \]
Step 2: Direction is given by right-hand rule (clockwise for current along \(-Z\)).
Step 3: Unit vector: \[ \hat{\phi}=\frac{y\hat{i}-x\hat{j}}{\sqrt{x^2+y^2}} \]
Step 4: Hence: \[ \vec{B}=\frac{\mu_0 I}{2\pi(x^2+y^2)}(y\hat{i}-x\hat{j}) \]