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

Options

A. `(mu_(0)I(yhat(i)-xhat(j)))/(2pi(x^(2)+y^(2)))`
B. `(mu_(0)I(xhat(i)+yhat(j)))/(2pi(x^(2)+y^(2)))`
C. `(mu_(0)I(xhat(j)-yhat(i)))/(2pi(x^(2)+y^(2)))`
D. `(mu_(0)I(xhat(i)-yhat(j)))/(2pi(x^(2)+y^(2)))`

Answer

To find the magnetic vector field \( \vec{B} \) at a point with coordinates \( (x, y) \) in the \( z = 0 \) plane due to a long straight wire carrying a current \( I \) in the negative \( z \)-direction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- The wire is oriented along the \( z \)-axis, and the current \( I \) flows in the negative \( z \)-direction.
- We want to find the magnetic field at a point \( P \) located at coordinates \( (x, y) \) in the \( z = 0 \) plane (the \( xy \)-plane).

2. **Use the Right-Hand Rule**:
- According to the right-hand rule, if you point your thumb in the direction of the current (negative \( z \)-direction), your curled fingers will point in the direction of the magnetic field lines.
- Therefore, at point \( P \), the magnetic field \( \vec{B} \) will be directed in the negative \( y \)-direction and positive \( x \)-direction.

3. **Calculate the Magnitude of the Magnetic Field**:
- The formula for the magnetic field \( \vec{B} \) due to a long straight wire at a distance \( r \) from the wire is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- Here, \( r \) is the distance from the wire to the point \( P \), which can be calculated as:
\[ r = \sqrt{x^2 + y^2} \]

4. **Determine the Components of the Magnetic Field**:
- The magnetic field can be expressed in terms of its components along the \( x \) and \( y \) axes.
- The angle \( \theta \) between the position vector \( \vec{r} \) and the axes can be used to find the components:
- \( \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} \)
- \( \sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \)

5. **Express the Magnetic Field in Vector Form**:
- The components of the magnetic field can be expressed as:
\[ B_x = B \cos \theta = \frac{\mu_0 I}{2 \pi \sqrt{x^2 + y^2}} \cdot \frac{x}{\sqrt{x^2 + y^2}} = \frac{\mu_0 I x}{2 \pi (x^2 + y^2)} \]
\[ B_y = -B \sin \theta = -\frac{\mu_0 I}{2 \pi \sqrt{x^2 + y^2}} \cdot \frac{y}{\sqrt{x^2 + y^2}} = -\frac{\mu_0 I y}{2 \pi (x^2 + y^2)} \]

6. **Combine the Components**:
- Thus, the magnetic field vector \( \vec{B} \) at point \( P \) can be written as:
\[ \vec{B} = B_x \hat{i} + B_y \hat{j} = \frac{\mu_0 I x}{2 \pi (x^2 + y^2)} \hat{i} - \frac{\mu_0 I y}{2 \pi (x^2 + y^2)} \hat{j} \]

### Final Result:
The magnetic vector field \( \vec{B} \) at point \( P \) is given by:
\[ \vec{B} = \frac{\mu_0 I}{2 \pi (x^2 + y^2)} (x \hat{i} - y \hat{j}) \]