Repeat the above problem, but with the current in both wires shown in fig. directed into the plane of the figure.
Derive the expression for the magnitude of `vecB` at any point on the x-axis in terms of the x-coordinate of the point. What is the direciton of `vecB`?

Options

A. `(sqrt2mu_0Ia)/(pi(x^2+a^2))`
B. `(3mu_0Ia)/(pi(x^2+a^2))`
C. `(mu_0Ia)/(2pi(x^2+a^2))`
D. `(mu_0Ix)/(pi(x^2+a^2))`

Explanation

To solve the problem of finding the magnetic field \( \vec{B} \) at any point on the x-axis due to two long parallel wires carrying current into the plane of the figure, we will follow these steps:

### Step 1: Understand the Configuration
We have two long parallel wires, both carrying current \( I \) into the plane of the figure. We need to find the magnetic field at a point \( P \) located on the x-axis at a distance \( x \) from the midpoint between the two wires.

### Step 2: Use the Biot-Savart Law
The magnetic field \( \vec{B} \) due to a long straight wire at a distance \( r \) from the wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 3: Determine the Distance from Each Wire
Let’s denote the distance from the point \( P \) to each wire as \( r_1 \) and \( r_2 \). If the wires are separated by a distance \( 2a \), then:
- The distance from the left wire to point \( P \) is \( r_1 = x + a \)
- The distance from the right wire to point \( P \) is \( r_2 = x - a \)

### Step 4: Calculate the Magnetic Field from Each Wire
Using the formula for the magnetic field:
- The magnetic field \( B_1 \) due to the left wire at point \( P \) is:
\[ B_1 = \frac{\mu_0 I}{2 \pi (x + a)} \]
- The magnetic field \( B_2 \) due to the right wire at point \( P \) is:
\[ B_2 = \frac{\mu_0 I}{2 \pi (x - a)} \]

### Step 5: Determine the Direction of the Magnetic Fields
Using the right-hand rule:
- For the left wire, the magnetic field \( B_1 \) at point \( P \) will be directed out of the plane (positive \( k \)-direction).
- For the right wire, the magnetic field \( B_2 \) at point \( P \) will be directed into the plane (negative \( k \)-direction).

### Step 6: Calculate the Net Magnetic Field
Since \( B_1 \) is directed out of the plane and \( B_2 \) is directed into the plane, the net magnetic field \( B_{\text{net}} \) at point \( P \) will be:
\[ B_{\text{net}} = B_1 - B_2 \]
Substituting the values:
\[ B_{\text{net}} = \frac{\mu_0 I}{2 \pi (x + a)} - \frac{\mu_0 I}{2 \pi (x - a)} \]

### Step 7: Simplify the Expression
To simplify, we can find a common denominator:
\[ B_{\text{net}} = \frac{\mu_0 I}{2 \pi} \left( \frac{(x - a) - (x + a)}{(x + a)(x - a)} \right) \]
This simplifies to:
\[ B_{\text{net}} = \frac{\mu_0 I}{2 \pi} \left( \frac{-2a}{(x + a)(x - a)} \right) \]
Thus:
\[ B_{\text{net}} = -\frac{\mu_0 I a}{\pi (x^2 - a^2)} \]

### Step 8: Final Result
The magnitude of the magnetic field \( \vec{B} \) at any point on the x-axis is:
\[ |\vec{B}| = \frac{\mu_0 I a}{\pi (x^2 - a^2)} \]
The direction of \( \vec{B} \) is in the negative \( k \)-direction (into the plane).