Repeat the above problem, but with the current in both wires shown in fig. directed into the plane of the figure.
At what value of x is the magnitude of `vecB` is maximum?

Options

A. `x=0`
B. `x=sqrt2a`
C. `x=+-a//2`
D. `x=+-a`

Explanation

To solve the problem of finding the value of \( x \) at which the magnitude of the magnetic field \( \vec{B} \) is maximum when the currents in both wires are directed into the plane of the figure, we can follow these steps:

### Step 1: Understand the Magnetic Field Contribution
The magnetic field \( \vec{B} \) at a point due to a long straight current-carrying wire is given by the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the distance from the wire to the point where the magnetic field is being calculated.

### Step 2: Set Up the Problem
Assuming we have two parallel wires separated by a distance \( 2a \) and carrying currents \( I \) into the plane, we need to find the total magnetic field at a point \( P \) located at a distance \( x \) from the midpoint between the two wires.

### Step 3: Calculate the Magnetic Field at Point P
The distance from point \( P \) to each wire will be:
- For the wire on the left: \( r_1 = x + a \)
- For the wire on the right: \( r_2 = x - a \)

The magnetic fields due to each wire at point \( P \) will be:
\[ B_1 = \frac{\mu_0 I}{2\pi (x + a)} \quad \text{(from the left wire)} \]
\[ B_2 = \frac{\mu_0 I}{2\pi (x - a)} \quad \text{(from the right wire)} \]

Since both currents are directed into the plane, the magnetic fields at point \( P \) will add up:
\[ B = B_1 + B_2 = \frac{\mu_0 I}{2\pi (x + a)} + \frac{\mu_0 I}{2\pi (x - a)} \]

### Step 4: Simplify the Expression for B
Combining the two terms:
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x + a} + \frac{1}{x - a} \right) \]
Finding a common denominator:
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{(x - a) + (x + a)}{(x + a)(x - a)} \right) = \frac{\mu_0 I}{2\pi} \left( \frac{2x}{x^2 - a^2} \right) \]
Thus,
\[ B = \frac{\mu_0 I x}{\pi (x^2 - a^2)} \]

### Step 5: Find the Maximum Value of B
To find the maximum value of \( B \), we need to differentiate \( B \) with respect to \( x \) and set the derivative equal to zero:
\[ \frac{dB}{dx} = 0 \]
Using the quotient rule:
\[ \frac{dB}{dx} = \frac{\pi (x^2 - a^2) \cdot \frac{\mu_0 I}{\pi} - \mu_0 I x \cdot 2x}{(x^2 - a^2)^2} \]
Setting the numerator equal to zero:
\[ (x^2 - a^2) - 2x^2 = 0 \implies -x^2 + a^2 = 0 \implies x^2 = a^2 \]
Thus,
\[ x = \pm a \]

### Step 6: Determine the Maximum Value
Since we are interested in the maximum value of \( B \), we evaluate at \( x = a \):
- At \( x = a \), the magnetic field will be maximum.

### Final Answer
The value of \( x \) at which the magnitude of \( \vec{B} \) is maximum is:
\[ \boxed{a} \]