Magnetic Field from a Single Wire

An infinitely long straight wire carrying current produces a magnetic field around it. The magnitude of the magnetic field at a distance \(r\) from the wire is given by the formula:

$$B = \frac{\mu_0 I}{2\pi r}$$

where \(\mu_0\) is the permeability of free space and \(I\) is the current in the wire.

For a wire placed along the z-axis carrying current \(I\) in the +z direction, the magnetic field in the xy-plane circulates around the wire in a counter-clockwise direction (according to the right-hand rule). At a point \((x, 0)\) on the x-axis, the magnetic field is directed along the y-axis.

If the wire is located at \(x_0\) on the x-axis (and extends infinitely in the z direction), the magnetic field at a point \((x, 0)\) on the x-axis due to this wire carrying current \(I\) in the +z direction has a y-component given by:

$$B_y(x) = \frac{\mu_0 I}{2\pi (x-x_0)}$$

This formula automatically accounts for the direction: if \(x > x_0\), \(x-x_0\) is positive and \(B_y\) is positive (field in +y direction); if \(x < x_0\), \(x-x_0\) is negative and \(B_y\) is negative (field in -y direction).

Total Magnetic Field on the x-axis

We have three infinitely long wires in the xy-plane along \(x = -d\), \(x = 0\), and \(x = +d\). Each wire carries an equal current \(I\). We assume the currents are all in the same direction, say +z.

The total magnetic field at a point \((x, 0)\) on the x-axis is the vector sum of the magnetic fields due to each wire. Since all wires are parallel to the z-axis and we are considering points on the x-axis, the magnetic field from each wire will be along the y-axis. Therefore, the total magnetic field at \((x, 0)\) is \(\vec{B}_{total}(x) = B_{total,y}(x) \hat{j}\).

The y-component of the total magnetic field is the sum of the y-components from the three wires:

$$B_{total,y}(x) = B_{-d,y}(x) + B_{0,y}(x) + B_{d,y}(x)$$

Using the formula for the y-component from a single wire:

• Wire at \(x_0 = -d\): \(B_{-d,y}(x) = \frac{\mu_0 I}{2\pi (x - (-d))} = \frac{\mu_0 I}{2\pi (x+d)}\)
• Wire at \(x_0 = 0\): \(B_{0,y}(x) = \frac{\mu_0 I}{2\pi (x - 0)} = \frac{\mu_0 I}{2\pi x}\)
• Wire at \(x_0 = +d\): \(B_{d,y}(x) = \frac{\mu_0 I}{2\pi (x - d)}\)

Summing these components gives the total y-component:

$$B_{total,y}(x) = \frac{\mu_0 I}{2\pi (x+d)} + \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi (x-d)}$$

$$B_{total,y}(x) = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x+d} + \frac{1}{x} + \frac{1}{x-d} \right)$$

This expression is valid for \(x \neq -d, 0, d\).

Finding Points Where Field Vanishes

The magnetic field vanishes at points where \(B_{total,y}(x) = 0\). This requires the term in the parenthesis to be zero:

$$\frac{1}{x+d} + \frac{1}{x} + \frac{1}{x-d} = 0$$

Solving the Equation

To solve the equation, we can find a common denominator:

$$\frac{x(x-d) + (x+d)(x-d) + x(x+d)}{x(x+d)(x-d)} = 0$$

For the fraction to be zero, the numerator must be zero (provided the denominator is not zero). The denominator \(x(x+d)(x-d)\) is non-zero at the points we are looking for, as they cannot coincide with the wire locations.

Set the numerator to zero:

$$x(x-d) + (x^2 - d^2) + x(x+d) = 0$$

Expand the terms:

$$x^2 - xd + x^2 - d^2 + x^2 + xd = 0$$

Combine like terms:

$$(x^2 + x^2 + x^2) + (-xd + xd) - d^2 = 0$$

$$3x^2 - d^2 = 0$$

Now, solve for \(x\):

$$3x^2 = d^2$$

$$x^2 = \frac{d^2}{3}$$

Take the square root of both sides:

$$x = \pm \sqrt{\frac{d^2}{3}}$$

$$x = \pm \frac{d}{\sqrt{3}}$$

These points \(x = \frac{d}{\sqrt{3}}\) and \(x = -\frac{d}{\sqrt{3}}\) are located between the wires at \(x=-d\), \(x=0\), and \(x=d\) (since \(0 < 1/\sqrt{3} < 1\)), specifically in the regions \(-d < x < 0\) and \(0 < x < d\), where field cancellation is possible.

Thus, the magnetic field vanishes at \(x = \pm \frac{d}{\sqrt{3}}\) on the xy-plane (specifically, on the x-axis).