Three infinitely long thin wires, each carrying current i in the same direction, are in the x-y plan

Q: Why do we only check the x-axis for zero magnetic field in part (a)?

By symmetry, the magnetic field components in the x and z directions can only perfectly cancel out on the axes. On the z-axis, the fields from the outer wires have x-components that add up, so the net field cannot be zero except at the origin.

Q: Why are the forces between the wires attractive?

According to Ampere's force law, parallel currents attract each other, while anti-parallel currents repel. Since all three wires carry current in the same direction, they pull on each other.

Q: How do we know the x-components of the forces cancel out in part (b)?

The setup is perfectly symmetric. The central wire is displaced along the z-axis, remaining equidistant from the two outer wires. Therefore, the leftward pull from wire 1 is exactly balanced by the rightward pull from wire 3.

Q: What justifies the approximation $z^2 + d^2 \approx d^2$?

The problem states the wire is displaced by a 'small amount'. In physics, when a quantity $z$ is much smaller than $d$, its square $z^2$ becomes negligibly small compared to $d^2$, allowing us to simplify the math significantly.

Q: What would happen if the central wire carried current in the opposite direction?

The forces would become repulsive. If displaced along the z-axis, the net force would push it further away, resulting in unstable equilibrium rather than simple harmonic motion.

Question

Three infinitely long thin wires, each carrying current $i$ in the same direction, are in the $x$-$y$ plane of a gravity free space. The central wire is along the $y$-axis while the other two are along $x = \pm d$.
(a) Find the locus of the points for which the magnetic field $B$ is zero.
(b) If the central wire is displaced along the $z$-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wires is $\lambda$, find the frequency of oscillation.

Sigma Percentile · Accepted Answer

### The Magnetic Null Points

Imagine the magnetic fields as invisible whirlpools of force swirling around each wire. To find the locus of points where the net magnetic field is zero, we must rely on the principle of superposition. By symmetry, the magnetic field components in the $x$ and $z$ directions can only perfectly cancel out on the axes. Let's analyze the fields on the $x$-axis.

Consider a point at a distance $x$ from the center, between wire 2 and wire 3. The magnetic field due to wires 1 and 2 will point in the negative $z$-direction, while the field from wire 3 points in the positive $z$-direction. For a zero net field, these must balance perfectly:

$$B_1 + B_2 = B_3$$

$$\frac{\mu_0 i}{2\pi (d+x)} + \frac{\mu_0 i}{2\pi x} = \frac{\mu_0 i}{2\pi (d-x)}$$

Canceling the common terms leaves us with a beautiful algebraic equation:

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

Solving this yields $x = \pm \frac{d}{\sqrt{3}}$. Also, exactly at the center, the fields cancel out, giving $x = 0$. Since the wires are infinitely long, these points form lines parallel to the $y$-axis. Thus, the locus is $x = 0, z = 0$ and $x = \pm \frac{d}{\sqrt{3}}, z = 0$.

### The Dynamics of Displacement

Now, let's explore the dynamics. Imagine we displace the central wire slightly along the $z$-axis and release it. The central wire is like a tightrope walker experiencing forces from its neighbors. Because parallel currents attract, the displaced central wire experiences an attractive pull from both outer wires.

Due to the perfect symmetry of the setup, the horizontal $x$-components of these forces will completely cancel each other out. The net force is simply the sum of their vertical $z$-components, pointing downwards towards the origin. Using the formula for the force per unit length between parallel wires, the net restoring force is:

$$F_{net} = -2F \cos	heta = -2 \left( \frac{\mu_0 i^2}{2\pi r} ight) \frac{z}{r}$$

$$F_{net} = -\frac{\mu_0 i^2}{\pi (z^2 + d^2)} z$$

### The Small Approximation and SHM

The beauty of physics lies in how complex forces simplify under small approximations. The problem states the displacement $z$ is very small. This allows us to approximate $z^2 + d^2$ as just $d^2$. The net force becomes directly proportional to the negative displacement:

$$F_{net} \approx -\left( \frac{\mu_0 i^2}{\pi d^2} ight) z$$

This linear restoring force is the hallmark of **Simple Harmonic Motion (SHM)**! Using Newton's second law, we equate the force to mass per unit length ($\lambda$) times acceleration:

$$\lambda a = -\left( \frac{\mu_0 i^2}{\pi d^2} ight) z$$

Comparing this to the standard SHM equation $a = -\omega^2 z$, we find the angular frequency $\omega$:

$$\omega^2 = \frac{\mu_0 i^2}{\pi \lambda d^2}$$

Finally, dividing by $2\pi$ gives the frequency of oscillation:

$$f = \frac{i}{2\pi d} \sqrt{\frac{\mu_0}{\pi \lambda}}$$

Visualized Solution

Visualizing the Setup

Locating Null Points

Balancing the Magnetic Fields

Solving for the Locus

Displacing the Central Wire

Analyzing the Magnetic Forces

Calculating the Net Restoring Force

The Small Displacement Approximation

Frequency of Oscillation

The Way Forward

The Sigma Insight: Magnetic Force on Current

The Magnetic Null Points

The Dynamics of Displacement

The Small Approximation and SHM

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