A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of

Q: Why do we consider force per unit length instead of total force?

The problem states that the wires are 'long', which in physics usually implies infinitely long. An infinitely long wire would have infinite mass and experience infinite total force. By considering the force and mass per unit length, we deal with finite, manageable quantities that yield the exact same physical dynamics.

Q: Why is the binomial approximation necessary here?

The problem states the wire is 'slightly depressed', meaning the displacement $x$ is much smaller than the equilibrium height $d$ ($x \ll d$). The true magnetic force is non-linear, but for very small displacements, we can linearize it using the binomial approximation. This linearity ($F \propto -x$) is the strict mathematical requirement for Simple Harmonic Motion.

Q: Does the current in the wires change when the wire moves?

The problem explicitly states that the wires carry a 'steady current'. This implies that the external power supply is actively maintaining a constant current of 20 A and 30 A, instantly compensating for any motional EMF generated by the wire's movement.

Q: Why is the time period independent of the currents and the mass of the wire?

This is a beautiful mathematical cancellation! The equilibrium condition ($F_m = mg$) intrinsically links the currents and mass to the equilibrium distance $d$. When we calculate the restoring force constant $k$, the currents and mass group together into the term $mg/d$. The mass $m$ then cancels out when calculating acceleration, leaving $\omega^2 = g/d$. Thus, the period depends only on gravity and the equilibrium height!

Question

A long horizontal wire $AB$, which is free to move in a vertical plane and carries a steady current of $20	ext{ A}$, is in equilibrium at a height of $0.01	ext{ m}$ over another parallel long wire $CD$ which is fixed in a horizontal plane and carries a steady current of $30	ext{ A}$, as shown in figure. Show that when $AB$ is slightly depressed, it executes simple harmonic motion. Find the period of oscillations.

Sigma Percentile · Accepted Answer

### The Magic of Levitating Wires

Imagine a wire floating in mid-air, defying gravity. This isn't magic; it's the beautiful interplay of electromagnetism and mechanics. When two parallel wires carry currents in the same direction, they generate magnetic fields that result in an attractive Lorentz force. If we fix the bottom wire and allow the top wire to move freely, we can find a sweet spot where the upward magnetic attraction perfectly balances the downward pull of gravity.

### Balancing Gravity and Magnetism

Let's denote the mass per unit length of the movable wire $AB$ as $m$. At an equilibrium height $d$ above the fixed wire $CD$, the upward magnetic force per unit length is given by Ampere's force law:

$$F_m = \frac{\mu_0 i_1 i_2}{2\pi d}$$

The downward gravitational force per unit length is simply its weight:

$$F_g = mg$$

For the wire to float in equilibrium, these two forces must be equal:

$$\frac{\mu_0 i_1 i_2}{2\pi d} = mg$$

### The Perturbation: What Happens When We Push It Down?

Equilibrium is great, but what happens if we disturb it? Let's slightly depress wire $AB$ by a tiny distance $x$ downwards. The new distance between the wires becomes $(d - x)$. Because the wires are closer, the magnetic attraction becomes stronger, while gravity remains constant. The new upward magnetic force is:

$$F_m' = \frac{\mu_0 i_1 i_2}{2\pi (d-x)}$$

Taking the upward direction as positive, the net force acting on the wire is:

$$F_{	ext{net}} = F_m' - F_g = \frac{\mu_0 i_1 i_2}{2\pi (d-x)} - mg$$

### The Math: Binomial Approximation and SHM

To see how this force behaves, let's factor out $d$ from the denominator:

$$F_{	ext{net}} = \frac{\mu_0 i_1 i_2}{2\pi d} \left(1 - \frac{x}{d}ight)^{-1} - mg$$

Since the displacement $x$ is very small compared to the equilibrium height $d$ ($x \ll d$), we can use the binomial approximation $(1 - z)^{-1} \approx 1 + z$:

$$F_{	ext{net}} \approx \frac{\mu_0 i_1 i_2}{2\pi d} \left(1 + \frac{x}{d}ight) - mg$$

Now, recall our equilibrium condition! The term $\frac{\mu_0 i_1 i_2}{2\pi d}$ is exactly equal to $mg$. Substituting this in, we get:

$$F_{	ext{net}} = mg \left(1 + \frac{x}{d}ight) - mg = \frac{mg}{d} x$$

This net force is directed upwards. However, our displacement $x$ was downwards. This means the force is acting opposite to the displacement, acting as a restoring force:

$$F_{	ext{restoring}} = -\frac{mg}{d} x$$

### Calculating the Time Period

Since the restoring force is directly proportional to the negative of the displacement, the wire executes Simple Harmonic Motion (SHM). The acceleration $a$ is:

$$a = \frac{F_{	ext{restoring}}}{m} = -\left(\frac{g}{d}ight) x$$

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

$$\omega^2 = \frac{g}{d}$$

The time period $T$ of the oscillation is therefore:

$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{d}{g}}$$

Plugging in the given values ($d = 0.01	ext{ m}$ and $g = 9.8	ext{ m/s}^2$):

$$T = 2\pi \sqrt{\frac{0.01}{9.8}} \approx 0.2	ext{ s}$$

And there we have it! The wire will bob up and down, completing one full cycle every 0.2 seconds.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Equilibrium Condition$

$Perturbing the System$

$Calculating Net Force$

$Binomial Approximation$

$Proving SHM$

$Time Period Calculation$

The Sigma Insight: Magnetic Force on Current

The Magic of Levitating Wires

Balancing Gravity and Magnetism

The Perturbation

The Math

Calculating the Time Period

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Perturbing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Net Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binomial Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Proving SHM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period Calculation