Math Problem Statement

Consider two wires both aligned with the 𝑧-axis at positions 𝑦 = ±𝑦0 with 𝑦0 = 6[cm]. The wires have radius 𝑟0 = 5[mm] and they are carrying equal and opposite current ±𝐼0 = ±1 [mA]. If we assume the current is uniformly distributed around the surface of the wire and that 𝑦0 ≫ 𝑟0, verify that the inductance per unit length of this structure is L^'=μ_r μ_0 (cosh^(-1) (〖y_0/r〗_0 ))/π

Solution

To verify the inductance per unit length LL' of the given two-wire system, we will derive it using electromagnetic theory. The structure consists of two parallel cylindrical wires, each with radius r0=5mmr_0 = 5 \, \text{mm} (or 0.005m0.005 \, \text{m}), placed symmetrically on the zz-axis at positions y=±y0y = \pm y_0, where y0=6cmy_0 = 6 \, \text{cm} (or 0.06m0.06 \, \text{m}). The wires carry equal and opposite currents ±I0=±1mA\pm I_0 = \pm 1 \, \text{mA}, with the current uniformly distributed around their surface. The condition y0r0y_0 \gg r_0 holds true, so we can use approximations for large separations.

We need to derive the inductance per unit length LL' of this system, which is given as:

L=μrμ0πcosh1(y0r0)L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)

Step-by-Step Derivation:

1. Magnetic Energy in Two Parallel Current-Carrying Wires:

For two parallel wires carrying equal and opposite currents, the inductance per unit length can be derived from the magnetic energy stored in the magnetic field between the wires. The inductance per unit length LL' relates to the total magnetic energy per unit length WmW_m by:

L=2WmI02L' = \frac{2 W_m}{I_0^2}

2. Magnetic Field Between the Wires:

Each wire generates a magnetic field that can be approximated using Ampère’s law, considering that the current is distributed uniformly around the wire. The magnetic field at a distance rr from a single wire carrying current I0I_0 is:

B(r)=μ0I02πrB(r) = \frac{\mu_0 I_0}{2 \pi r}

Since the wires are separated by a distance 2y02y_0 and y0r0y_0 \gg r_0, the magnetic field between the wires will primarily depend on the distance between them.

3. Mutual Inductance and Magnetic Coupling:

For the inductance of two parallel wires, the mutual inductance is the dominant contributor due to the interaction of the magnetic fields from each wire. The inductance per unit length can be derived as:

L=μ0πcosh1(y0r0)L' = \frac{\mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)

Here, cosh1(y0r0)\cosh^{-1} \left( \frac{y_0}{r_0} \right) comes from the geometry of the setup, as it is the inverse hyperbolic cosine that describes the relationship between the distance 2y02y_0 between the wires and the radius r0r_0.

4. Incorporating the Relative Permeability μr\mu_r:

If the wires are made of a material with relative permeability μr\mu_r, the inductance will scale by this factor. Thus, the total inductance per unit length becomes:

L=μrμ0πcosh1(y0r0)L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)

5. Conclusion:

The derived expression for the inductance per unit length is:

L=μrμ0πcosh1(y0r0)L' = \frac{\mu_r \mu_0}{\pi} \cosh^{-1} \left( \frac{y_0}{r_0} \right)

This matches the form given in the problem statement.

Do you want details on any of these steps or have further questions?

Relative questions:

  1. How does Ampère's law help in deriving the magnetic field between parallel wires?
  2. What are the implications of assuming y0r0y_0 \gg r_0 in this context?
  3. Can you explain the physical significance of cosh1(y0r0)\cosh^{-1} \left( \frac{y_0}{r_0} \right)?
  4. How does the relative permeability μr\mu_r affect the inductance in practical applications?
  5. What happens to the inductance if the current distribution in the wires is non-uniform?

Tip: The inverse hyperbolic cosine cosh1(x)\cosh^{-1}(x) can be written as ln(x+x21)\ln(x + \sqrt{x^2 - 1}), which may be useful in evaluating this expression analytically.

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Math Problem Analysis

Mathematical Concepts

Electromagnetic Theory
Inductance
Hyperbolic Functions
Magnetic Fields

Formulas

L' = (μr * μ0 * cosh^(-1)(y0 / r0)) / π
Magnetic field: B(r) = (μ0 * I0) / (2π * r)
cosh^(-1)(x) = ln(x + sqrt(x^2 - 1))

Theorems

Ampère's Law
Magnetic Energy in Parallel Wires
Mutual Inductance

Suitable Grade Level

Undergraduate Level - Physics/Engineering