Question

A very long co-axial cable with inner conductor as a cylindrical shell of radius a and outer conductor as a cylindrical shell of radius 2a carrying equal current each in opposite direction. The region between inner and outer conductor is being filled with material of permeability $\mu = \frac{2\mu_0 a}{a+r}$(r is the distance from central axis). The self inductance per unit length of this cable is $\frac{\mu_0}{\pi}ln(\frac{2}{N})$. Find the value of N.

Answer 1

1.5

Answer 2

To find the self-inductance per unit length of the coaxial cable, we need to calculate the magnetic flux per unit length ($\Phi$) through the region between the conductors and then divide it by the current ($I$).

1. Magnetic Field (B) between the conductors:
   We use Ampere's Law for a circular Amperian loop of radius $r$ (where $a < r < 2a$) concentric with the cable. The current enclosed by this loop is $I$. $\oint \vec{B} \cdot d\vec{l} = \mu I_{enc}$
   Due to symmetry, $\vec{B}$ is tangential and has a constant magnitude $B$ at radius $r$. $B (2\pi r) = \mu I$
   So, $B = \frac{\mu I}{2\pi r}$

2. Substitute the given variable permeability:
   The permeability is given as $\mu = \frac{2\mu_0 a}{a+r}$.
   Substitute this into the expression for $B$:
   $B = \frac{\left(\frac{2\mu_0 a}{a+r}\right) I}{2\pi r} = \frac{2\mu_0 a I}{2\pi r(a+r)} = \frac{\mu_0 a I}{\pi r(a+r)}$

3. Calculate the Magnetic Flux (dΦ) through an elemental area:
   Consider a rectangular strip of length $l$ (for self-inductance per unit length, we take $l=1$) and thickness $dr$ at a distance $r$ from the central axis. The area of this strip is $dA = l dr = dr$.
   The magnetic flux $d\Phi$ through this elemental area is:
   $d\Phi = B \cdot dA = \frac{\mu_0 a I}{\pi r(a+r)} dr$

4. Integrate to find the total magnetic flux (Φ) per unit length:
   The magnetic flux exists in the region between the inner conductor (radius $a$) and the outer conductor (radius $2a$). So, we integrate $d\Phi$ from $r=a$ to $r=2a$: $\Phi = \int_{a}^{2a} \frac{\mu_0 a I}{\pi r(a+r)} dr$
   $\Phi = \frac{\mu_0 a I}{\pi} \int_{a}^{2a} \frac{1}{r(a+r)} dr$

5. Perform the integral using partial fraction decomposition:
   First, decompose the term $\frac{1}{r(a+r)}$:
   $\frac{1}{r(a+r)} = \frac{A}{r} + \frac{B}{a+r}$
   $1 = A(a+r) + Br$
   Setting $r=0 \implies 1 = Aa \implies A = \frac{1}{a}$
   Setting $r=-a \implies 1 = B(-a) \implies B = -\frac{1}{a}$
   So, $\frac{1}{r(a+r)} = \frac{1}{a} \left(\frac{1}{r} - \frac{1}{a+r}\right)$

   Substitute this back into the integral for $\Phi$:
   $\Phi = \frac{\mu_0 a I}{\pi} \int_{a}^{2a} \frac{1}{a} \left(\frac{1}{r} - \frac{1}{a+r}\right) dr$
   $\Phi = \frac{\mu_0 I}{\pi} \int_{a}^{2a} \left(\frac{1}{r} - \frac{1}{a+r}\right) dr$
   Now, integrate:
   $\Phi = \frac{\mu_0 I}{\pi} \left[ \ln|r| - \ln|a+r| \right]_{a}^{2a}$
   $\Phi = \frac{\mu_0 I}{\pi} \left[ \ln\left|\frac{r}{a+r}\right| \right]_{a}^{2a}$

   Evaluate the expression at the limits:
   $\Phi = \frac{\mu_0 I}{\pi} \left[ \ln\left(\frac{2a}{a+2a}\right) - \ln\left(\frac{a}{a+a}\right) \right]$
   $\Phi = \frac{\mu_0 I}{\pi} \left[ \ln\left(\frac{2a}{3a}\right) - \ln\left(\frac{a}{2a}\right) \right]$
   $\Phi = \frac{\mu_0 I}{\pi} \left[ \ln\left(\frac{2}{3}\right) - \ln\left(\frac{1}{2}\right) \right]$

   Using the logarithm property $\ln x - \ln y = \ln(x/y)$:
   $\Phi = \frac{\mu_0 I}{\pi} \ln\left(\frac{2/3}{1/2}\right)$
   $\Phi = \frac{\mu_0 I}{\pi} \ln\left(\frac{2}{3} \times 2\right)$
   $\Phi = \frac{\mu_0 I}{\pi} \ln\left(\frac{4}{3}\right)$

6. Calculate the self-inductance per unit length (L/l):
   The self-inductance per unit length is $L/l = \frac{\Phi}{I}$.
   $L/l = \frac{1}{I} \left( \frac{\mu_0 I}{\pi} \ln\left(\frac{4}{3}\right) \right)$
   $L/l = \frac{\mu_0}{\pi} \ln\left(\frac{4}{3}\right)$

7. Compare with the given expression:
   The problem states that the self-inductance per unit length is $\frac{\mu_0}{\pi}ln(\frac{2}{N})$.
   Comparing our result with the given expression:
   $\frac{\mu_0}{\pi} \ln\left(\frac{4}{3}\right) = \frac{\mu_0}{\pi} \ln\left(\frac{2}{N}\right)$
   This implies:
   $\ln\left(\frac{4}{3}\right) = \ln\left(\frac{2}{N}\right)$
   Therefore,
   $\frac{4}{3} = \frac{2}{N}$
   $4N = 2 \times 3$
   $4N = 6$
   $N = \frac{6}{4} = \frac{3}{2}$

The value of N is 1.5.