Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity `sigma` and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is

Options

A. `(4pi sigma)/(ln (b//a) )V `
B. `(4pi sigma)/((b+a) ) V`
C. `(2pi sigma)/(ln (b//a))V`
D. `(2pi sigma)/((b+a))V`

Answer

To find the current per unit length flowing between two coaxial conducting cylinders separated by a material of conductivity \(\sigma\), we can follow these steps:

### Step 1: Understand the Setup
We have two long coaxial cylinders:
- Inner cylinder radius = \(a\)
- Outer cylinder radius = \(b\)
- Conductivity of the material between the cylinders = \(\sigma\)
- Potential difference maintained between the cylinders = \(V\)

### Step 2: Define the Resistance
The resistance \(R\) of the cylindrical shell between the two cylinders can be calculated using the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Where:
- \(\rho\) is the resistivity of the material, which is given by \(\rho = \frac{1}{\sigma}\)
- \(L\) is the length of the cylinder
- \(A\) is the cross-sectional area

### Step 3: Calculate the Cross-Sectional Area
For a cylindrical shell at a distance \(x\) from the center, the thickness is \(dx\) and the area \(A\) is given by the circumference of the cylinder multiplied by the thickness:
\[ A = 2\pi x \cdot L \]

### Step 4: Calculate the Resistance of a Differential Element
The differential resistance \(dR\) of a thin cylindrical shell of thickness \(dx\) at radius \(x\) is:
\[ dR = \frac{\rho \cdot dx}{2\pi x L} \]
Substituting \(\rho = \frac{1}{\sigma}\):
\[ dR = \frac{1}{\sigma} \cdot \frac{dx}{2\pi x L} \]

### Step 5: Integrate to Find Total Resistance
To find the total resistance \(R\) from \(x = a\) to \(x = b\), we integrate:
\[ R = \int_{a}^{b} dR = \int_{a}^{b} \frac{1}{\sigma} \cdot \frac{dx}{2\pi x L} \]
This simplifies to:
\[ R = \frac{1}{\sigma \cdot 2\pi L} \int_{a}^{b} \frac{dx}{x} \]
The integral \(\int \frac{dx}{x} = \ln x\), so:
\[ R = \frac{1}{\sigma \cdot 2\pi L} \left[ \ln b - \ln a \right] = \frac{1}{\sigma \cdot 2\pi L} \ln \frac{b}{a} \]

### Step 6: Calculate the Current
Using Ohm's law, the current \(I\) flowing through the resistance \(R\) when a potential difference \(V\) is applied is:
\[ I = \frac{V}{R} \]
Substituting the expression for \(R\):
\[ I = V \cdot \sigma \cdot 2\pi L \cdot \frac{1}{\ln \frac{b}{a}} \]

### Step 7: Calculate Current per Unit Length
To find the current per unit length \( \frac{I}{L} \):
\[ \frac{I}{L} = \frac{V \cdot \sigma \cdot 2\pi}{\ln \frac{b}{a}} \]

### Final Result
Thus, the current per unit length flowing from one cylinder to the other is:
\[ \frac{I}{L} = \frac{V \cdot \sigma \cdot 2\pi}{\ln \frac{b}{a}} \]