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. `(4pisigma)/(ln(b//a))V `
B. `(4pisigma)/(b//a) V`
C. `(2pisigma)/(ln(b//a)) V `
D. `(2pisigma)/(b+a) V`

Explanation

To find the current per unit length flowing from one coaxial conducting cylinder to another, we can follow these steps:

### Step 1: Understand the Setup
We have two long coaxial conducting cylinders with inner radius \( a \) and outer radius \( b \). They are separated by a material with conductivity \( \sigma \), and a constant potential difference \( V \) is maintained between them.

### Step 2: Consider a Differential Element
To analyze the current flow, we can consider a thin cylindrical shell of radius \( x \) and thickness \( dx \) between the two cylinders. The length of the cylinder is \( L \).

### Step 3: Calculate the Resistance of the Differential Element
The resistance \( dR \) of the differential element can be expressed as:
\[ dR = \frac{dx}{\sigma \cdot A} \]
where \( A \) is the cross-sectional area through which the current flows. For a cylindrical shell, the area \( A \) is given by:
\[ A = 2 \pi x L \]
Thus, the resistance becomes:
\[ dR = \frac{dx}{\sigma \cdot (2 \pi x L)} = \frac{1}{2 \pi \sigma L} \cdot \frac{dx}{x} \]

### Step 4: Integrate to Find Total Resistance
To find the total resistance \( R \) between the two cylinders, we integrate \( dR \) from \( a \) to \( b \):
\[ R = \int_a^b dR = \int_a^b \frac{1}{2 \pi \sigma L} \cdot \frac{dx}{x} \]
This integral evaluates to:
\[ R = \frac{1}{2 \pi \sigma L} \left[ \ln x \right]_a^b = \frac{1}{2 \pi \sigma L} \left( \ln b - \ln a \right) = \frac{1}{2 \pi \sigma L} \ln \frac{b}{a} \]

### Step 5: Use Ohm's Law to Find Current
According to Ohm's Law, the current \( I \) flowing through the resistance \( R \) with a potential difference \( V \) is given by:
\[ I = \frac{V}{R} \]
Substituting the expression for \( R \):
\[ I = V \cdot \left( \frac{2 \pi \sigma L}{\ln \frac{b}{a}} \right) \]

### Step 6: Find Current per Unit Length
To find the current per unit length \( I' \), we divide the total current \( I \) by the length \( L \):
\[ I' = \frac{I}{L} = \frac{V \cdot 2 \pi \sigma}{\ln \frac{b}{a}} \]

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