Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of magnetic field at distance `a/3` and 2a from axis of wire is.

Options

A. `3/5`
B. `2/3`
C. `1/2`
D. `4/3`

Explanation

To solve the problem of finding the ratio of the magnetic field at distances \( \frac{a}{3} \) and \( 2a \) from the axis of an infinitely long current-carrying cylindrical wire, we can follow these steps:

### Step 1: Understand the magnetic field equations
For an infinitely long cylindrical wire:
- If the distance \( r \) is less than the radius \( a \) of the wire, the magnetic field \( B \) is given by:
\[ B = \frac{\mu_0 I r}{2 \pi a^2} \]
- If the distance \( r \) is greater than the radius \( a \), the magnetic field \( B \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]

### Step 2: Calculate the magnetic field at \( r = \frac{a}{3} \)
Since \( \frac{a}{3} < a \), we use the first equation:
\[ B_1 = \frac{\mu_0 I \left(\frac{a}{3}\right)}{2 \pi a^2} \]
Simplifying this:
\[ B_1 = \frac{\mu_0 I}{2 \pi a^2} \cdot \frac{a}{3} = \frac{\mu_0 I}{6 \pi a} \]

### Step 3: Calculate the magnetic field at \( r = 2a \)
Since \( 2a > a \), we use the second equation:
\[ B_2 = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a} \]

### Step 4: Find the ratio of the magnetic fields
Now, we need to find the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{6 \pi a}}{\frac{\mu_0 I}{4 \pi a}} \]
This simplifies to:
\[ \frac{B_1}{B_2} = \frac{1}{6} \cdot \frac{4}{1} = \frac{4}{6} = \frac{2}{3} \]

### Conclusion
The ratio of the magnetic field at distance \( \frac{a}{3} \) to the magnetic field at distance \( 2a \) is:
\[ \frac{B_1}{B_2} = \frac{2}{3} \]