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. `2//3`
B. `1//2`
C. `3//2`
D. 2

Answer

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 Inside and Outside the Wire
For a long straight wire carrying current \( I \):
- Inside the wire (for \( r < a \)), the magnetic field \( B \) is given by:
\[ B = \frac{\mu_0 I}{2\pi a^2} r \]
where \( r \) is the distance from the center of the wire.

- Outside the wire (for \( r \geq a \)), 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 formula for the magnetic field inside the wire:
\[ B_1 = \frac{\mu_0 I}{2\pi a^2} \left(\frac{a}{3}\right) = \frac{\mu_0 I}{2\pi a^2} \cdot \frac{a}{3} = \frac{\mu_0 I}{6\pi a} \]

### Step 3: Calculate the Magnetic Field at \( r = 2a \)
Since \( 2a > a \), we use the formula for the magnetic field outside the wire:
\[ B_2 = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a} \]

### Step 4: Find the Ratio of the Magnetic Fields
Now, we can find the ratio of \( B_1 \) to \( B_2 \):
\[ \text{Ratio} = \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{6\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1/6}{1/4} = \frac{4}{6} = \frac{2}{3} \]

### Final Answer
Thus, the ratio of the magnetic field at a distance \( \frac{a}{3} \) from the axis of the wire to the magnetic field at a distance \( 2a \) is:
\[ \frac{B_1}{B_2} = \frac{2}{3} \]