A circular loop of radius r carrying current `IA`. The radio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its aixs is :

Options

A. `1:3sqrt2`
B. `2sqrt 2:1`
C. `sqrt2:2`
D. `1:sqrt2`

Answer

To solve the problem of finding the ratio of the magnetic field at the center of a circular loop and at a distance \( r \) from the center of the loop on its axis, we will follow these steps:

### Step 1: Magnetic Field at the Center of the Loop
The magnetic field \( B_1 \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B_1 = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space and \( r \) is the radius of the loop.

### Step 2: Magnetic Field on the Axis of the Loop
Next, we calculate the magnetic field \( B_2 \) at a distance \( r \) from the center of the loop along its axis. The formula for the magnetic field at a distance \( z \) (where \( z = r \) in this case) from the center of the loop on its axis is:
\[ B_2 = \frac{\mu_0 I r^2}{2(r^2 + z^2)^{3/2}} \]
Substituting \( z = r \):
\[ B_2 = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2^{3/2} r^3)} = \frac{\mu_0 I r^2}{2 \cdot 2\sqrt{2} r^3} \]
This simplifies to:
\[ B_2 = \frac{\mu_0 I}{4\sqrt{2} r} \]

### Step 3: Finding the Ratio of Magnetic Fields
Now, we find the ratio of \( B_1 \) to \( B_2 \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2} r}} = \frac{\frac{1}{2}}{\frac{1}{4\sqrt{2}}} \]
This simplifies to:
\[ \frac{B_1}{B_2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Final Answer
Thus, the ratio of the magnetic field at the center of the circular loop to the magnetic field at a distance \( r \) from the center on its axis is:
\[ \frac{B_1}{B_2} = 2\sqrt{2} \]