Magnetic fields at two points on the axis of a circular coil at a distance of `0.05m` and `0.2 m` from the centre are in the ratio `8:1`. The radius of the coil is

Options

A. `1.0 m`
B. `0.1 m`
C. `0.15 m`
D. `0.2 m`

Explanation

To solve the problem, we need to determine the radius of a circular coil given the magnetic field strengths at two points along its axis. The magnetic fields at distances of 0.05 m and 0.2 m from the center of the coil are in the ratio of 8:1.

### Step-by-Step Solution:

1. **Understand the Magnetic Field Formula**:
The magnetic field \( B \) at a point on the axis of a circular coil is given by the formula:
\[ B = \frac{\mu_0 I}{2} \cdot \frac{r^2}{(r^2 + x^2)^{3/2}} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( r \) is the radius of the coil, and \( x \) is the distance from the center of the coil.

2. **Set Up the Magnetic Field Equations**:
Let \( B_1 \) be the magnetic field at \( x_1 = 0.05 \, m \) and \( B_2 \) be the magnetic field at \( x_2 = 0.2 \, m \).
\[ B_1 = \frac{\mu_0 I}{2} \cdot \frac{r^2}{(r^2 + (0.05)^2)^{3/2}} \]
\[ B_2 = \frac{\mu_0 I}{2} \cdot \frac{r^2}{(r^2 + (0.2)^2)^{3/2}} \]

3. **Use the Given Ratio**:
The ratio of the magnetic fields is given as:
\[ \frac{B_1}{B_2} = \frac{8}{1} \]
Substituting the expressions for \( B_1 \) and \( B_2 \):
\[ \frac{\frac{\mu_0 I}{2} \cdot \frac{r^2}{(r^2 + (0.05)^2)^{3/2}}}{\frac{\mu_0 I}{2} \cdot \frac{r^2}{(r^2 + (0.2)^2)^{3/2}}} = 8 \]
Simplifying this gives:
\[ \frac{(r^2 + (0.2)^2)^{3/2}}{(r^2 + (0.05)^2)^{3/2}} = 8 \]

4. **Cross-Multiply and Simplify**:
Taking the cube root of both sides:
\[ \frac{(r^2 + (0.2)^2)}{(r^2 + (0.05)^2)} = 2 \]
Cross-multiplying gives:
\[ r^2 + (0.2)^2 = 2(r^2 + (0.05)^2) \]

5. **Expand and Rearrange**:
Expanding the right side:
\[ r^2 + 0.04 = 2r^2 + 2(0.0025) \]
\[ r^2 + 0.04 = 2r^2 + 0.005 \]
Rearranging gives:
\[ 0.04 - 0.005 = 2r^2 - r^2 \]
\[ 0.035 = r^2 \]

6. **Solve for \( r \)**:
\[ r^2 = 0.035 \implies r = \sqrt{0.035} \approx 0.187 \, m \]

### Final Answer:
The radius of the coil is approximately \( 0.187 \, m \).