The magnetic induction at the centre of a current carrying circular coil of radius `10cm` is `5sqrt(5)` times the magnetic induction at a point on its axis. The distance of the point from the centre of the soild in `cm` is

Answer

To solve the problem, we need to find the distance \( x \) from the center of a current-carrying circular coil where the magnetic induction is given in relation to the magnetic induction at the center of the coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the coil, \( R = 10 \, \text{cm} \)
- Magnetic induction at the center, \( B_{\text{center}} \)
- Magnetic induction at a point on the axis, \( B_{\text{axis}} \)
- Relation: \( B_{\text{center}} = 5\sqrt{5} \times B_{\text{axis}} \)

2. **Write the formula for magnetic induction at the center of the coil:**
\[ B_{\text{center}} = \frac{\mu_0 I}{2R} \]

3. **Write the formula for magnetic induction at a distance \( x \) on the axis of the coil:**
\[ B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

4. **Set up the equation using the given relation:**
\[ \frac{\mu_0 I}{2R} = 5\sqrt{5} \cdot \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

5. **Cancel common terms:**
- Cancel \( \mu_0 I \) and \( 2 \) from both sides:
\[ \frac{1}{R} = 5\sqrt{5} \cdot \frac{R^2}{(R^2 + x^2)^{3/2}} \]

6. **Cross-multiply to eliminate the fraction:**
\[ (R^2 + x^2)^{3/2} = 5\sqrt{5} R^3 \]

7. **Raise both sides to the power of \( \frac{2}{3} \):**
\[ R^2 + x^2 = (5\sqrt{5})^{\frac{2}{3}} R^2 \]

8. **Calculate \( (5\sqrt{5})^{\frac{2}{3}} \):**
- \( (5\sqrt{5})^{\frac{2}{3}} = 5^{\frac{2}{3}} \cdot 5^{\frac{1}{3}} = 5^{1} = 5 \)
\[ R^2 + x^2 = 5R^2 \]

9. **Rearranging gives:**
\[ x^2 = 5R^2 - R^2 = 4R^2 \]

10. **Substituting \( R = 10 \, \text{cm} \):**
\[ x^2 = 4(10)^2 = 400 \]
\[ x = \sqrt{400} = 20 \, \text{cm} \]

### Final Answer:
The distance of the point from the center of the circular coil is \( \boxed{20 \, \text{cm}} \).