At very close point on the axis of a current carrying circular coil `(xltltltR)` of radius 'R', the value of magnetic field decreses by a fraction of `5%` with respect to centre value. The position of the point from the centre of the coil is :-

Options

A. `(R)/(sqrt(10))`
B. `(R)/(sqrt(30))`
C. `(R)/(sqrt(50))`
D. `(R)/(sqrt(150))`

Answer

To solve the problem, we need to determine the position \( x \) from the center of a current-carrying circular coil where the magnetic field decreases by 5% compared to its value at the center.

### Step-by-Step Solution:

1. **Understanding Magnetic Field at the Center and on the Axis**:
- The magnetic field \( B_0 \) at the center of a circular coil of radius \( R \) carrying current \( I \) is given by:
\[ B_0 = \frac{\mu_0 I}{2R} \]
- The magnetic field \( B \) at a point on the axis of the coil at a distance \( x \) from the center is given by:
\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

2. **Setting Up the Equation for 5% Decrease**:
- According to the problem, the magnetic field decreases by 5% at the point \( x \):
\[ B = B_0 \times (1 - 0.05) = 0.95 B_0 \]

3. **Substituting the Expressions**:
- Substituting the expressions for \( B \) and \( B_0 \):
\[ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = 0.95 \left(\frac{\mu_0 I}{2R}\right) \]

4. **Canceling Common Terms**:
- Cancel \( \frac{\mu_0 I}{2} \) from both sides:
\[ \frac{R^2}{(R^2 + x^2)^{3/2}} = 0.95 \cdot \frac{1}{R} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ R^3 = 0.95 (R^2 + x^2)^{3/2} \]

6. **Cubing Both Sides**:
- Cubing both sides to eliminate the cube root:
\[ R^9 = 0.857375 (R^2 + x^2)^3 \]

7. **Solving for \( x^2 \)**:
- To simplify, we can express \( R^2 + x^2 \) in terms of \( R \):
\[ R^2 + x^2 = \left(\frac{R^3}{0.857375}\right)^{1/3} \]
- This leads to:
\[ x^2 = \left(\frac{R^3}{0.857375}\right)^{1/3} - R^2 \]

8. **Finding the Value of \( x \)**:
- After calculating, we find:
\[ x = \frac{R}{\sqrt{30}} \]

### Final Answer:
The position of the point from the center of the coil is:
\[ x = \frac{R}{\sqrt{30}} \]