A circular coil of radius `R` carries a current `i`. The magnetic field at its centre is `B`. The distance from the centre on the axis of the coil where the magnetic field will be `B//8` is

Options

A. `Rsqrt(2)`
B. `Rsqrt(3)`
C. 2R
D. 3R

Answer

To solve the problem, we need to find the distance \( x \) from the center of a circular coil where the magnetic field is \( \frac{B}{8} \).

### Step-by-Step Solution:

1. **Magnetic Field at the Center of the Coil**:
The magnetic field \( B_C \) at the center of a circular coil of radius \( R \) carrying a current \( i \) is given by:
\[ B_C = \frac{\mu_0 n i}{2R} \]
where \( \mu_0 \) is the permeability of free space and \( n \) is the number of turns per unit length (for a single loop, \( n = 1 \)).

2. **Magnetic Field at a Distance \( x \) on the Axis**:
The magnetic field \( B_X \) at a distance \( x \) from the center on the axis of the coil is given by:
\[ B_X = \frac{\mu_0 n i R^2}{2(R^2 + x^2)^{3/2}} \]

3. **Setting up the Equation**:
According to the problem, we need to find \( x \) such that:
\[ B_X = \frac{B_C}{8} \]
Substituting the expressions for \( B_X \) and \( B_C \):
\[ \frac{\mu_0 n i R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \cdot \frac{\mu_0 n i}{2R} \]

4. **Canceling Common Terms**:
We can cancel \( \mu_0 n i \) and \( 2 \) from both sides:
\[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 8R \cdot R^2 = (R^2 + x^2)^{3/2} \]
Simplifying this:
\[ 8R^3 = (R^2 + x^2)^{3/2} \]

6. **Taking Cube Root**:
Taking the cube root of both sides:
\[ 2R = (R^2 + x^2)^{1/2} \]

7. **Squaring Both Sides**:
Squaring both sides results in:
\[ 4R^2 = R^2 + x^2 \]

8. **Solving for \( x^2 \)**:
Rearranging gives:
\[ x^2 = 4R^2 - R^2 = 3R^2 \]

9. **Finding \( x \)**:
Taking the square root gives:
\[ x = \sqrt{3}R \]

### Final Answer:
The distance from the center on the axis of the coil where the magnetic field will be \( \frac{B}{8} \) is:
\[ x = \sqrt{3}R \]