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. `sqrt(2)R`
B. `sqrt(3)R`
C. `2R`
D. `3R`

Explanation

To solve the problem of finding the distance from the center on the axis of a circular coil where the magnetic field is \( \frac{B}{8} \), we can follow these steps:

### 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 the formula:
\[ B_c = \frac{\mu_0 n i}{2R} \]
where \( \mu_0 \) is the permeability of free space and \( n \) is the number of turns per unit length.

2. **Magnetic Field at a Distance \( x \) from the Center**:
The magnetic field \( B_x \) at a distance \( x \) along 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**:
We need to find the distance \( x \) where the magnetic field \( B_x \) is \( \frac{B_c}{8} \):
\[ B_x = \frac{1}{8} B_c \]
Substituting the expressions for \( B_x \) and \( B_c \):
\[ \frac{\mu_0 n i R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \left(\frac{\mu_0 n i}{2R}\right) \]

4. **Canceling Common Terms**:
We can cancel \( \mu_0 n i \) from both sides and simplify:
\[ \frac{R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{16R} \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 16R \cdot R^2 = 2(R^2 + x^2)^{3/2} \]
Simplifying this yields:
\[ 16R^3 = 2(R^2 + x^2)^{3/2} \]

6. **Dividing by 2**:
Dividing both sides by 2:
\[ 8R^3 = (R^2 + x^2)^{3/2} \]

7. **Cubing Both Sides**:
Cubing both sides to eliminate the exponent:
\[ (8R^3)^2 = R^2 + x^2 \]
This simplifies to:
\[ 64R^6 = R^2 + x^2 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ x^2 = 64R^6 - R^2 \]

9. **Taking the Square Root**:
Taking the square root of both sides:
\[ x = \sqrt{64R^6 - R^2} = R\sqrt{64R^4 - 1} \]

10. **Final Result**:
The distance \( x \) from the center on the axis of the coil where the magnetic field is \( \frac{B}{8} \) is:
\[ x = R\sqrt{64R^4 - 1} \]