The intensity of magnetic induction at the centre of a current `-` carrying circular coil is `B_(1)` and at a point on its axis at a distance equal to its radius from the centre is `B_(2)` , then `B_(1)//B_(2)` is

Options

A. `2sqrt(2)`
B. `(1)/(2sqrt(2))`
C. `(1)/(sqrt(2))`
D. `sqrt(2)`

Explanation

To solve the problem, we need to find the relationship between the magnetic induction at the center of a current-carrying circular coil (denoted as \( B_1 \)) and the magnetic induction at a point on its axis at a distance equal to its radius (denoted as \( B_2 \)).

### Step-by-Step Solution:

1. **Magnetic Field at the Center of the Coil (\( B_1 \))**:
The magnetic field at the center of a circular coil carrying current \( I \) and having radius \( r \) is given by the formula:
\[ B_1 = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

2. **Magnetic Field on the Axis of the Coil at Distance \( r \) (\( B_2 \))**:
The magnetic field at a point on the axis of the coil at a distance equal to its radius \( r \) is given by:
\[ B_2 = \frac{\mu_0 I}{2r \sqrt{2}} \]
This formula accounts for the geometry of the coil and the distance from the center.

3. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
Now we need to find the ratio of \( B_1 \) to \( B_2 \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{2r \sqrt{2}}} \]
Simplifying this expression:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{2r \sqrt{2}}} = \frac{2r \sqrt{2}}{2r} = \sqrt{2} \]

4. **Conclusion**:
Therefore, the relationship between \( B_1 \) and \( B_2 \) is:
\[ B_1 = \sqrt{2} B_2 \]

### Final Answer:
\[ B_1 / B_2 = \sqrt{2} \]