A circular coil of wire of radius 'r' has 'n' turns and carries a current 'I'. The magnetic induction (B) at a point on the axis of the coil at a distance `sqrt3r` from its centre is

Options

A. `(mu_(0)In)/(4r)`
B. `(mu_(0)In)/(8r)`
C. `(mu_(0)nI)/(16r)`
D. `(mu_(0)In)/(32r)`

Answer

To find the magnetic induction (B) at a point on the axis of a circular coil of wire with radius 'r', 'n' turns, and carrying a current 'I', we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a circular coil with radius \( r \), \( n \) turns, and current \( I \). We need to find the magnetic induction at a distance \( x = \sqrt{3}r \) from the center of the coil along its axis.

2. **Use the Formula for Magnetic Field on the Axis of a Coil**: The magnetic field \( B \) at a distance \( x \) from the center of a circular coil can be calculated using the formula:
\[ B = \frac{n \mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]
where \( \mu_0 \) is the permeability of free space.

3. **Substitute the Value of \( x \)**: In our case, \( x = \sqrt{3}r \). We substitute this into the formula:
\[ B = \frac{n \mu_0 I r^2}{2(r^2 + (\sqrt{3}r)^2)^{3/2}} \]

4. **Simplify the Expression**: Calculate \( (\sqrt{3}r)^2 \):
\[ (\sqrt{3}r)^2 = 3r^2 \]
So, we have:
\[ B = \frac{n \mu_0 I r^2}{2(r^2 + 3r^2)^{3/2}} = \frac{n \mu_0 I r^2}{2(4r^2)^{3/2}} \]

5. **Calculate \( (4r^2)^{3/2} \)**:
\[ (4r^2)^{3/2} = 4^{3/2} \cdot (r^2)^{3/2} = 8r^3 \]
Therefore, we can rewrite \( B \):
\[ B = \frac{n \mu_0 I r^2}{2 \cdot 8r^3} = \frac{n \mu_0 I r^2}{16r^3} \]

6. **Final Simplification**: Cancel \( r^2 \) in the numerator with \( r^3 \) in the denominator:
\[ B = \frac{n \mu_0 I}{16r} \]

7. **Conclusion**: The magnetic induction at the point on the axis of the coil at a distance \( \sqrt{3}r \) from its center is:
\[ B = \frac{n \mu_0 I}{16r} \]

### Final Answer:
\[ B = \frac{n \mu_0 I}{16r} \]