The field normal to the plane of a wire of `n` turns and radis `r` which carriers `i` is measured on the axis of the coil at a small distance `h` from the centre of the coil. This is smaller than the field at the centre by the fraction.

Options

A. `3/2 (h^(2))/(r^(2))`
B. `2/3 (h^(2))/(r^(2))`
C. `3/2 (r^(2))/(h^(2))`
D. `2/3 (r^(2))/(h^(2))`

Explanation

To solve the problem, we need to find the magnetic field at a distance \( h \) from the center of a circular coil with \( n \) turns, radius \( r \), and carrying a current \( i \). We will compare this with the magnetic field at the center of the coil and determine the fraction by which the field at distance \( h \) is smaller than that at the center.

### 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 with \( n \) turns, radius \( r \), and 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.

2. **Magnetic Field at a Distance \( h \) from the Center**:
The magnetic field \( B_h \) at a distance \( h \) along the axis of the coil is given by:
\[ B_h = \frac{\mu_0 n i r^2}{2(r^2 + h^2)^{3/2}} \]

3. **Assumption for Small \( h \)**:
Since \( h \) is much smaller than \( r \) (i.e., \( h \ll r \)), we can simplify the expression for \( B_h \). We can approximate \( (r^2 + h^2)^{3/2} \) using a binomial expansion:
\[ (r^2 + h^2)^{3/2} \approx r^3 \left(1 + \frac{3h^2}{2r^2}\right) \]
Therefore, we can rewrite \( B_h \) as:
\[ B_h \approx \frac{\mu_0 n i r^2}{2r^3 \left(1 + \frac{3h^2}{2r^2}\right)} = \frac{\mu_0 n i}{2r} \cdot \frac{1}{1 + \frac{3h^2}{2r^2}} \]

4. **Using Binomial Approximation**:
For small \( h \), we can use the binomial approximation:
\[ \frac{1}{1 + x} \approx 1 - x \quad \text{for small } x \]
Thus,
\[ B_h \approx \frac{\mu_0 n i}{2r} \left(1 - \frac{3h^2}{2r^2}\right) \]

5. **Finding the Difference**:
Now, we can find the difference between the magnetic field at the center and at distance \( h \):
\[ B_c - B_h = \frac{\mu_0 n i}{2r} - \left(\frac{\mu_0 n i}{2r} \left(1 - \frac{3h^2}{2r^2}\right)\right) \]
Simplifying this gives:
\[ B_c - B_h = \frac{\mu_0 n i}{2r} \cdot \frac{3h^2}{2r^2} = \frac{3\mu_0 n i h^2}{4r^3} \]

6. **Finding the Fraction**:
To find the fraction by which \( B_h \) is smaller than \( B_c \):
\[ \text{Fraction} = \frac{B_c - B_h}{B_c} = \frac{\frac{3\mu_0 n i h^2}{4r^3}}{\frac{\mu_0 n i}{2r}} = \frac{3h^2}{2r^2} \]

### Final Result:
The fraction by which the magnetic field at distance \( h \) is smaller than that at the center is:
\[ \frac{3h^2}{2r^2} \]