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. `(3)/(2)(r^(2))/(h^(2))`

Answer

To solve the problem, we need to find the fraction by which the magnetic field at a small distance \( h \) from the center of a coil is smaller than the magnetic field at the center of the coil.

### Step-by-Step Solution:

1. **Magnetic Field on the Axis of the Coil**:
The magnetic field \( B \) at a distance \( h \) from the center of a coil with \( n \) turns, carrying a current \( i \), and having a radius \( r \) is given by the formula:
\[ B = \frac{\mu_0 n i}{2} \cdot \frac{1}{r^2 + h^2} \cdot \frac{1}{\sqrt{r^2 + h^2}} \]
This can be simplified to:
\[ B = \frac{\mu_0 n i}{2 \pi} \cdot \frac{1}{(r^2 + h^2)^{3/2}} \]

2. **Magnetic Field at the Center of the Coil**:
The magnetic field \( B_c \) at the center of the coil is given by:
\[ B_c = \frac{\mu_0 n i}{2 r} \]

3. **Finding the Ratio of the Two Fields**:
To find how much smaller the field at distance \( h \) is compared to the field at the center, we take the ratio:
\[ \frac{B}{B_c} = \frac{\frac{\mu_0 n i}{2 \pi (r^2 + h^2)^{3/2}}}{\frac{\mu_0 n i}{2 r}} = \frac{r}{(r^2 + h^2)^{3/2}} \]

4. **Using Binomial Approximation**:
Since \( h \) is small compared to \( r \) (i.e., \( h \ll r \)), we can use the binomial approximation:
\[ (1 + \frac{h^2}{r^2})^{-3/2} \approx 1 - \frac{3}{2} \cdot \frac{h^2}{r^2} \]
Therefore, we can rewrite the ratio as:
\[ \frac{B}{B_c} \approx \frac{r}{r^3} \left(1 - \frac{3}{2} \cdot \frac{h^2}{r^2}\right) = \frac{1}{r^2} \left(1 - \frac{3}{2} \cdot \frac{h^2}{r^2}\right) \]

5. **Finding the Fraction**:
The fraction by which the magnetic field at the distance \( h \) is smaller than the magnetic field at the center is:
\[ 1 - \frac{B}{B_c} \approx 1 - \left(1 - \frac{3}{2} \cdot \frac{h^2}{r^2}\right) = \frac{3}{2} \cdot \frac{h^2}{r^2} \]

Thus, the fraction by which the magnetic field at a distance \( h \) is smaller than that at the center is:
\[ \frac{3}{2} \cdot \frac{h^2}{r^2} \]

### Final Answer:
The fraction by which the magnetic field at a distance \( h \) from the center of the coil is smaller than the magnetic field at the center is:
\[ \frac{3}{2} \cdot \frac{h^2}{r^2} \]