The field due to a wire of n turns and radius r which carries a current I is measure on the axis of the coil at a small distance h form 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 fraction by which the magnetic field at a point on the axis of a coil with n turns and radius r carrying a current I is smaller than the magnetic field at the center of the coil. We will denote the magnetic field at point P (a small distance h from the center) as \( B_P \) and the magnetic field at the center as \( B_C \).

### Step-by-Step Solution:

1. **Magnetic Field at Point P**:
The magnetic field at a point on the axis of a current-carrying coil at a distance h from the center can be expressed as:
\[ B_P = \frac{\mu_0}{4\pi} \cdot \frac{2\pi n I r^2}{(r^2 + h^2)^{3/2}} \]
where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns, \( I \) is the current, and \( r \) is the radius of the coil.

2. **Magnetic Field at the Center of the Coil**:
The magnetic field at the center of the coil can be given as:
\[ B_C = \frac{\mu_0}{4\pi} \cdot \frac{2\pi n I}{r} \]

3. **Finding the Ratio \( \frac{B_P}{B_C} \)**:
To compare the two magnetic fields, we divide \( B_P \) by \( B_C \):
\[ \frac{B_P}{B_C} = \frac{\frac{\mu_0}{4\pi} \cdot \frac{2\pi n I r^2}{(r^2 + h^2)^{3/2}}}{\frac{\mu_0}{4\pi} \cdot \frac{2\pi n I}{r}} \]
Simplifying this expression, we find:
\[ \frac{B_P}{B_C} = \frac{r^3}{(r^2 + h^2)^{3/2}} \]

4. **Simplifying the Denominator**:
We can factor out \( r^2 \) from the denominator:
\[ \frac{B_P}{B_C} = \frac{r^3}{r^3 \left(1 + \frac{h^2}{r^2}\right)^{3/2}} = \frac{1}{\left(1 + \frac{h^2}{r^2}\right)^{3/2}} \]

5. **Using Binomial Expansion**:
For small \( h \) compared to \( r \), we can use the binomial expansion:
\[ \left(1 + x\right)^{-n} \approx 1 - nx \quad \text{for small } x \]
Applying this to our expression:
\[ \left(1 + \frac{h^2}{r^2}\right)^{-3/2} \approx 1 - \frac{3}{2} \cdot \frac{h^2}{r^2} \]

6. **Finding the Fraction**:
Now we can express \( \frac{B_P}{B_C} \):
\[ \frac{B_P}{B_C} \approx 1 - \frac{3}{2} \cdot \frac{h^2}{r^2} \]
Therefore, the fraction by which \( B_C \) is greater than \( B_P \) is:
\[ \text{Fraction} = 1 - \frac{B_P}{B_C} = \frac{3}{2} \cdot \frac{h^2}{r^2} \]

### Final Answer:
The magnetic field at point P is smaller than the magnetic field at the center by the fraction:
\[ \frac{3}{2} \cdot \frac{h^2}{r^2} \]