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

Answer

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

### Step 1: Magnetic Field at the Center of the Coil
The magnetic field \( B_C \) at the center of a 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.

### Step 2: Magnetic Field on the Axis at Distance \( h \)
The magnetic field \( B_h \) at a distance \( h \) from the center of the coil along the axis is given by:
\[ B_h = \frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}} \]

### Step 3: Express \( B_h \) in Terms of \( B_C \)
We can express \( B_h \) in terms of \( B_C \):
\[ B_h = B_C \cdot \frac{1}{1 + \frac{h^2}{r^2}}^{3/2} \]

### Step 4: Approximation for Small \( h \)
For small \( h \) compared to \( r \) (i.e., \( h \ll r \)), we can use the binomial approximation:
\[ \left(1 + \frac{h^2}{r^2}\right)^{-3/2} \approx 1 - \frac{3}{2} \frac{h^2}{r^2} \]
Thus,
\[ B_h \approx B_C \left(1 - \frac{3}{2} \frac{h^2}{r^2}\right) \]

### Step 5: Finding the Fraction
Now, we need to find the fraction by which \( B_h \) is smaller than \( B_C \):
\[ \text{Fraction} = \frac{B_C - B_h}{B_C} \]
Substituting the expression for \( B_h \):
\[ \text{Fraction} = \frac{B_C - B_C \left(1 - \frac{3}{2} \frac{h^2}{r^2}\right)}{B_C} \]
This simplifies to:
\[ \text{Fraction} = \frac{3}{2} \frac{h^2}{r^2} \]

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