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Step-by-Step Solution
Step 1: Identify the Magnetic Field at the Center of the Coil
For a current-carrying circular coil of radius $a$ through which a current $i$ flows, the magnetic field at its center is given by:
$B_{\text{centre}} \;=\; \frac{\mu_0\,i}{2\,a}.$
Step 2: Write the Magnetic Field on the Axis of the Coil
At an axial distance $r$ from the center of the coil (along its axis), the magnetic field is:
$B_{\text{axis}} \;=\; \frac{\mu_0 \, i \, a^2}{2 \,\bigl(a^2 + r^2\bigr)^{3/2}}.$
Here, $a$ is the radius of the coil, and $r$ is the distance from the center along the axis.
Step 3: Express the Fractional Change in the Magnetic Field
The fractional change we need is:
$\displaystyle \text{Fractional change} \;=\; \frac{B_{\text{centre}} - B_{\text{axis}}}{B_{\text{centre}}}.$
Substitute the expressions for $B_{\text{centre}}$ and $B_{\text{axis}}$:
$
\frac{\bigl(\tfrac{\mu_0 \, i}{2\,a}\bigr) - \bigl(\tfrac{\mu_0 \, i\, a^2}{2\,\bigl(a^2 + r^2\bigr)^{3/2}}\bigr)}{\tfrac{\mu_0 \, i}{2\,a}}
\;=\; 1 \;-\; \frac{\tfrac{\mu_0 \, i\, a^2}{2\,\bigl(a^2 + r^2\bigr)^{3/2}}}{\tfrac{\mu_0 \, i}{2\,a}}
\;=\; 1 \;-\; \frac{1}{\bigl(1 + \tfrac{r^2}{a^2}\bigr)^{3/2}}.
$
Step 4: Use Binomial Approximation for Small r/a
When $r \ll a$, $\frac{r^2}{a^2}$ is very small, so we can use the binomial expansion:
$\displaystyle
\bigl(1 + x\bigr)^{-3/2} \approx 1 - \frac{3}{2}\,x
\quad
\text{for small } x.
$
Here, $x = \frac{r^2}{a^2}$. Therefore:
$\displaystyle
\frac{1}{\bigl(1 + \tfrac{r^2}{a^2}\bigr)^{3/2}}
\;\approx\;
1 - \frac{3}{2}\,\frac{r^2}{a^2}.
$
Step 5: Calculate the Fractional Change
Using this approximation in our expression for the fractional change:
$\displaystyle
1 \;-\;\Bigl[\,1 - \frac{3}{2}\,\frac{r^2}{a^2}\Bigr]
\;=\;
\frac{3}{2}\,\frac{r^2}{a^2}.
$
Hence, the fractional change in the magnetic field at a distance $r$ from the center (on the axis) relative to the field at the center is:
$\displaystyle \frac{3}{2}\,\frac{r^2}{a^2}.$
Step 6: Identify the Correct Option
From the given choices, the expression $\frac{3}{2}\,\frac{r^2}{a^2}$ corresponds to Option 4. Therefore, the correct answer is:
$\displaystyle \boxed{\frac{3}{2}\,\frac{r^2}{a^2}}.
$
