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Step-by-Step Solution
Step 1: Express the total length of wire for each coil configuration
Let the total length of the wire be $L$. When bent into $N$ turns, each turn has radius $r_1$. Thus:
$L = 2\pi r_1 \times N.$
When the same wire is bent to form $n$ turns (each turn has radius $r_2$):
$L = 2\pi r_2 \times n.$
Step 2: Relate the radii of the two coils
Because the length of the wire is the same in both cases, set the two expressions equal:
$2\pi r_1 \, N = 2\pi r_2 \, n.$
Canceling $2\pi$ from both sides:
$r_1 \, N = r_2 \, n \quad \Longrightarrow \quad r_1 = \frac{r_2 \, n}{N}.$
Step 3: Formula for the magnetic field at the center of a circular coil
For a single-turn circular coil of radius $r$ carrying current $I$, the magnetic field at its center is:
$B_{\text{single turn}} = \frac{\mu_0 \, I}{2\,r}.$
With $k$ turns, the magnetic field becomes:
$B = k \times \frac{\mu_0 \, I}{2\,r}.$
Step 4: Magnetic field in each configuration
Coil of $N$ turns and radius $r_1$:
$B_1 = N \times \frac{\mu_0 \, I}{2\,r_1}.$
Coil of $n$ turns and radius $r_2$:
$B_2 = n \times \frac{\mu_0 \, I}{2\,r_2}.$
Step 5: Express $B_1$ in terms of $r_2$
From $r_1 = \frac{r_2 \, n}{N}$, we get
$\displaystyle \frac{1}{r_1} = \frac{N}{n \, r_2}.$
Hence,
$\displaystyle B_1 = N \times \frac{\mu_0 \, I}{2\,r_1}
= N \times \frac{\mu_0 \, I}{2} \times \frac{N}{n \, r_2}
= \frac{\mu_0 \, I}{2} \times \frac{N^2}{n} \times \frac{1}{r_2}.$
Step 6: Form the ratio $B_1/B_2$
Divide $B_1$ by $B_2$:
$\displaystyle \frac{B_1}{B_2}
= \frac{\dfrac{\mu_0 \, I}{2} \,\dfrac{N^2}{n \, r_2}}
{n \times \dfrac{\mu_0 \, I}{2 \, r_2}}
= \frac{N^2}{n} \times \frac{1}{n}
= \frac{N^2}{n^2}.$
Step 7: State the final result
Thus, the ratio of the magnetic field in the first case (with $N$ turns) to that in the second case (with $n$ turns) is
$\displaystyle \boxed{\frac{N^2}{n^2}}.$
