Step 1: Understanding the Concept:
The magnetic field at the center of a circular coil depends on the current, the radius of the coil, and the number of turns.
When a single piece of wire is rewound from 1 turn to $n$ turns, its length remains constant, but the radius of the new loops must be smaller to accommodate the extra turns.
Step 2: Key Formula or Approach:
1. Length constraint: The total length of the wire $L$ is constant. For 1 turn of radius $R$: $L = 2\pi R$. For $n$ turns of radius $r_1$: $L = n \times (2\pi r_1)$.
2. Magnetic field formula: The magnetic field at the center of a coil with $N$ turns and radius $r$ is $B = \frac{\mu_0 N I}{2r}$.
Step 3: Detailed Explanation:
Let the initial single-turn loop have a radius $R$.
Its magnetic field at the center is given as $B$:
\[ B = \frac{\mu_0 (1) I}{2R} = \frac{\mu_0 I}{2R} \]
The total length of the wire is $L = 2\pi R$.
Now, the same wire is bent into $n$ turns of radius $r_1$.
Conserving the length of the wire:
\[ L = n \times 2\pi r_1 \]
\[ 2\pi R = n \times 2\pi r_1 \]
\[ R = n r_1 \implies r_1 = \frac{R}{n} \]
Now, calculate the new magnetic field $B'$ at the center of this new $n$-turn coil:
\[ B' = \frac{\mu_0 n I}{2r_1} \]
Substitute $r_1 = R/n$ into the equation:
\[ B' = \frac{\mu_0 n I}{2(R/n)} \]
The $1/n$ in the denominator moves to the numerator:
\[ B' = \frac{\mu_0 n^2 I}{2R} \]
Separate the $n^2$ term from the rest:
\[ B' = n^2 \left( \frac{\mu_0 I}{2R} \right) \]
Since the term in parentheses is the original magnetic field $B$:
\[ B' = n^2 B \]
Step 4: Final Answer:
The new magnetic field at the center is $n^2 B$.