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Step-by-Step Solution
Step 1: Understand the Physical Situation
A long straight wire carrying a steady current is first bent into a circular loop of radius $R$, making one turn. Then, the same wire is rebent into a circular coil with $n$ turns, each having a smaller radius. We need to compare the magnetic field at the center of the single-turn loop with that of the $n$-turn coil.
Step 2: Magnetic Field at the Center of the Single-Turn Loop
For a single circular loop of radius $R$ carrying a current $i$, the magnetic field at the center is given by:
$ B = \frac{\mu_0 \, i}{2R} $
This field is denoted as $B$ in the problem statement.
Step 3: Determine the Radius for Each Turn in the $n$-Turn Coil
If the total length of the wire remains the same, then when it is bent into $n$ turns, each turn will have a smaller radius. Suppose the new radius of each turn is $r$. For simplicity, let us assume this new radius is $\frac{R}{n}$ (because the total circumference is divided among $n$ loops, approximating that the total length is roughly unchanged except for slight bending considerations). Thus, we take:
$ r = \frac{R}{n}. $
Step 4: Magnetic Field at the Center of the $n$-Turn Coil
The magnetic field at the center of a coil with $n$ turns, each of radius $r$, carrying current $i$, is given by:
$ B' = n \times \frac{\mu_0 \, i}{2r}. $
Each turn contributes $\frac{\mu_0 i}{2r}$, and since there are $n$ such turns, we multiply by $n$.
Step 5: Substitute $r = \frac{R}{n}$
$ B' = n \times \frac{\mu_0 \, i}{2 \left(\frac{R}{n}\right)}
= n \times \frac{\mu_0 \, i}{\frac{2R}{n}}
= n \times \frac{\mu_0 \, i \, n}{2R}
= \frac{\mu_0 \, i \, n^2}{2R}.
$
We observe that this expression can be compared to the original single-turn loop's field $B = \frac{\mu_0 \, i}{2R}.$
Step 6: Compare with Original Magnetic Field $B$
Since $B = \frac{\mu_0 \, i}{2R}$, substituting we get:
$ B' = n^2 \times \left(\frac{\mu_0 \, i}{2R}\right) = n^2 \, B.
$
Step 7: Conclusion
The magnetic field at the center of the $n$-turn loop is $n^2 B$. Hence, the correct answer is:
$ n^2 B $
