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Step-by-Step Solution
Step 1: Recall the Magnetic Field Formula for a Single-Turn Coil
For a circular coil of radius $R$ carrying a steady current $i$,
the magnetic field at its center is given by:
$B = \frac{\mu_0 \, i}{2R}.$
Step 2: Relate the Radius of the Multi-Turn Coil to the Original Length of the Wire
The original wire is first bent into one loop of radius $R$. This wire is then re-bent into
$n$ turns, each having a smaller radius $r'$. Since the total length of the wire remains
the same in both cases:
$n \times (2\pi r') = 2\pi R \quad \Longrightarrow \quad n\,r' = R.$
Step 3: Write the Magnetic Field for the $n$-Turn Coil
For $n$ turns, each of radius $r'$, carrying the same current $i$, the magnetic field
at the center (assuming tightly wound turns) is:
$B' = n \cdot \frac{\mu_0 \, i}{2\,r'}.$
The factor of $n$ in front appears because the contributions from each of the $n$ turns
are added at the center.
Step 4: Substitute $r'$ in Terms of $R$
From $n\,r' = R,$ we have $r' = \frac{R}{n}.$ Substituting this into
$B' = n \cdot \frac{\mu_0 \, i}{2\,r'}$:
$B'
= n \times \frac{\mu_0 \, i}{2\,\left(\frac{R}{n}\right)}
= n \times \frac{\mu_0 \, i \, n}{2\,R}
= \frac{n^2 \mu_0 \, i}{2\,R}.$
Step 5: Compare with the Original Magnetic Field $B$
Recall that the original magnetic field for the single turn was $B = \frac{\mu_0 \, i}{2R}.$
Thus,
$B' = n^2 \times \left(\frac{\mu_0 \, i}{2\,R}\right) = n^2\, B.$
Final Answer:
The magnetic field at the center of the coil when the wire is bent into $n$ turns is
$n^2 B.$
