A single-turn coil of radius R produces a magnetic field B at its centre. If the same wire is wound into a coil of n turns (each of radius R/n) carrying the same current I, what will be the value of the new field at the centre?
\(n^2B\)
The magnetic field at the centre of a circular coil is \(B = \dfrac{\mu_0 N I}{2 r}\), where N is the number of turns and r is the radius.
For the single turn: \(B = \dfrac{\mu_0 I}{2R}\).
When the same wire is rewound into n turns, each turn has radius \(r = \dfrac{R}{n}\) and N = n.
New field \(B' = \dfrac{\mu_0 (n) I}{2 (R/n)} = \dfrac{\mu_0 n^2 I}{2R}\).
Comparing, \(B' = n^2 \cdot \dfrac{\mu_0 I}{2R} = n^2 B\).
Hence, the new field at the centre is \(n^2 B\).
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1. Plastic
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3. Aluminium
4. Stainless Steel
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