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Consider a tightly wound  \(100\)  turn coil of radius \(10 \text{ cm}\) carrying a current of  \(1\;A\) . What is the magnitude of the magnetic field at the centre of the coil?

Answer

\(6.28\times 10^{-4}\;T\)
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Solution

Since the coil is tightly wound. We may take each circular element to have the same radius  \(R=10 \text{ cm} = 0.1\text{ m}\)
The number of turns  \(N=100\)
The magnitude of the magnetic field is.
 \(B=\frac{\mu _o\mathit{NI}}{2R}\) 
 \(=\frac{4\Pi \times 10^{-7}\times 10^2\times 1}{2\times 10^{-1}}\) 
 \(=6.28\times 10^{-4}\;T\)  
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