Magnetic Field Due to Circular Current Carrying Wire: Practice Problem & Solution
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the center of this coil of n turns will be: (2016 - II)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Magnetic Field Due to Circular Current Carrying Wire. Understanding the underlying formula is key to arriving at the correct answer below:
For a single turn loop, radius $R = \frac{L}{2\pi}$ and $B = \frac{\mu_0 i}{2R}$. When bent into $n$ turns, the new radius is $R' = \frac{R}{n}$. The magnetic field becomes $B' = \frac{\mu_0 n i}{2R'} = n^2 B$.
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