Rankers Physics

Magnetic Field Due to Circular Current Carrying Wire: Practice Problem & Solution

The magnetic field of given length of wire for single turn coil at its centre is 'B' then its value for two turns coil for the same wire is: (2002)
$\frac{B}{4}$
$\frac{B}{2}$
$4B$
$2$

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 wire of length $L$, radius for single turn is $R = L/(2\pi)$ and for two turns is $R' = L/(4\pi) = R/2$. Magnetic field $B = \frac{\mu_0 N I}{2 R}$. For 2 turns, $B' = \frac{\mu_0 (2) I}{2 (R/2)} = 4 \left(\frac{\mu_0 I}{2R}\right) = 4B$.

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