Question:
A wire of certain length carrying current $I$, when bent into a circular coil of single turn produces a magnetic field $B$ at its centre. If the same wire is bent into a circular coil of 3 turns and it carries the same current, then the magnetic field at the centre of the coil is
A wire of certain length carrying current $I$, when bent into a circular coil of single turn produces a magnetic field $B$ at its centre. If the same wire is bent into a circular coil of 3 turns and it carries the same current, then the magnetic field at the centre of the coil is
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Same wire → radius decreases as turns increase.
Updated On: Apr 24, 2026
- $12B$
- $3B$
- $6B$
- $9B$
- $15B$
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The Correct Option is D
Solution and Explanation
Concept:
• Magnetic field at centre: \[ B = \frac{\mu_0 NI}{2R} \]
Step 1: Same wire length
Length constant $\Rightarrow N \cdot 2\pi R = constant$ \[ R \propto \frac{1}{N} \]
Step 2: Substitute in formula
\[ B \propto \frac{N}{R} \propto N^2 \]
Step 3: For $N=3$
\[ B' = 3^2 B = 9B \] Final Conclusion:
Option (D)
• Magnetic field at centre: \[ B = \frac{\mu_0 NI}{2R} \]
Step 1: Same wire length
Length constant $\Rightarrow N \cdot 2\pi R = constant$ \[ R \propto \frac{1}{N} \]
Step 2: Substitute in formula
\[ B \propto \frac{N}{R} \propto N^2 \]
Step 3: For $N=3$
\[ B' = 3^2 B = 9B \] Final Conclusion:
Option (D)
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