The magnetic field at the center of a circular coil of radius R carrying current I is:
A. μ₀NI/(4πR)
B. μ₀I/(2R)
C. μ₀NI/(2R)
D. μ₀I/(4R)
Answer: Option C
Solution (By JKSSB Mock Tests)
For a circular coil with N turns, magnetic field at center B = μ₀NI/(2R). For single turn (N=1), B = μ₀I/(2R). This derives from Biot-Savart law. Option A is for single turn; B and D have incorrect constants. Memory tip: 'Coil center: B = μ₀NI/(2R); straight wire: B = μ₀I/(2πr)'. This formula application is frequently tested in magnetism sections of competitive exams. Always note number of turns N; competitive exams often include it to test attention to detail.
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