Question
Download Solution PDFA circular coil of wire consisting of 500 turns, each of radius 6.0 cm carries a current of 0.30 A. Find the magnitude of the magnetic field B at the center of the coil?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Using the formula for the magnetic field at the center of a current carrying a circular coil
\(B = \frac{μ_0}{4π}× \frac{2π N I}{r}\)
where μ0 = 4π × 10-7Wb/Am, N is the number of turns, I is the current intensity and r is the radius of the coil
Calculation:
Given N = 500, I = 0.30A and r = 0.06m
\(B = 10^{-7} × \frac{2\pi × 500 × 0.30}{0.06}\)
B = 1.57 × 10-3T
Last updated on Jun 19, 2025
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