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An equilateral triangular loop of side 2L carries a current i. What is the magnitude of the magnetic field at the centroid (center) of the loop?
- 9*mu0*i / (4*pi*L)
- 3*sqrt(3)*mu0*i / (4*pi*L)
- 2*sqrt(3)*mu0*i / (pi*L)
- 3*mu0*i / (4*pi*L)
Correct answer: 3*sqrt(3)*mu0*i / (4*pi*L)
Solution
For an equilateral triangle of side a=2L: perpendicular distance from center to each side (apothem) = a/(2*sqrt(3)) = 2L/(2*sqrt(3)) = L/sqrt(3). For a finite wire of length 2L at distance d=L/sqrt(3), the angle subtended at the center from each end is 60 deg (since it's an equilateral triangle). B per side = (mu0*i/(4*pi*d)) * (sin(60)+sin(60)) = (mu0*i/(4*pi*(L/sqrt(3)))) * (2*sqrt(3)/2) = (mu0*i*sqrt(3)/(4*pi*L))*sqrt(3) = 3*mu0*i/(4*pi*L). Total = 3 sides * 3*mu0*i/(4*pi*L) = 9*mu0*i/(4*pi*L). Let me redo with correct angles.
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