Asked by James help
A square loop of side "a" carries current "I". It creates a magnetic field "B-square" at its center. The same loop is then reshaped into a circle, which lowers the magnetic field "B-circle" at the center. What is the ratio B-square/B-circle?
Answers
Answered by
Elena
circle:
L=2πR
R=L/2π
B₀=μ₀I/2R= μ₀I2 π /2L=
=μ₀πI/L
square:
L=4a=> a=L/4
The distance from the center of the square to the midpoint of the side of the square is
b=a/2tan45=a/2
B⋄=4B₁=4B=
=4(μ₀/4π)I/b∫cosφdφ {limits: (- π/4) (+π/4)}
=8(μ₀/4π)I/a•√2=2√2 μ₀I/πa=
=8√2 μ₀I/πL
B⋄/B₀={8√2 μ₀I/πL}/{ μ₀πI/L}=8√2/π²=1.15
L=2πR
R=L/2π
B₀=μ₀I/2R= μ₀I2 π /2L=
=μ₀πI/L
square:
L=4a=> a=L/4
The distance from the center of the square to the midpoint of the side of the square is
b=a/2tan45=a/2
B⋄=4B₁=4B=
=4(μ₀/4π)I/b∫cosφdφ {limits: (- π/4) (+π/4)}
=8(μ₀/4π)I/a•√2=2√2 μ₀I/πa=
=8√2 μ₀I/πL
B⋄/B₀={8√2 μ₀I/πL}/{ μ₀πI/L}=8√2/π²=1.15
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