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二. 毕萨伐尔定律: B=∫dB = ∫(μ₀I dl × eᵣ)/(4πr²)

1. 载流长直导线. [图示：三角形，角度θ₁、θ₂，电流方向向里（叉号）] B = (μ₀I)/(4πr₀) (cosθ₁ - cosθ₂) = (μ₀I)/(4πr₀) [cosθ₁ + cos(π - θ₂)] 无限长直导线: B = μ₀I/(2πr) 半无限长直导线: B = μ₀I/(4πr₀)

推导: dB = (μ₀)/(4π) · (Idz sinθ)/r² B = ∫(μ₀I dz sinθ)/(4πr²) 由于各电流元产生的磁场方向相同 z = r₀ ctg(π - θ) = -r₀ ctgθ dz = r₀ dθ/sin²θ r = r₀/sin(π - θ) = r₀/sinθ B = ∫(μ₀I r₀ dθ sinθ)/(4π r₀²/sin²θ · sin²θ) = ∫(μ₀I sinθ dθ)/(4πr₀) = (μ₀I)/(4πr₀) ∫_{θ₁}^{θ₂} sinθ dθ = (μ₀I)/(4πr₀) (cosθ₁ - cosθ₂) = (μ₀I)/(4πr₀) [cosθ₁ + cos(π - θ₂)]

载流圆线圈在轴线上: 注 若有N匝则 需要乘以N [图示：圆线圈，轴线上点P，半径R，距离x，电流元Idl，磁场方向沿x轴] Bₓ = (μ₀p²I)/[2(p² + x²)^(3/2)] ① x=0, 即电流环中心的磁感应强度: B = μ₀I/(2R) ② x≫a, (p² + x²)^(3/2) ≈ x³, B = (μ₀Ip²)/(2x³) 引入磁矩, \(\vec{m} = I\vec{S} = IS\vec{e}_n\) 对任意形状的平面载流线圈都适 轴线上的磁感应强度, \(\vec{B} = \frac{\mu_0 I p^2 \vec{e}_n}{2x^3} = \frac{\mu_0}{2\pi} \frac{\vec{m}}{x^3}\)

推导: 由图和右手定则可以判断出磁场只有沿x的分量 dB = (μ₀I)/(4πr²) dl cosα = p/r = p/√(p² + x²) Bₓ = (μ₀I cosα)/(4πr²) ∮dl = (μ₀I p)/(4πr² r) · 2πp = (μ₀I p²)/(2r⁴) = (μ₀I p²)/[2(p² + x²)^(3/2)]