Ответ:
В=2*10⁻⁴ Тл
Объяснение:
АО=АВ=5 см
В=В₁+В₂+В₃+В₄
В₁=μ₀μ*І₁/(4π*r₀₁)*(cosφ₁-cosφ₂)
В₁=1.257*10⁻⁶*1*10/(4π*1.96*10⁻²)*(cos23.13°-cos73.06°)=0.32*10⁻⁴ Tл
В₂=1.257*10⁻⁶*1*10/(4π*0.6*10⁻²)*(cos163.06°-cos173.13°)=0.17*10⁻⁴ Tл
В₃=1.257*10⁻⁶*1*10/(4π*4.96*10⁻²)*(cos47.19°-cos83.13°)=0.93*10⁻⁴ Tл
В₄=1.257*10⁻⁶*1*10/(4π*4.6*10⁻²)*(cos113.13°-cos137.19°)=0.57*10⁻⁴ Tл
В=(0.32+0.17+0.93+0.57)*10⁻⁴=2*10⁻⁴ Тл
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Ответ:
В=2*10⁻⁴ Тл
Объяснение:
АО=АВ=5 см
В=В₁+В₂+В₃+В₄
В₁=μ₀μ*І₁/(4π*r₀₁)*(cosφ₁-cosφ₂)
В₁=1.257*10⁻⁶*1*10/(4π*1.96*10⁻²)*(cos23.13°-cos73.06°)=0.32*10⁻⁴ Tл
В₂=1.257*10⁻⁶*1*10/(4π*0.6*10⁻²)*(cos163.06°-cos173.13°)=0.17*10⁻⁴ Tл
В₃=1.257*10⁻⁶*1*10/(4π*4.96*10⁻²)*(cos47.19°-cos83.13°)=0.93*10⁻⁴ Tл
В₄=1.257*10⁻⁶*1*10/(4π*4.6*10⁻²)*(cos113.13°-cos137.19°)=0.57*10⁻⁴ Tл
В=(0.32+0.17+0.93+0.57)*10⁻⁴=2*10⁻⁴ Тл