For toroid
"B=\\mu\\mu_0\\frac{NI}{\\pi d}=\\mu\\mu_0\\frac{NU}{\\pi dR}"
So, we have
"\\Phi_c=BS\\cos\\alpha=\\mu\\mu_0\\frac{NU}{\\pi dR}S"
Finally
"\\Psi=NB=\\mu\\mu_0\\frac{N^2 U}{\\pi dR}S=250\\cdot4\\cdot3.14\\cdot10^{-7} \\cdot\\frac{500^2\\cdot240}{3.14\\cdot0.25\\cdot474}\\cdot400\\cdot10^{-6 }=0.02Wb"
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