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Question Number 147554 by jlewis last updated on 21/Jul/21
$$\mathrm{A}\:\mathrm{toroid}\:\mathrm{core}\:\mathrm{has}\:\mathrm{N}=\mathrm{1200}\:\mathrm{turns}, \\ $$$$\mathrm{length}\:\mathrm{L}=\mathrm{80cm},\mathrm{cross}-\mathrm{section}\:\mathrm{area} \\ $$$$\mathrm{A}=\mathrm{60cm}^{\mathrm{2}} ,\mathrm{current}\:\mathrm{I}=\mathrm{1}.\mathrm{5A}. \\ $$$$\:\:\mathrm{Compute}\:\mathrm{B}\:\mathrm{and}\:\mathrm{H}.\mathrm{Assume}\:\mathrm{an} \\ $$$$\:\mathrm{empty}\:\mathrm{core} \\ $$
Answered by Olaf_Thorendsen last updated on 22/Jul/21
$$\mathrm{B}\:=\:\frac{\mu_{\mathrm{0}} \mathrm{NI}}{\:\sqrt{\mathrm{L}^{\mathrm{2}} +\mathrm{D}^{\mathrm{2}} }}\:=\:\frac{\mu_{\mathrm{0}} \mathrm{NI}}{\:\sqrt{\mathrm{L}^{\mathrm{2}} +\frac{\mathrm{4A}}{\pi}}}\: \\ $$$$\mathrm{B}\:=\:\frac{\mathrm{4}\pi.\mathrm{10}^{−\mathrm{7}} ×\mathrm{1200}×\mathrm{1},\mathrm{5}}{\:\sqrt{\mathrm{0},\mathrm{8}^{\mathrm{2}} +\frac{\mathrm{4}×\mathrm{60}.\mathrm{10}^{−\mathrm{4}} }{\pi}}}\:\approx\:\mathrm{2},\mathrm{8}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{T} \\ $$$$ \\ $$$$\:\mathrm{H}=\:\frac{\mu_{\mathrm{0}} \mathrm{N}^{\mathrm{2}} \mathrm{A}}{\mathrm{L}} \\ $$$$\:\mathrm{H}=\:\frac{\mathrm{4}\pi.\mathrm{10}^{−\mathrm{7}} ×\mathrm{1200}^{\mathrm{2}} ×\mathrm{60}.\mathrm{10}^{−\mathrm{4}} }{\mathrm{0},\mathrm{8}} \\ $$$$\:\mathrm{H}\:\approx\:\mathrm{13},\mathrm{6}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{H} \\ $$
Commented by jlewis last updated on 22/Jul/21
$$\mathrm{hello}\:\mathrm{sir},\mathrm{i}\:\mathrm{think}\:\mathrm{D}^{\mathrm{2}} \:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{substituted}\:\mathrm{with}\:\mathrm{4A}/\pi\:\mathrm{not}\:\mathrm{A}/\pi \\ $$
Commented by Olaf_Thorendsen last updated on 22/Jul/21
$${You}\:{are}\:{right}\:{sir}.\:{Sorry}. \\ $$
Commented by jlewis last updated on 22/Jul/21
$$\mathrm{well},\mathrm{thank}\:\mathrm{you}\:\mathrm{anyway} \\ $$