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Question Number 147554 by jlewis last updated on 21/Jul/21

$$\mathrm{A}\mathrm{toroid}\mathrm{core}\mathrm{has}\mathrm{N}=1200\mathrm{turns},$$$$\mathrm{length}\mathrm{L}=80\mathrm{cm},\mathrm{cross}-\mathrm{section}\mathrm{area}$$$$\mathrm{A}={60\mathrm{cm}}^2,\mathrm{current}\mathrm{I}=1.5\mathrm{A}.$$$$\mathrm{Compute}\mathrm{B}\mathrm{and}\mathrm{H}.\mathrm{Assume}\mathrm{an}$$$$\mathrm{empty}\mathrm{core}$$

Answered by Olaf_Thorendsen last updated on 22/Jul/21

$$\mathrm{B}=\frac{{\mu }_0\mathrm{NI}}{\sqrt{{\mathrm{L}}^2+{\mathrm{D}}^2}}=\frac{{\mu }_0\mathrm{NI}}{\sqrt{{\mathrm{L}}^2+\frac{4\mathrm{A}}{\pi }}}$$$$\mathrm{B}=\frac{4\pi .{10}^{-7}\times 1200\times 1,5}{\sqrt{0,8^2+\frac{4\times 60.{10}^{-4}}{\pi }}}\approx 2,8\times {10}^{-3}\mathrm{T}$$$$$$$$\mathrm{H}=\frac{{\mu }_0{\mathrm{N}}^2\mathrm{A}}{\mathrm{L}}$$$$\mathrm{H}=\frac{4\pi .{10}^{-7}\times {1200}^2\times 60.{10}^{-4}}{0,8}$$$$\mathrm{H}\approx 13,6\times {10}^{-3}\mathrm{H}$$

Commented by jlewis last updated on 22/Jul/21

$$\mathrm{hello}\mathrm{sir},\mathrm{i}\mathrm{think}{\mathrm{D}}^2\mathrm{should}\mathrm{be}$$$$\mathrm{substituted}\mathrm{with}4\mathrm{A}/\pi \mathrm{not}\mathrm{A}/\pi$$

Commented by Olaf_Thorendsen last updated on 22/Jul/21

$$Youarerightsir.Sorry.$$

Commented by jlewis last updated on 22/Jul/21

$$\mathrm{well},\mathrm{thank}\mathrm{you}\mathrm{anyway}$$

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