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Question Number 9779 by tawakalitu last updated on 03/Jan/17

Answered by sandy_suhendra last updated on 04/Jan/17

A=5 cm^2 =5×10^(−4) m^2 L=25 cm=0.25 m N=120 I_1 =1.5 A ⇒ Φ_1 =0.3 mWb=3×10^(−4) Wb I_2 =5 A ⇒ Φ_2 =0.6 mWb=6×10^(−4) Wb (a) (i) H_1 =((N I_1 )/L)=((120×1.5)/(0.25))=720 Ampere/meter (ii) H_2 =((N I_2 )/L)=((120×5)/(0.25))=2400 Ampere/meter (b) (i) B_1 =(Φ_1 /A)=((3×10^(−4) )/(5×10^(−4) ))=0.6 Wb/m^2 μ=(B_1 /H_1 ) μ_(r ) μ_o =((0.6)/(720)) μ_r ×4π×10^(−7) =8.33×10^(−4) μ_r =((8.33×10^(−4) )/(4π×10^(−7) ))=((2082.5)/π)≈663.2 (ii) B_2 =(Φ_2 /A)=((6×10^(−4) )/(5×10^(−4) ))=1.2 Wb/m^2 μ=(B_2 /H_2 ) μ_r ×4π×10^(−7) =((1.2)/(2400)) μ_r =((1250)/π)≈398.1

$$\mathrm{A}=\mathrm{5}\:\mathrm{cm}^{\mathrm{2}} =\mathrm{5}×\mathrm{10}^{−\mathrm{4}} \:\mathrm{m}^{\mathrm{2}} \\ $$$$\mathrm{L}=\mathrm{25}\:\mathrm{cm}=\mathrm{0}.\mathrm{25}\:\mathrm{m} \\ $$$$\mathrm{N}=\mathrm{120} \\ $$$$\mathrm{I}_{\mathrm{1}} =\mathrm{1}.\mathrm{5}\:\mathrm{A}\:\Rightarrow\:\Phi_{\mathrm{1}} =\mathrm{0}.\mathrm{3}\:\mathrm{mWb}=\mathrm{3}×\mathrm{10}^{−\mathrm{4}} \:\mathrm{Wb} \\ $$$$\mathrm{I}_{\mathrm{2}} =\mathrm{5}\:\mathrm{A}\:\Rightarrow\:\Phi_{\mathrm{2}} =\mathrm{0}.\mathrm{6}\:\mathrm{mWb}=\mathrm{6}×\mathrm{10}^{−\mathrm{4}} \:\mathrm{Wb} \\ $$$$\left(\mathrm{a}\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{H}_{\mathrm{1}} =\frac{\mathrm{N}\:\mathrm{I}_{\mathrm{1}} }{\mathrm{L}}=\frac{\mathrm{120}×\mathrm{1}.\mathrm{5}}{\mathrm{0}.\mathrm{25}}=\mathrm{720}\:\mathrm{Ampere}/\mathrm{meter} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{H}_{\mathrm{2}} =\frac{\mathrm{N}\:\mathrm{I}_{\mathrm{2}} }{\mathrm{L}}=\frac{\mathrm{120}×\mathrm{5}}{\mathrm{0}.\mathrm{25}}=\mathrm{2400}\:\mathrm{Ampere}/\mathrm{meter} \\ $$$$\left(\mathrm{b}\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{B}_{\mathrm{1}} =\frac{\Phi_{\mathrm{1}} }{\mathrm{A}}=\frac{\mathrm{3}×\mathrm{10}^{−\mathrm{4}} }{\mathrm{5}×\mathrm{10}^{−\mathrm{4}} }=\mathrm{0}.\mathrm{6}\:\mathrm{Wb}/\mathrm{m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mu=\frac{\mathrm{B}_{\mathrm{1}} }{\mathrm{H}_{\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\mu_{\mathrm{r}\:} \:\mu_{\mathrm{o}} =\frac{\mathrm{0}.\mathrm{6}}{\mathrm{720}} \\ $$$$\:\:\:\:\:\:\mu_{\mathrm{r}} ×\mathrm{4}\pi×\mathrm{10}^{−\mathrm{7}} =\mathrm{8}.\mathrm{33}×\mathrm{10}^{−\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mu_{\mathrm{r}} =\frac{\mathrm{8}.\mathrm{33}×\mathrm{10}^{−\mathrm{4}} }{\mathrm{4}\pi×\mathrm{10}^{−\mathrm{7}} }=\frac{\mathrm{2082}.\mathrm{5}}{\pi}\approx\mathrm{663}.\mathrm{2} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{B}_{\mathrm{2}} =\frac{\Phi_{\mathrm{2}} }{\mathrm{A}}=\frac{\mathrm{6}×\mathrm{10}^{−\mathrm{4}} }{\mathrm{5}×\mathrm{10}^{−\mathrm{4}} }=\mathrm{1}.\mathrm{2}\:\mathrm{Wb}/\mathrm{m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mu=\frac{\mathrm{B}_{\mathrm{2}} }{\mathrm{H}_{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\mu_{\mathrm{r}} ×\mathrm{4}\pi×\mathrm{10}^{−\mathrm{7}} =\frac{\mathrm{1}.\mathrm{2}}{\mathrm{2400}} \\ $$$$\:\:\:\:\:\:\:\mu_{\mathrm{r}} =\frac{\mathrm{1250}}{\pi}\approx\mathrm{398}.\mathrm{1} \\ $$

Commented by tawakalitu last updated on 04/Jan/17

Thank you sir. God bless you.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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