91影视

Relationship between the applied magnetic flux, the grommet initiates a current. The torsion that a rotor emits was given by,

$B=\frac{{渭}_o{N}_{s}I}{l}$

The current,$I\text{by}I=I_o\sin 鈦�\left(2蟺ft\right)$.

$B=\frac{{渭}_o{N}_{s}I}{l}=\frac{\left(4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}\right)\left(200\right)}{0.20\mathrm{m}}\left(I_o\sin 鈦�\left(2蟺ft\right)\right)$

$=12.56脳{10}^{鈭�4}{I}_o\sin 鈦�\left(2蟺ft\right)$

Solenoid Area,

$A=蟺{\left(\frac{d}{2}\right)}^2=蟺{\left(\frac{0.03\mathrm{m}}{2}\right)}^2=7.06脳{10}^{鈭�4}{\mathrm{m}}^2$

Faraday's Law,

$蔚=N_{\text{coil}}\frac{d{\mathrm{桅}}_{\mathrm{m}}}{dt}=N_{\text{coil}}\frac{d12.56脳{10}^{3}A{I}_o\sin 鈦�\left(2蟺ft\right)}{dt}$

$=12.56脳{10}^{鈭�4}{N}_{\text{coil}}A{I}_o\frac{d\sin 鈦�\left(2蟺ft\right)}{dt}$

$=\left(25.12脳{10}^{鈭�4}\right)蟺f{N}_{\text{coil}}A{I}_o\cos 鈦�\left(2蟺ft\right)$

$=\left(25.12脳{10}^{鈭�4}\right)蟺f{N}_{\text{coil}}A{I}_o\cos 鈦�\left(2蟺ft\right)$

The angle $\cos 鈦�\left(2蟺ft\right)=1$, the induced emf

$蔚=\left(25.12脳{10}^{鈭�4}\right)蟺f{N}_{\text{coil}}A{I}_o$

The Ohm's law calculation we using the induced current through the coil

$I_{\text{induced}}=\frac{蔚}{R}$

$I_{\text{induced}}=\frac{\left(25.12脳{10}^{鈭�4}\right)蟺f{N}_{\text{coil}}A{I}_o}{R}$

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12脳{10}^{鈭�4}\right)蟺fA{N}_{\text{coil}}}$

we plug the values for $R,f,I_{\text{induced}}\text{and}{N}_{\text{coil}}$

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12脳{10}^{鈭�4}\right)蟺fA{N}_{\text{coil}}}$

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12脳{10}^{鈭�4}\right)蟺fA{N}_{\text{coil}}}$

$=\frac{\left(0.40\mathrm{惟}\right)\left(0.20\mathrm{A}\right)}{\left(25.12脳{10}^{鈭�4}\right)蟺\left(60\mathrm{Hz}\right)\left(7.06脳{10}^{鈭�4}{\mathrm{m}}^2\right)\left(40\right)}$

$=6\mathrm{A}$