A 40-turn, 4.0-cm-diameter coil with $R=0.40\mathrm{\Omega }$ surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The

60 $\mathrm{Hz}$current through the solenoid is I = $I_0\sin \left(2\pi ft\right)$. What is $I_0$ if the maximum induced current in the coil is $0.20\mathrm{A}?$?

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

$B=\frac{{\mu }_o{N}_{s}I}{l}$

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

$B=\frac{{\mu }_o{N}_{s}I}{l}=\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(200\right)}{0.20\mathrm{m}}\left(I_o\sin \left(2\pi ft\right)\right)$

$=12.56\times {10}^{-4}{I}_o\sin \left(2\pi ft\right)$

Solenoid Area,

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.03\mathrm{m}}{2}\right)}^2=7.06\times {10}^{-4}{\mathrm{m}}^2$

Faraday's Law,

$\epsilon =N_{\text{coil}}\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}=N_{\text{coil}}\frac{d12.56\times {10}^{3}A{I}_o\sin \left(2\pi ft\right)}{dt}$

$=12.56\times {10}^{-4}{N}_{\text{coil}}A{I}_o\frac{d\sin \left(2\pi ft\right)}{dt}$

$=\left(25.12\times {10}^{-4}\right)\pi f{N}_{\text{coil}}A{I}_o\cos \left(2\pi ft\right)$

$=\left(25.12\times {10}^{-4}\right)\pi f{N}_{\text{coil}}A{I}_o\cos \left(2\pi ft\right)$

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

$\epsilon =\left(25.12\times {10}^{-4}\right)\pi f{N}_{\text{coil}}A{I}_o$

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

$I_{\text{induced}}=\frac{\epsilon }{R}$

$I_{\text{induced}}=\frac{\left(25.12\times {10}^{-4}\right)\pi f{N}_{\text{coil}}A{I}_o}{R}$

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12\times {10}^{-4}\right)\pi fA{N}_{\text{coil}}}$

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

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12\times {10}^{-4}\right)\pi fA{N}_{\text{coil}}}$

$I_o=\frac{R{I}_{\text{induced}}}{\left(25.12\times {10}^{-4}\right)\pi fA{N}_{\text{coil}}}$

$=\frac{\left(0.40\mathrm{\Omega }\right)\left(0.20\mathrm{A}\right)}{\left(25.12\times {10}^{-4}\right)\pi \left(60\mathrm{Hz}\right)\left(7.06\times {10}^{-4}{\mathrm{m}}^2\right)\left(40\right)}$

$=6\mathrm{A}$