An alternating current I(t)$\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{\omega }\mathit{t}\mathbf{\right)}$ (amplitude 0.5 A, frequency ) flows down a straight wire, which runs along the axis of a toroidal coil with rectangular cross section (inner radius 1cm , outer radius 2 cm , height 1 cm, 1000 turns). The coil is connected to a 500$\mathbf{\Omega }$ resistor.

(a) In the quasistatic approximation, what emf is induced in the toroid? Find the current, ${\mathit{I}}_{\mathbf{R}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, in the resistor.

(b) Calculate the back emf in the coil, due to the current ${\mathit{I}}_{\mathbf{R}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ . What is the ratio of the amplitudes of this back emf and the "direct" emf in (a)?

Alternating current, $I=I_0\cos \left(\omega t\right)$

The amplitude of current,$I_0=0.5A$

The frequency,$f=60\mathrm{Hz}$

Number of the turns, $N=1000$

The inner radius, a= 1cm

The outer radius, b=2cm

The height, h= 1cm

The resistance, R=500$\mathrm{\Omega }$

The angular frequency is given by,

$\omega =2\pi f$

The expression for the magnetic field inside the toroid is given by,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\tau \tau s}}$

The expression for the magnetic flux through the number of the turns is given by,

$\varphi =N{\int }_a^{b}B.dA$

Substitute $\frac{{\mu }_{0}I}{2\pi s}$ for B and hds for dA into above equation.

$$\begin{gathered} \varphi =N{\int }_a^b\frac{{\mu }_{0}I}{2\pi s}.hds \\ \varphi =\frac{N{\mu }_{0}Ih}{2\pi }{\int }_a^b\frac{1}{s}ds \\ \varphi =\frac{N{\mu }_{0}Ih}{2\pi }{\left(\mathrm{In}\right)}_a^b \\ \varphi =\frac{N{\mu }_{0}Ih}{2\pi }\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right) \end{gathered}$$

Substitute $I_0$$\cos \left(\omega t\right)$for I into above equation.

$\varphi =\frac{N{\mu }_{0}Ih}{2\pi }\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)I_0\cos \left(\omega t\right)$

The induced emf is the negative of the rate of change of flux.

localid="1658144129709" $\epsilon =-\frac{d\varphi }{dt}$

Substitute localid="1658144324086" $\left[\frac{N{\mu }_{0}h}{2\pi }In\left(\frac{b}{a}\right)I_0\cos \left(\omega t\right)\right]I_0$for $\varphi$into above equation.

$\epsilon =-\frac{d}{dt}\left[\frac{N{\mu }_{0}h}{2\pi }In\left(\frac{b}{a}\right)I_0\cos \left(\omega t\right)\right]$

$\epsilon =\frac{N{\mu }_{0}h}{2\pi }In\left(\frac{b}{a}\right)I_0\omega \sin \left(\omega t\right)$

Substitute $2\pi f$for $\omega$into above equation.

$\epsilon =\frac{N{\mu }_{0}h}{2\pi }In\left(\frac{b}{a}\right)I_0\times 2\pi f\sin \left(2\pi ft\right)$

Substitute 0.5 A for $I_0$, 2cm for b , 1cm for a , 60 Hz for , f 1cm for h, $4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}$ and 1000 for N into above equation.

$\epsilon =\frac{1000\times 4\pi \times {10}^{-7}\times 1\times {10}^{-2}}{2\pi }\mathrm{In}\left(\frac{2\times {10}^{-7}}{1\times {10}^{-7}}\right)\times 0.5\times 2\mathrm{\pi }\times 60\sin \left(2\mathrm{\pi }\times 60\mathrm{t}\right)$

$$\begin{gathered} \epsilon =12566.370\times {10}^{-9}\mathrm{In}\left(2\right)\times 30\sin \left(120\pi t\right) \\ \mathrm{\epsilon }=3.769\times {10}^{-4}\times 0.6931\times \sin \left(120\pi t\right) \\ \mathrm{\epsilon }=2.612\times {10}^{-4}\sin \left(120\pi t\right)\mathrm{V} \end{gathered}$$

The expression for the current is given by,

$I=\frac{\epsilon }{R}$

Substitute $2.612\times {10}^{-4}\sin \left(120\pi t\right)\mathrm{V}$for $\epsilon$and $500\mathrm{\Omega }$ for Rinto above equation.

$$\begin{gathered} I=\frac{2.612\times {10}^{-4}\sin \left(120\pi t\right)\mathrm{V}}{500} \\ I=5.224\times {10}^{-7}\sin \left(120\pi t\right)A \end{gathered}$$

Hence the induced emf is $2.612\times {10}^{-4}\sin \left(120\pi t\right)\mathrm{V}$ and current in the resistor is $5.224\times {10}^{-7}\sin \left(120\pi t\right)A$.

The expression to calculate the inductance is given by,

$L=\frac{{\mu }_0{N}^{2}h}{2\pi }\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$

Substitute 2cm for b , 1cm for a , 1cm for h , $4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}$ and 1000 for N into above equation.

$L=\frac{4\pi \times {10}^{-7}{1000}^2\times 1\times {10}^{-2}}{2\pi }\mathrm{In}\left(\frac{2\times {10}^{-2}}{1\times {10}^{-2}}\right)$

$$\begin{gathered} L=2\times {10}^{-3}\times 0.6931 \\ L=1.386\times {10}^{-3}\mathrm{H} \end{gathered}$$

The expression for the back emf is given by,

$\epsilon =-L\frac{d{l}_R}{dt}$

Substitute 60Hz for f , $1.386\times {10}^{-3}\mathrm{H}$for L and $5.22\times {10}^{-7}\cos \left(120\pi t\right)$A/s for $d{l}_R/dt$ into above equation.

$$\begin{gathered} {\epsilon }_b=-1.386\times {10}^{-3}\times 5.22\times {10}^{-7}\times 2\pi \times 60\times \cos \left(120\pi t\right) \\ {\epsilon }_b=-2.735\times {10}^{-7}\cos \left(\omega t\right)\mathrm{V} \end{gathered}$$

The ratio of the back emf and direct emf can be calculated as,

$$\begin{gathered} \frac{{\epsilon }_b}{\epsilon }=\frac{2.74\times {10}^{-7}}{2.61\times {10}^{-4}} \\ \frac{{\epsilon }_b}{\epsilon }=1.05\times {10}^{-3} \end{gathered}$$

Hence the value of the back emf is $-2.735\times {10}^{-7}\cos \left(\omega t\right)\mathrm{V}$ and ratio of the back and direct emf is $1.05\times {10}^{-3}$.