91Ó°ÊÓ

Alternating current, $I=I_0\cos \left(\mathrm{Ó\neg }t\right)$

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$\mathrm{Î\copyright }$

The angular frequency is given by,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$

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

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{Ï„Ï„²õ}}$

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

$\mathrm{ϕ}=N{∫}_a^{b}B.dA$

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

$$\begin{gathered} \mathrm{ϕ}=N{∫}_a^b\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}.hds \\ \mathrm{ϕ}=\frac{N{\mathrm{μ}}_{0}Ih}{2\mathrm{π}}{∫}_a^b\frac{1}{s}ds \\ \mathrm{ϕ}=\frac{N{\mathrm{μ}}_{0}Ih}{2\mathrm{π}}{\left(\mathrm{In}\right)}_a^b \\ \mathrm{ϕ}=\frac{N{\mathrm{μ}}_{0}Ih}{2\mathrm{π}}\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right) \end{gathered}$$

Substitute $I_0$$\cos \left(\mathrm{Ó\neg }t\right)$for I into above equation.

$\mathrm{Ï•}=\frac{N{\mathrm{μ}}_{0}Ih}{2\mathrm{Ï€}}\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)I_0\cos \left(\mathrm{Ó\neg }t\right)$

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

localid="1658144129709" $\mathrm{Î\mu }=-\frac{d\mathrm{Ï•}}{dt}$

Substitute localid="1658144324086" $\left[\frac{N{\mathrm{μ}}_{0}h}{2\mathrm{Ï€}}In\left(\frac{b}{a}\right)I_0\cos \left(\mathrm{Ó\neg }t\right)\right]I_0$for $\mathrm{Ï•}$into above equation.

$\mathrm{Î\mu }=-\frac{d}{dt}\left[\frac{N{\mathrm{μ}}_{0}h}{2\mathrm{Ï€}}In\left(\frac{b}{a}\right)I_0\cos \left(\mathrm{Ó\neg }t\right)\right]$

$\mathrm{Î\mu }=\frac{N{\mathrm{μ}}_{0}h}{2\mathrm{Ï€}}In\left(\frac{b}{a}\right)I_0\mathrm{Ó\neg }\sin \left(\mathrm{Ó\neg }t\right)$

Substitute $2\mathrm{Ï€}f$for $\mathrm{Ó\neg }$into above equation.

$\mathrm{Î\mu }=\frac{N{\mathrm{μ}}_{0}h}{2\mathrm{Ï€}}In\left(\frac{b}{a}\right)I_0×2\mathrm{Ï€}f\sin \left(2\mathrm{Ï€}ft\right)$

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

$\mathrm{Î\mu }=\frac{1000×4\mathrm{Ï€}×{10}^{-7}×1×{10}^{-2}}{2\mathrm{Ï€}}\mathrm{In}\left(\frac{2×{10}^{-7}}{1×{10}^{-7}}\right)×0.5×2\mathrm{Ï€}×60\sin \left(2\mathrm{Ï€}×60\mathrm{t}\right)$

$$\begin{gathered} \mathrm{Î\mu }=12566.370×{10}^{-9}\mathrm{In}\left(2\right)×30\sin \left(120\mathrm{Ï€}t\right) \\ \mathrm{Î\mu }=3.769×{10}^{-4}×0.6931×\sin \left(120\mathrm{Ï€}t\right) \\ \mathrm{Î\mu }=2.612×{10}^{-4}\sin \left(120\mathrm{Ï€}t\right)\mathrm{V} \end{gathered}$$

The expression for the current is given by,

$I=\frac{\mathrm{Î\mu }}{R}$

Substitute $2.612×{10}^{-4}\sin \left(120\mathrm{Ï€}t\right)\mathrm{V}$for $\mathrm{Î\mu }$and $500\mathrm{Î\copyright }$ for Rinto above equation.

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

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

The expression to calculate the inductance is given by,

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

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

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

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

The expression for the back emf is given by,

$\mathrm{Î\mu }=-L\frac{d{l}_R}{dt}$

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

$$\begin{gathered} {\mathrm{Î\mu }}_b=-1.386×{10}^{-3}×5.22×{10}^{-7}×2\mathrm{Ï€}×60×\cos \left(120\mathrm{Ï€}t\right) \\ {\mathrm{Î\mu }}_b=-2.735×{10}^{-7}\cos \left(\mathrm{Ó\neg }t\right)\mathrm{V} \end{gathered}$$

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

$$\begin{gathered} \frac{{\mathrm{Î\mu }}_b}{\mathrm{Î\mu }}=\frac{2.74×{10}^{-7}}{2.61×{10}^{-4}} \\ \frac{{\mathrm{Î\mu }}_b}{\mathrm{Î\mu }}=1.05×{10}^{-3} \end{gathered}$$

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