# Example problem of induced emf in a 100 turn coil by a 100 turn coil which has 90kHz sin wave input voltage

Main Question: Let's say there are two coils each 100 turn and diameter 30 mm, And the second coil is perpendicular to the first coil's magnetic field (placed both coils next to each other; we can see through the holes of both coils). And we want to induce emf in second coil by the first coil which is connected to a function generator which is producing 90 kHz sine wave, 12 V peak to peak voltage. The circuit resistance of the first coil is 48 Ω hence current through the first coil is 0.25 A. Calculate the induced emf in the second coil as a function of t. Edit: coil's height(width) is 20mm,

Answer:

$$Induced \space emf = -N \times \frac{d(\phi)}{dt} \space volts$$

To calculate induced emf in second coil we need to find the equation for flux first.

\begin{align} \phi &= BA \times cos(\theta) \space Wb \\ \phi &= B \times \pi(0.03)^2 \times cos(0) \\ \phi &= \pi(0.03)^2 \times B \\ \end{align}

Now we need to calculate the magnetic field strength B,

$$B = (100) \frac{\mu \times I}{2\pi r} \space T$$ where μ = permeability of the air, I = current , r = distance, 100 = number of turns.

Question 1: How should I express the current I? Because they say in an inductor voltage and current has 90 degree phase shift. So voltage is oscillating by 0.25sin(θ) with 90 kHz frequency, Hence should I write current as 0.25sin(θ) OR 0.25cos(θ)? Did I wrote the equation for current right? Below in the equation I wrote 0.25sin(ωt) because θ = ω x t !???

$$B = (100)\frac{\mu}{2\pi r} \times (0.25sin(\omega t) )$$ $$\omega = 2\pi f$$ $$\omega = 2\pi \times 90e+3$$ $$\omega = 180\pi e+3$$

$$B = (100)\frac{\mu}{2\pi r} \times (0.25sin((180\pi e+3)t))$$

Now let's put the value of B into the flux equation,

$$\phi = (0.03)^2\pi \times [\frac{100\mu \times 0.25sin((180\pi e+3)t)}{2\pi r}]$$

Now we need to calculate the rate of change of flux φ with respect to time,

\begin{align} \frac{d(\phi)}{dt} &= \frac{(0.03)^2\pi \times 100\mu}{2\pi r} \times \frac{d[0.25sin((180\pi e+3)t)]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(0.03)^2\pi \times 100\mu \times 0.25}{2\pi r} \times \frac{d[sin((180\pi e+3)t)]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(0.03)^2\pi \times 100\mu \times 0.25}{2\pi r} \times cos((180\pi e+3)t) \times \frac{d[(180\pi e+3)t]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(0.03)^2\pi \times 100\mu \times 0.25 \times (180\pi e+3)}{2\pi} \times \frac{cos((180\pi e+3)t)}{r} \\\\ \frac{d(\phi)}{dt} &= (7.994e-3) \times \frac{cos((180\pi e+3)t)}{r} \\\\ \end{align}

Now induced emf is, \begin{align} Induced \space emf &= -(100) \times (7.994e-3) \times \frac{cos((180\pi e+3)t)}{r} \\ Induced \space emf &= -(0.7994) \times \frac{cos((180\pi e+3)t)}{r} volt \\ \end{align}

Question 2: I don't know why but when I plot this graph the output seems wired! Desmos graph for induced emf . I mean when r is 0.02 meter (2 cm close) the voltage seems to go all the way up to 40 V peak to peak!!! Is the answer correct?

Question 3: Should I take coil's height in consideration too?

• placed both coils next to each other ... do you mean that they form a figure eight? Commented Apr 1 at 19:09
• @jsotola No, I mean they placed so as it seems like it is a single coil. and then we can increase or decrease the distance only in one direction - forwards or backwards. Commented Apr 1 at 19:30
• then your post should contains something like placed two coils on a common axis Commented Apr 1 at 19:36

## 2 Answers

we want to induce emf in second coil by the first coil which is connected to a function generator which is producing 90kHz sine wave, 12v peak to pesk voltage. The circuit resistance of the first coil is 48ohm hence current through the first coil is 0.25amps.

Incorrect. You have not considered the inductive reactance of the 100 turn primary coil. That will limit the current to much much less than 0.25 amps at 90 kHz.

Question 2: I don't why but when I plot this graph the output seems wired! Desmos graph for induced emf . I mean when r is 0.02 meter (2cm close) the voltage seems to go all the way up to 40 volts peak to peak

That's a knock-on effect of your incorrect assumption that current was 0.25 amps.

Question 3: Should I take coil's height in consideration too?

If by height you mean the net width of 100 turns then it won't be too far wrong if you assume all the 100 wires exist at one point.

This image may summarize what to do to solve your problem: -

Image taken from here and here.

Question 1: How should I express the current I? Because they say in an inductor voltage and current has 90 degree phase shift. So voltage is oscillating by 0.25sin(θ) with 90 kHz frequency, Hence should I write current as 0.25sin(θ) OR 0.25cos(θ)? Did I wrote the equation for current right? Below in the equation I wrote 0.25sin(ωt) because θ = ω x t !???

We know Voltage = sin(θ) , oscillating at 90kHz , 12 volt peak to peak, And to write all this in one single equation, $$V(t) = 12\sin( \omega t + phase \space shift )$$ $$Angular \space frequency \space = \omega = 2\pi f$$ $$\omega = 2\pi(90e+3)$$ $$V(t) = 12\sin( (2\pi(90e+3))t + 0)$$

To use the formula, $$I=\frac{V}{R}$$

First we need to calculate total impedance and it's phase shift with respect to voltage's phase shift, \begin{align} Resistance &= 48 \Omega \\\\ Reactance &= 2\pi fL \\ Reactance &= 2\pi (90e+3)(2.6e-4) \Omega \\ Reactance &= 147 \Omega \end{align} Now, to calculate the total impedance we have to write it as complex number in rectangular form,

\begin{align} Z &= (Real + j*Imaginary) \Omega \\ Z &= (48 + j0) + (0 + j147) \Omega \\ Z &= 48 + j147 \Omega \end{align} To find actual impedance value and phase shift we can write this complex number in polar form , $$Z = 154.64 \angle 71.92 ^\circ$$ So total Impedance Z is 154.64 Ω with the phase shift of 71.92 degree.

Now, We can use Ohm's law as, \begin{align} I(t) &= \frac{V(t)}{Z} \\\\ I(t) &= \frac{12\sin( (2\pi(90e+3))t) \angle 0 ^\circ}{154.64 \angle 71.92 ^\circ} \end{align}

After performing complex division, $$I(t) = 0.078\sin( (2\pi(90e+3))t) \angle -71.92 ^\circ$$ $$\angle -71.92 ^\circ = -1.26 \space rad.$$ $$I(t) = 0.078\sin( (2\pi(90e+3))t - (-1.26))$$

Note: The phase shift is written inside the sine function. And negative sign is because current lags behind voltage in an inductor. And voltage is the reference for phase shift of the current.

Final equation of Current, $$I(t) = 0.078\sin( (2\pi(90e+3))t + 1.26) Ampere$$

Now, Let's solve the main question,

Main Question: Let's say there are two coils each 100 turn and diameter 30 mm and the second coil is perpendicular to the first coil's magnetic field (placed both coils next to each other; we can see through the holes of both coils). And we want to induce emf in second coil by the first coil which is connected to a function generator which is producing 90 kHz sine wave, 12 V peak to peak voltage. The circuit resistance of the first coil is 48 Ω hence current through the first coil is 0.25 A. Calculate the induced emf in the second coil as a function of t.

\begin{align} B &= (100) \frac{\mu \times I}{2\pi r} \space T \\\\ B &= (100)\frac{\mu}{2\pi r} \times 0.078\sin( (2\pi(90e+3))t + 1.26) ) \\\\ \end{align}

Now let's put the value of B into the flux equation,

$$\phi = BA \times cos(\theta) \space Wb$$ The dimeter of the coil is 30mm hence radius is 15mm, and height(width) of the coil is 20mm, \begin{align} A &= \pi (r)^2h \\ A &= \pi \cdot (0.015)^2 \cdot (0.020) \\ A &= 14.13e-6 \\\\ \end{align}

\begin{align} \phi &= B \times (14.13e-6) \times cos(0) \\\\ \phi &= (14.13e-6) \times B \\\\ \phi &= (14.13e-6) \times [\frac{100\mu \times 0.078\sin( (2\pi(90e+3))t + 1.26) )}{2\pi r}] \\\\ \phi &= \frac{(14.13e-6)(100(1.25663753e−6))}{2\pi r} \times 0.078\sin((2\pi(90e+3))t + 1.26) \\\\ \phi &= \frac{(2.83e-10)}{r} \times 0.078\sin((2\pi(90e+3))t + 1.26) \\\\ \end{align}

Now we need to calculate the rate of change of flux Φ with respect to time,

\begin{align} \frac{d(\phi)}{dt} &= \frac{(2.83e-10)}{r} \times \frac{d[0.078\sin((2\pi(90e+3))t + 1.26)]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(2.83e-10)(0.078)}{r} \times (\cos((2\pi(90e+3))t + 1.26)) \times \frac{d[(2\pi(90e+3)t+1.26)]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(2.21e-11)}{r} \times (\cos((2\pi(90e+3))t + 1.26)) \times \frac{d[(2\pi(90e+3)t + 1.26)]}{dt} \\\\ \frac{d(\phi)}{dt} &= \frac{(2.21e-11)}{r} \times (\cos((2\pi(90e+3))t + 1.26)) \times (2\pi(90e+3)) \\\\ \frac{d(\phi)}{dt} &= \frac{(12.47e-6)}{r} \times (\cos((2\pi(90e+3))t + 1.26))\\\\ \end{align}

Hence, Induced emf in the second coil is,

\begin{align} Induced \space emf &= -(100) \times \frac{(12.47e-6)}{r} \times (\cos((2\pi(90e+3))t + 1.26)) \\\\ Induced \space emf &= \frac{-(0.001247)}{r} \times (\cos((2\pi(90e+3))t + 1.26)) volt \end{align}

Note: r is the radius(distance between two coils) and we could use it as a variable while plotting graph, Induced emf graph