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My friend and I were trying out a circuit with 6 volts and found that when the motor was spinning slower, the voltage shown on an analog meter was greater. I vaguely guessed that this was caused by the position of the solenoid or wire in the motor, which affected EMF. Could someone please explain?

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    $\begingroup$ What circuit? What voltage? Please clarify your question. $\endgroup$
    – Puk
    Jun 5 at 6:11
  • $\begingroup$ This question cannot really be answered with some degree of certainty unless more information is given. Is it an simple dc motor or an ac motor? Is the voltage source ac or dc? If the voltmeter is positioned across the terminals of a dc voltage source then the reading on it will be the source voltage minus the potential drop across the source resistance $R_{\rm source}\, I$. Note the the back emf produced by the motor affects the value of the current , $I$, in the circuit. $\endgroup$
    – Farcher
    Jun 5 at 8:01

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When the motor rotates, the magnetic flux through the loop changes, resulting in an induced emf. Consistent with Lenz's law, this emf always acts to reduce the current in the coil (and is hence called the back emf).

The voltage available to supply current equals the difference between the supply voltage and back emf, that is: $$\varepsilon_{\text{available}} = \varepsilon - \varepsilon_{\text{back}}$$

Now, the back emf can be calculated by using Faraday's law: $$\Phi_B = BA \text{ cos}\theta = BA \text{ cos } \omega t$$ $$\varepsilon_{\text{back}} = -N\frac{\text{d}\Phi_B}{\text{dt}} = N\omega BA \text{ sin } \omega t$$

Clearly $\varepsilon_{\text{back}}$ varies as $\omega \text{ sin } \omega t$ and hence decreases with a decrease in rotation rate ($\omega$). It then follows that $\varepsilon_{\text{available}} = \varepsilon - \varepsilon_{\text{back}}$ increases, just as you have measured.

You may also have everyday experiences with motors becoming warm when they are prevented from turning (for example when operating a saw, if the blade becomes jammed it's temperature increases rapidly). This is because $\varepsilon_{\text{back}}$ becomes zero, and hence $\varepsilon_{\text{available}}$ becomes dangerously high.

Hope this helps.

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  • $\begingroup$ Thank you for the detailed explanation! $\endgroup$ Jun 5 at 7:24
  • $\begingroup$ $\sin \omega t$ does not decrease if $\omega$ decreases. It is a function which varies between $-1$ and $+1$. You have made a mistake differentiating the $\cos \omega t$ function. $\endgroup$
    – Farcher
    Jun 5 at 15:10
  • $\begingroup$ @Farcher, Thanks for pointing it out, I fixed it now. $\endgroup$
    – Cross
    Jun 5 at 15:32
  • $\begingroup$ A voltmeter across the terminals of the voltage supply does not measure $\mathcal E_{\rm available}$ rather it measures $\mathcal E$ or a smaller value due the voltage source having an internal resistance. $\endgroup$
    – Farcher
    Jun 5 at 19:51

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