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  1. The induced voltage of the secondary coil of a transformer, is caused by the changing magnetic flux of the primary coil...

  2. ...and by decreasing the number of turns of the primary coil, the flux decreases and the voltage across the secondary coil should also decrease...

  3. ...right, but when I was experimenting, I lowered the primary turns, and that leads to an increase in the secondary voltage. Why?

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2 Answers 2

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The magnitude of the voltage induced in the secondary is the magnitude of the rate of change of flux in the secondary, not the primary.

Neglecting leakage flux, if the flux in the transformer core is $\Phi$, the fluxes through the primary and secondary are $N_1\Phi$ and $N_2\Phi$, respectively. When you apply a voltage on the primary, you set $$V_1 = N_1 \frac{d\Phi}{dt} $$ and the secondary voltage is $$V_2 = N_2 \frac{d\Phi}{dt} = \frac {N_2}{N_1} V_1$$ which is inversely proportional to $N_1$. I've left out the signs here to not obfuscate the point, i.e. all quantities are magnitudes.

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It's not clear enough what you've been doing.

Induction is caused by varying the flux phi, which is proportional to the area encircled and, if you break it down, the current flowing (more precisely: its change).

Assuming, your measurements are reliable, then your observation means: you increased phi somehow (more precisely: its change d_phi/dt).

So, here are a few things that might have gone wrong:

  • applied a different, i.e. larger, change rate d/dt
  • used a second coil with fewer windings (with the same wire material, then your resistance is lower, hence your current I and its change dI/dt will be larger)
  • and so on.
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