Physics Asked by Chris Bolig on February 14, 2021
Say you have some $textbf{B}$ field from an alternating current, $textbf B(t) = textbf{B}_0cos{omega ,t}$. And using Faraday law, e.m.f. $varepsilon$ = turns of a coil $times$ the time derivative of the flux of the $textbf{B}$ field going through the coil area. $$varepsilon = -N cdot frac{dphi_B}{dt}.$$
If my $textbf B(t)$ max has a amplitude of $0.0005$ T and a frequency of $500,000$ kHz, then does the derivative of $B(t)$ have a $1000000pi$ constant pop out front? This seems wrong, like you can make the $varepsilon$ voltage really high just from the frequency?
I feel like I’m missing something simple, Thank you.
We have the proof mathematically. It is clear that increasing the frequency increases the current induced.
However if we try analysing in the physical sense, we see that a magnetic field that varies with a large frequency oscillates faster than a one with smaller frequency. The principle of electromagnetic induction states that to generate current we should have a varying magnetic field, and not so much about how large the field is.
So yes, we can make the current very large by increasing the frequency.
Answered by Phy_Amatuer on February 14, 2021
Related point of interest: The low voltage DC power supplies in many modern devices (like DVD players) rectify the incoming AC, and use that (high voltage) DC to power a high frequency oscillator. The high frequency drives a small step down transformer (with a high self inductance) and that output is rectified to give the low voltage output. This greatly reduces the size and weight of the transformer and the filter capacitors.
Answered by R.W. Bird on February 14, 2021
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