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I have tried to find in the webpage some answers about measuring AC magnetic field strength in the range above 50 kHz and more with no luck. Sorry, but if someone could help me will be great.

I need to measure intensity or strength of the magnetic field (1 to 50 mT) between 50khz to 300khz aprox. I have some like "induction heater", with different LC ! L= workcoil., Radius = 2 cm copper tube.

I have used a sensing coil in the middle and i have a strong signal, but iI don't know how to calculate the mT i have. The small sensing coil is 10 turn of awg32 cooper wire with radius = 2,5mm. The voltage is about 0.5 volt, and I can see the out put in the osciloscope(sine wave), I cannot measure the currrent of the sensing coil, and at this frequency my ammeter does not work.

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  • \$\begingroup\$ you mean using something like these magneticsciences.com/magnetic-field-sensors \$\endgroup\$ – JIm Dearden Jan 1 '17 at 19:48
  • \$\begingroup\$ Thank Jim, you are right, but these sensor give 1 mV/mG or V/G, and i have more than 10 mT in my circuit. About 100V output. \$\endgroup\$ – Guillermo Jan 2 '17 at 8:49
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As with any coil, induced voltage is N\$\dfrac{d\Phi}{dt}\$.

So, armed with the number of turns (N), the voltage, the frequency and the area of your loop, you can calculate the average magnetic flux density seen by the coil.

The coil should be measured open circuit and the fewer turns the better because coil parasitic capacitance can easily resonate the circuit and give big errors.

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  • \$\begingroup\$ VThank you Andy, but i am still confused. I mean, i have 10 turn of 0.2 mm of copper wire as a searching coil, with 5 mm diameter and 2 mm in lenght. The induced voltage is 0.5 Volts. I can not measure the current. I know the frequency. I Also build different coils with 5, 20 and 30 turns, and have obtained different voltage. How can i calculate de the magnetic field strength with this experimental value, i know if i have N turn and I current, the B is more or less equal to (N/s)*I. Many Thank \$\endgroup\$ – Guillermo Jan 2 '17 at 16:48
  • \$\begingroup\$ I'm not sure why you are confused, the formula is easy enough and you don't need to measure current - a change in flux induces a voltage in a coil and the number of turns (N) magnifies that voltage by N times. So calculate \$\frac{d\Phi}{dt}\$ based on voltage and N. Then you need to turn this to total flux by using frequency. Then turn to flux density knowing the area of the coil. \$\endgroup\$ – Andy aka Jan 2 '17 at 16:58
  • \$\begingroup\$ Hi Andy, I agree that a change in flux induce a voltage and increase by N (turn) times. But I missing The next two steps And The relations usind The frequency. In my experiment I have 10 turn and the induced voltage is 0.5 volts. The frequency is 120 kHz, and the area of the coil is 80 mm2. \$\endgroup\$ – Guillermo Jan 3 '17 at 21:14
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    \$\begingroup\$ At the peak of your voltage, flux rate of change is maximum. If the frequency were 1 rad/sec there is no conversion factor; with ten turns and (say) 500 mV peak, the flux also hits a peak of +50 mWb when the voltage waveform rises through zero and hits a trough of -50 mWb when the voltage waveform falls through zero. But your frequency is a lot bigger than 1 rad/sec hence the peak flux has to be proportionally smaller. So, you get your peak flux (in webers) and divide that by the area of your coil in sq metres to get flux density in teslas. Then your RMS flux density is peak/\$\sqrt2\$. \$\endgroup\$ – Andy aka Jan 3 '17 at 21:31
  • \$\begingroup\$ Using faraday law, could be like this?. B= V(NA/f). I mean in my experiment A= 80e-5, them A/f= 80e-5*120000. Them B= 0.5 / (10 80e-5*120000) = 0.053 T, 53 mT, is ok? \$\endgroup\$ – Guillermo Jan 3 '17 at 21:36

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