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Why is my electro magnet switching on and off! Answered

I've recently built an electromagnet with a DC power supply off 5v 4a but when I turn it on it goes on and off . I'v checked this by downloading a sensor on my phone and surely enough every second the magnetic field goes higher than my phone's range and then returns to normal can you please explain why and how to keep a constant field

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Downunder35m

7 weeks ago

By supplying a constant power supply.
I am just assuming that you use some switchmode power supply that is not really rated for such a big inductive load...

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Jack A LopezDownunder35m

Reply 7 weeks ago

Inductance does not matter for DC, except at those moments in time when the voltage is switched on or off.

But then you have to admit those moments, those transient moments, are not actually DC, because in those moments the voltage (and current) is changing with time.

I claim, in the steady state condition,

https://en.wikipedia.org/wiki/Steady_state_(electr...

an electromagnet powered by DC, looks just like a resistor, specifically the resistance present in that length of copper magnet wire.

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Downunder35mJack A Lopez

Reply 7 weeks ago

That is all true but a switchmode supply does not care about it.
All it can see is the inrush current is higher than what it can supply, so it craps out and tries again and again....

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zaintareen

7 weeks ago

Does completely turn off or does the magnetic field just become weaker?
If it emits a strong magnetic field at first and then becomes weaker then it might be because you're driving an inductive load (a coil of wire) and inductive loads (in your case coiled wire) draws in a lot of current and if you refer to the magnetic field formula here http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html You'll see that the magnetic field depends on the current. So when you turn the magnet on, a lot of current is drawn in and the result is that the magnet shows a very high magnetic field but the magnet only does this for a small amount of time after which the current goes back to normal and hence the magnetic field reduces in magnitude. You can reduce this effect by putting a capacitor (preferably of a high value) in parallel with the electromagnet.
However, if your magnet turns off completely, it might be due to a faulty power supply. You should try checking with another power supply.

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Jack A Lopez

7 weeks ago

Another guess is, that the application reading your phone's magnetometer is doing something goofy. Perhaps this is what it does when it detects too high, aka out of range, magnetic field?

By the way, one easy way to make the field (at the sensor) a little weaker, and thus bring it into range, is by simply moving the sensor a few centimeters away from the electromagnet.

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Jack A Lopez

7 weeks ago

My guess is your DC power supply is turning itself on and off.

Moreover I am guessing the reason why your DC power supply is turning itself off, is because you have overloaded it. That is to say your electromagnet draws more current than the power supply can comfortably supply.

By the way, it turns out the DC steady state current of an electromagnet is easy to predict, using Ohm's law. The coil just looks like a length of wire with resistance R. So the DC current is I = V/R.

Thus if the DC resistance of your electromagnet is less than about 1.25 ohm, then you are probably overloading your power supply; i.e. drawing more than 4.0 A from it (since 5.0/1.25 = 4.0 A)

Also you can calculate the expected resistance a length of copper magnet wire, just from its geometry; i.e. its length and its cross section area or gauge number., if you happen to know those numbers.

Also it might be handy if you had some test equipment besides just your phone's magnetometer.

A voltmeter (or multimeter configured as voltmeter) placed in parallel with the electromagnet, will tell you the voltage across it.

An ammeter (or multimeter configured as ammeter) placed in series with the electromagnet, will tell you the current flowing through the electromagnet.

Also if you had a variable DC supply, you could try supply voltages other than 5 volts DC, in particular lower voltages, which should give lower currents (since I=V/R)