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I've created a large electromagnet using a 12" long, 3" diameter mild steel core. I'm using 18 awg magnet wire. Wire length is 3000 ft which gives me approximately 3600 turns with resistance at 18 ohms.

Case1: Using a variable transformer I dial up the voltage from 0 VAC to 120 VAC and the magnet barely picks up paper clips. The wire gets warm pretty fast.

Case2: Using a 9-volt battery the magnet is easily 10 times stronger than in case1.

Case3: Using a variable DC power supply I dial up the voltage from 0 VDC to 30 VDC. The magnet pulls nails out of my hand. It is easily 100 times stronger than case1.

Can anyone please help me understand why this is happening?

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The strength of the magnetic field is proportional to the magnitude of the current. For case 1, the magnitude of the ac current varies sinusoidally. The current is also limited due to inductive reactance. Consequently, the strength of the magnetic field also varies being a maximum at the peak values of current, and zero when the current is zero.

For cases 2 and 3, the current is constant so the strength of the magnetic field is also constant. The greater the dc voltage, the greater the current and strength of the magnetic field. Also there is no inductive reactance to limit the magnitude of the current.

I understand what you are saying but I'm perplexed by my current measurements. in Case3, current is 1.89 amps. This adds up for me since E=IR (30=I*18) gives me I=30/18=1.67. But in case1, with 120 VAC the measured current is only .9 amps. E=IR gives me a resistance of R=133.33 ohms. My assumption is that I am getting impedance from the steel core that makes the resistance feel like 133.33 ohms when the actual static resistance is 18 ohms.

In Case 1 you have, in addition to the ohmic resistance of 18 ohms, the inductive reactance of the coil which is

$$X_{L}=2πfL$$

Where $f$ is the ac frequency, or 60 Hz in your case.

So for Case 1 the magnitude of the total equivalent impedance, in ohms, is not $R$, but

$$Z_{equiv}=(X_{L}^{2}+R^2)^{1/2}$$

In cases 2 and 3, there is no inductive reactance ($f=0$) so the only impedance is the ohmic resistance of 18 ohms.

Hope this helps.

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  • $\begingroup$ Thank you Bob D. I understand what you are saying but I'm perplexed by my current measurements. in Case3, current is 1.89 amps. This adds up for me since E=IR (30=I*18) gives me I=30/18=1.67. But in case1, with 120 VAC the measured current is only .9 amps. E=IR gives me a resistance of R=133.33 ohms. My assumption is that I am getting impedance from the steel core that makes the resistance feel like 133.33 ohms when the actual static resistance is 18 ohms. $\endgroup$
    – Bradman
    Commented Feb 6, 2020 at 13:51
  • $\begingroup$ @Bradman I've addressed this in an update to my answer. Hope it helps. $\endgroup$
    – Bob D
    Commented Feb 6, 2020 at 14:32
  • $\begingroup$ It does help. Thank you. Now my issue is that I am having a hard time calculating L. The formula I believe is N^2μ0μr(D/2)[ln(8D/d)-2]. N=3600, μ0 is .000126 for carbon steel, μr is 100??, D is .0762 meters (3 inches), d is .0010236 meters (.0403 inches). But when I run these number it gives me L=39732. This gives me 15 million ohms impedance. haha yikes. Any ideas on what i'm doing wrong? $\endgroup$
    – Bradman
    Commented Feb 6, 2020 at 14:40
  • $\begingroup$ @Bradman Sorry I can't follow your math (learn to use MathJax in future). But if I use your calculation of 133 Ohm, square it and set it equal to my $Z_{equiv}$ (assuming no core losses), I theoretically get $L$ = 0.35 H. $\endgroup$
    – Bob D
    Commented Feb 6, 2020 at 14:53
  • $\begingroup$ @Bradman Maybe you have a scaling error of a factor of 5? $\endgroup$
    – Bob D
    Commented Feb 6, 2020 at 15:08

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