Clamp Meter – Why is a Clamp Meter Unaffected by Nearby Wires?

clampelectromagneticmeter

I understand a clamp meter works by measuring the magnetic field generated by current flowing through a wire.

It's well known that clamping 2 wires carrying the same current in opposite directions will show a null reading because one magnetic field is cancelling the other.

What I find surprising though is that if the second wire carrying the opposite current is placed immediately outside the clamp, it has no measurable effect. I was expecting its magnetic field to couple with the clamp to some degree but that seems not to be the case. Why?

Best Answer

Imagine that the return conductor for the red-wire (the blue-wire in your picture) is a million miles away. You would probably agree that it has negligible effect.

Now as the blue return wire is brought closer, the localized magnetic field it produces helps the red wire's magnetic field (on the left) but it also produces an opposing magnetic field passing through the "ring" on the right.

The two opposing fields always cancel to produce the same net effect: -

The magnetic field from the red-wire not drawn for reasons of clarity.

Related Solutions

Electrical – Question for increasing the Flux of an electromagnet

Iron is a poor conductor compared to copper so making an electromagnet from iron wire is off to a bad start. Next, an electromagnet has basically a massive air gap and this is the bit that attracts i.e. the working end of an electromagnet is the air gap. This is generally why the attractive force produced by a solenoid electromagnet has nothing to do with the permeability of the core: -

Force = \$(N\cdot I)^2\cdot 4\pi 10^{-7}\cdot \dfrac{A}{2g^2}\$

F = Force
I = Current
N = Number of turns
g = Length of the gap between the solenoid and the magnetizable metal
A = Area

Electronic – I am measuring a 9W LED with a clamp on ammeter. Why does it only draw 7.62W

That user manual (which you should link to in your question) shows the following:

Figure 1. Just because it's digital, doesn't mean it's accurate.

You are measuring at the bottom end of the range and if it were an analog meter you would be squinting at it to try to make out the reading.

Figure 2. Reading position on an analog scale.

0        0.4       0.8       1.2       1.6       2.0 A
|---------|---------|---------|---------|---------|
 ^-- 0.033 A

The manual shows that that accuracy is only for readings > 5% of full scale, 100 mA on the 2 A range. I have no idea what is meant by "<20 residue readings".
The manual doesn't make any claims about true RMS.

Figure 3. The crest factor of an AC current waveform is the ratio of waveform's peak value to its rms value. Source: Ametek.

Your LED lamp will probably have a high crest factor (peak current to RMS value due to the rectification action of the diodes. The meter doesn't handle this well with a further 7% error possible.

Multiplying V_RMS by I_RMS gives you the VA and not the watts. To calculate the power consumed is more difficult and involves integration of the power curve. Digital power meters sample the voltage and current waveform many times per cycle, multiply the instantaneous readings together to get the instantaneous power, sum them (integration) and average the readings to give the average power.

In short, it's the wrong meter for a true power calculation.

Best Answer

Related Solutions

Electrical – Question for increasing the Flux of an electromagnet

Electronic – I am measuring a 9W LED with a clamp on ammeter. Why does it only draw 7.62W

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