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| Funny little problem - can you answer? |
| << < (4/5) > >> |
| Benta:
Answered in #4. |
| cdev:
So, yes! The Van Allen belt |
| magic:
--- Quote from: FriedMule on May 04, 2022, 09:22:56 pm ---The question is mostly about what you would imagine your Ohm-meter will say when it returns a measuring result before the Ohm-meter's signal can have time to travel all the way. :-) --- End quote --- If the shield is connected to ground of the DMM on both ends then you will read 50Ω, which is the same thing you will read if the remote end (shield and core) are left floating. With no shield and no externally induced currents in the cable, two ends of the cable coming back to you form a twin-wire transmission line. You will read its wave impedance or something approximating it, because if the distance between the two halves isn't constant (like in a round loop) then the wave impedance isn't too. With a floating shield, not 100% sure, but still something similar. |
| Benta:
--- Quote from: magic on May 05, 2022, 05:19:19 am --- --- Quote from: FriedMule on May 04, 2022, 09:22:56 pm ---The question is mostly about what you would imagine your Ohm-meter will say when it returns a measuring result before the Ohm-meter's signal can have time to travel all the way. :-) --- End quote --- If the shield is connected to ground of the DMM on both ends then you will read 50Ω, which is the same thing you will read if the remote end (shield and core) are left floating. --- End quote --- Why would you read 50 ohms? It's a DC measurement. You're talking AC. |
| T3sl4co1l:
Yes, destruction, shield grounded or no, because it's not nearly good enough of a shield: it's only effective above modest frequencies (10s, 100s kHz?), and even still not very (RG-58 is the cheapest 50 ohm for a reason). Such a condition will generate ~mHz frequencies (corresponding to the relative motion of Earth's field, the loop, and sure let's include solar wind in that too), the shield might as well not exist. Or, the shield is just another conductor on a parallel path, no different from twisted pair if one of the conductors had much higher resistance. Note that Earth's rotation by itself doesn't cause induction, because it's on a similar axis to the magnetic field. There's no such thing as a "rotating magnetic field" on an axis parallel to the lines of flux -- one of the conceits of the image of flux lines that can be overly confusing. But the axes don't match exactly, and there will be orbital or dynamic motion (the loop flopping around in space relative to its attachment point), so it will still move up and down through varying field strength and thus generate voltage. If it were a superconducting shield, including the means to keep it superconducting as it passes through the atmosphere, and enough cross-section to keep it from exceeding critical field strength internally (so, doubly so very much NOT RG-58), the center conductor can be effectively shielded and measure approximately nothing. (Likely it would still measure something, but due to stress/flex of the cable itself.) It would also short-circuit the ionosphere to ground, which must draw... I have no idea how much current, but it's going to get HOT from all that plasma it's literally lightning-rod-ing through. Tim |
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