In lecture, you may have seen a video of a frog being levitated by an extremely strong magnetic field. It takes magnetic fields
in the neighborhood of 12 Teslas to accomplish this. We've learned that we can make magnetic fields simply by passing currents through coils of wire, so let's find out what happens if we try to make a homemade magnetic levitator We're going to take lots of copper wire and wrap it into a coil. The bigger the coil is, the harder this project will be, so let's give it a radius of 10 cm, which is big enough to comfortably fit a frog but not unmanageable. Realistically, we're not going to be able to get more than 10,000 turns of the wire into one place, so let's take that to be our N. How much current will we need to pass through the wire to get the necessary magnetic field in the center of the coil? 1.91x102 A Opts You are correct. Submissions to practice problems are not permanently reco That's kind of a lot, but it's not necessarily a deal-breaker. A more important consideration will be the power dissipated in all that wire. That means we need to know the total resistance coil? of the wire. I can look up some information online and discover that 8-gauge copper wire has 0.002 Ohms of resistance for every meter of wire. What will be the total resistance of our 1.26x101 Ohm 0pts You are correct. Previous Tries Submissions to practice problems are not permanently recorded Now, regardless of what your numbers gave you, let's assume that in the first part you found a current of 600 A and in the second part you found a resistance of 13 Ohms. How much power will be dissipated by our coil? 4.68x106 w Opts You are correct. Submissions to practice problems are not permanently recorded
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