You want to construct a solenoid 20cm long that has an interior magnetic field S

Question

You want to construct a solenoid 20cm long that has an interior magnetic field Strength B-bar of about 1.5 MN/C. The coil has to be wound as a single layer of wire around a form whose diameter is 3.0cm. You have two spools of wire handy. #18 wire has a diameter of about 1.02mm and can carry a current of 6.0A before overheating. #26 wire has a diameter of 0.41mm and can carry up to 1.0A. Which kind of wire should you use and why? what potential difference do you want to put across the coils end? B bar is = cB c is the speed of light

Explanation / Answer

2. Relevant equations

B=NI/(cLE0)...or simplified as about 377NI/L.

3. The attempt at a solution

377NI/L must equal about 1.5X10^6N/C.......so NI/L must equal 3979. I Have an I....and don't really know where to go from here...and not even sure if this is the right path to take. Not really understanding the question? So I unwind the wires and see how many of them I'd need to fill the 3cm diameter? Any help would be appreciated.

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