Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
What is the magnitude of the magnetic field?
A long solenoid with 70.0 turns/cm and a radius of 5.00 cm carries a current of 16.0 mA. A current of 2.00 A exists in a straight conductor located along the central axis of the solenoid.
(a) At what radial distance from the axis will the direction of the resulting magnetic field be at 45.0° to the axial direction?
d=I2/(2pi * n * I1) = 2.00A / (2 * pi * 70turns/cm * 0.0160A) = 0.2842m
Now they want to know what is the magnitude of the magnetic field there.
I thought I had some ideas on how to calculate this question, but none of my calculations are correct. Any help would be much appreciated. Thanks for your time!
1 Answer
- biire2uLv 71 decade agoFavorite Answer
I think you use a different equation when you figure for a long solenoid. Your solenoid only specifies 70 turns per cm but doesn't give a length. Plus it says it is a long solenoid which is the clue you use the equation: B = unl,
B = magnetic field
u = permeability
n = turn density
l = current
When I plug in the numbers (using core permeability of 200)
B = 3.518 Tesla or 35185.8 gauss
Look at this interactive link on solenoids called "Solenoid Magnetic Field Calculation", scroll down to third box down:
