Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 32 - The Magnetic Field - Exercises and Problems - Page 959: 50

Answer

Not feasible.

Work Step by Step

We should find the number of turns and then multiply it by the diameter of the wire to find the length of the coil. If it has the same length (or less) than that of the Alnico magnet, we can do it. But if the length needed for the coil is greater than that of the Alnico magnet, then we can't. Assuming that we will wound the coils as close together as possible. This means that the length of the coil is given by $$l=ND\tag 1$$ where $D$ is the diameter of the wire. The needed solenoid has the same length as the Alnico magnet. Recalling that the magnetic field strength inside a solenoid is given by $$B=\dfrac{\mu_0NI}{l}$$ So the number of turns is given by $$N=\dfrac{lB}{\mu_0 I}$$ Plug the known; $$N=\dfrac{(0.08)(0.10)}{(4\pi \times 10^{-7})(2)}=\bf 3183\;\rm turn\tag 2$$ We know, from the previous problem, that the wire of diameter 0.41 mm can carry a current of 1 A as a maximum, and the wire of diameter of 1.02 mm can carry a current of 6 A as a maximum. Assuming that this is a linear relation where we can increase the diameter of the wire that increases the maximum current it can hold. $$\rm 1.02\;mm-0.41\;mm=0.61$$ The wire diameter increased by 0.61 mm to increase the current 5 times. So for increasing one extra 1 A, $$\dfrac{0.61\;\rm mm}{5}= 0.122\;\rm mm$$ Add this extra width to the original wire of 0.41-mm diameter, we can use a wire of $0.41+0.122=\bf 0.532\;\rm mm$ Assuming we can use a wire of 0.55 mm diameter. The one layer of the coil will take turns of $N_1$ which is given by (1). $$N_1=\dfrac{l}{D}=\dfrac{0.08}{0.55\times 10^{-3}}=\bf 145\;\rm turn$$ So the number of layers is then given by $$N_{\rm layer}=\dfrac{N}{N_1}=\dfrac{3183}{145}\approx 22\;\rm Layer$$ This number of layers will make the total diameter of the solenoid $$D_{\rm solenoid }=2\times (22D)=22\times 0.55=\bf 24.2\;\rm mm$$ where $D$ is the diameter of the wire used, and 2 is due to the wire being wrapped around twice times. That's if we assumed that the wrapping of the wire starts from zero radii. This is the minimum diameter for the solenoid. Recalling that the diameter of the Alnico magnet is 20 mm. So it is too hard to reach to the same result but this is not impossible. In other words, it seems not feasible.
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