Magnetic Fields In Semi-Infinite Solenoid's

[abdullahkiran]

Homework Statement

Suppose your physics lab class lasts for a very long time - long enough for you to wind a semi-infinite solenoid. (That spool of copper wire is the gift that keeps on giving.) What is the on-axis magnetic field at the end of the solenoid closest to you (ie., not at infinity)?

Homework Equations

1. i think: http://www.netdenizen.com/emagnet/solenoids/Image34.gif
2. mu(0)*n*i
3. http://img.sparknotes.com/figures/2/288a4611d51a3a7ce874c4a906855ac9/latex_img33.gif

The Attempt at a Solution

- well i thought that i could use formula 1, but there is a L term in that, and i thought that it wouldn't work because this is a semi-infinite solenoid.
- the formula from spark notes (4pi*i*n)/(c) , i had never ever seen before so i was scared to go for it.

i don't really know how to apply any of these formulas to the question.

+++++++

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heres a problem from spark notes that's practically the same except it asks to find the other end:
A semi-infinite solenoid is a solenoid which starts at a point, and is infinite in length in one direction. What is the strength of the magnetic field on the axis of the solenoid at the end of a semi-infinite solenoid?

Solution for Problem
To solve this problem, we use the superposition principle. If we put two semi- infinite solenoids end to end, we have an infinite solenoid, and the field strength at any point in the infinite solenoid is (4pi*i*n)/(c) . By symmetry, the contribution of each semi-infinite solenoid is equal, so the contribution of one semi-infinite solenoid must be exactly one half of the magnetic field in an infinite solenoid, or

B = (2pi*i*n)/(c)

This problem displays the power of the superposition principle, which simplifies what would be a complex calculation.

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[Spinnor]

So you have an answer and everything is OK? The magnetic field is one half that of an infinite solenoid.

 

[abdullahkiran]

so really the answer would be 0.5*mu(0)*i*n

i think i get it. superposition of two semi-infinite solenoids counts as a infinite solenoid. so i guess if you find the magnetic field at the non infinite end of a semi-infinite solenoid then it would be just half of the formula.

sweet. i get it now :D

 

[Spinnor]

Another way to think of it which is similar. Set up an integral from plus to minus infinity for the sum magnetic field of the individual coils of the solenoid at some point on the axis. By symmetry of the integral the integral from zero to infinity will be half the integral from plus to minus infinity.

 

[rwisz]

I would first clarify the terminology used in the problem. A semi-infinite solenoid is not a physical object, but rather a mathematical concept used to simplify calculations. It is essentially an infinitely long solenoid with one end cut off. Also, the term "on-axis" refers to the center of the solenoid, not the end closest to the observer.

To find the on-axis magnetic field at the end of a semi-infinite solenoid, we can use the formula B = (2pi*i*n)/(c), as shown in the example from Spark Notes. Here, n represents the number of turns per unit length of the solenoid, and i is the current passing through it. The value of c is the speed of light in a vacuum, and it is included to convert the units of current from amperes to teslas.

Alternatively, we can also use the formula B = mu(0)*n*i, as you suggested. In this case, mu(0) represents the permeability of free space, which is a fundamental constant. The value of L, which represents the length of the solenoid, is not needed in this case because we are only interested in the field at the end of the solenoid.

In summary, the on-axis magnetic field at the end of a semi-infinite solenoid can be calculated using either the formula B = (2pi*i*n)/(c) or B = mu(0)*n*i, depending on which is more convenient for the given information. It is important to carefully consider the terminology and units when applying these formulas to ensure accurate results.