Field outside a solenoid on its axis

In summary, the conversation discusses the calculation of the magnetic field at a point P on the axis of a tightly wound solenoid with n turns of wire per unit length and carrying a current I. The magnitude of the field is to be calculated at a distance y from one end of the solenoid. Formulas for solenoids and other relevant equations are provided. The first question asks for the current dI in a narrow section of the solenoid at a distance x from the left end. The answer is found to be I*n*dx. The second question asks for the contribution dB to the magnetic field at P due to this narrow section, and the third question asks for the magnitude of B at P. The solution involves using
  • #1
DD31
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Homework Statement


A tightly wound solenoid of length L and radius a that has n turns of wire per unit length carries a current I. Calculate the the magnitude of the magnetic field at a point P on the axis of the solenoid, outside it and a distance y from one end, as shown in the figure.
Note: other than the solenoid being tightly wound, we are not making any approximations regarding the relative sizes of a, L, and y. You answer should be valid anywhere on the axis, even, with some care with signs, inside the solenoid.

http://img256.imageshack.us/img256/8978/lg4xii.jpg

All three of the below questions are looking for formulas:

a): What is the current dI in a narrow section of the solenoid of width dx a distance x from the left end of the solenoid?

b): What is the contribution dB to the magnetic field at P due to this narrow section of the solenoid? Notice that the center of this section is a distance (x+y) from point P. Use your result from a), not dI, in your answer.

c): Find the magnitude of B at P. Note: you may look up the integral.


Homework Equations


Not sure if all these are relevant, but here are the ones I've been trying to use:
Solenoid formulas:
B = (mu_0)nI, where n = number of turns per length

Other formulas:
B = [tex]\frac{(mu_0)IR^2}{2(z^2+R^2)^3/2}[/tex]
where z = distance from the center of a ring of charge and R = the radius of the ring.



The Attempt at a Solution


I really haven't made much of an attempt, because I don't even know where to begin fully.

For a, I first figured that the current would just still be I because the current in any given segment of the wire should be the same throughout the wire...but that's incorrect.

Then, I thought that maybe it wanted the current as a portion of the length, so I found the total length of the wire to be 2pi*a*n*L, or the amount of coil around one loop times the number of loops per length times the length. I divided I by this, but that was incorrect as well.

I feel like the bottom formula I listed in the "relevant equations" section may have something to do with the final integration, as it seems like, at point P, the solenoid may "look like" a simple ring of charge, but if I can't find dI, I don't know how I can even start. I think I'm trying to get something slightly different than what they're asking for right now.

Any help is greatly appreciated...this seems like a very difficult problem so I'll be all the more grateful to anybody who takes the time to help me figure it out.
 
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  • #2
Sorry for the bump, but I'm still flat-footed on this one. I'm going to look at it more tonight extensively, but I'm not quite sure where to even start to find dI.
 
  • #3
Alright...so I've found part a. dI = I*n*dx; I was the current in the wire, so the current from some bundle of wires dx wide is I*n*dx. Makes sense.
Now, for b, I'm trying to find some way to use coils or rings and find the field "above" one of those, but I'm searching for a place to use this dI and I'm not finding.

Thanks in advance for any help.
 

1. What is a solenoid?

A solenoid is a cylindrical coil of wire that produces a magnetic field when an electric current is passed through it. It is commonly used in electromagnets, transformers, and other electronic devices.

2. How is the magnetic field outside a solenoid on its axis calculated?

The magnetic field outside a solenoid on its axis can be calculated using the equation B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current passing through the solenoid.

3. Why is the magnetic field outside a solenoid on its axis stronger than the field inside?

The magnetic field outside a solenoid on its axis is stronger because the magnetic field lines are more spread out and less concentrated compared to the field inside the solenoid. This is due to the fact that the current only flows through the wire, creating a more concentrated field inside, while the field outside is affected by the current in multiple turns of the solenoid.

4. How does the number of turns in a solenoid affect the magnetic field outside on its axis?

The number of turns in a solenoid directly affects the strength of the magnetic field outside on its axis. The more turns there are, the stronger the magnetic field will be.

5. What factors can affect the magnetic field outside a solenoid on its axis?

The magnetic field outside a solenoid on its axis can be affected by the strength of the current passing through the solenoid, the number of turns in the solenoid, and the permeability of the material surrounding the solenoid. Additionally, the distance from the solenoid and the angle of measurement can also affect the strength of the field.

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