Integrate the on axis field of finite solenoid from a thin shell one, HELP

In summary, the conversation discusses a problem with finding the on axis field of a finite solenoid using the thin shell equation and the on axis finite solenoid equation. The solution involves integrating over dr from r1 to r2 and expressing di in terms of dr using current density. The answer is found using this method.
  • #1
demon_samuel
3
0
Dear all,

i met a problem that i couldn't get the on axis field of a finite solenoid that integrating from a thin shell solenoid equation.

The thin shell equation link:
http://www.netdenizen.com/emagnet/solenoids/thinsolenoid.htm

And the on axis finite solenoid equation link:
http://www.netdenizen.com/emagnet/solenoids/solenoidonaxis.htm

If i was right, the integral of (a/sqrt (a*a+r*r)) dr should be a*ln(r+sqrt(r*r+a*a)), if so i couldn't figure out how to get the (r2-r1) term in the second link equation.

Thank you for your attention

Samuel
 
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  • #2
To get the second expression from the first, you need to integrate over dr from r1 to r2. You do this by adding "solenoids" of radius r, thickness dr, and length l carrying current di. According to the first expression, the contribution from one such solenoid is

[tex]dB = \frac{\mu_{0}N di}{2 l}[\frac{x_{2}}{\sqrt{x^{2}_{2}+r^{2}}}-\frac{x_{1}}{\sqrt{x^{2}_{1}+r^{2}}}] [/tex]

Before you integrate, you need to express di in terms of dr. Hint: Think in terms of current density.
 
  • #3
Thank you very much, i got the answer.
 

1. What is a finite solenoid and how is it different from a thin shell one?

A finite solenoid is a coil of wire with a finite length and a finite number of turns, while a thin shell solenoid is a hypothetical coil with an infinite number of turns and negligible thickness. The main difference between the two is that a finite solenoid has a finite magnetic field, while a thin shell one has an infinitely strong magnetic field.

2. What is the on-axis field of a finite solenoid?

The on-axis field of a finite solenoid is the magnetic field strength at a point on the central axis of the solenoid, which is also known as the z-axis. It is calculated using the equation B = μ0nI, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current through the solenoid.

3. How do you integrate the on-axis field of a finite solenoid from a thin shell one?

To integrate the on-axis field, you need to divide the solenoid into small segments and calculate the magnetic field strength at each segment. Then, you can use the formula for the magnetic field due to a current-carrying wire to find the contribution of each segment. Finally, you can add up all the contributions to get the total on-axis field.

4. Why is it important to integrate the on-axis field of a finite solenoid from a thin shell one?

Integrating the on-axis field allows us to understand the behavior of a real finite solenoid by approximating it as a thin shell solenoid with an infinite number of turns. It also helps us to calculate the magnetic field strength at any point along the axis, which is important in many practical applications such as designing electromagnets or understanding the behavior of magnetic materials.

5. Can you provide an example of integrating the on-axis field of a finite solenoid from a thin shell one?

Sure, let's say we have a finite solenoid with a length of 10 cm and 500 turns, carrying a current of 2 amps. We want to find the on-axis field at a point 5 cm away from the center of the solenoid. First, we divide the solenoid into small segments of length Δz. Then, we use the formula B = μ0nI to calculate the magnetic field strength at each segment. Finally, we integrate the contributions from each segment to find the total on-axis field. In this case, the answer would be approximately 1.26 x 10^-4 Tesla.

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