How Do You Calculate the Magnetic Field at a Point Along the Axis of a Solenoid?

In summary: B}= \frac{\mu_o I}{2} \int_{x=l+a}^{x=a} \frac{dx}{(R^2 +x^2)^{3/2}}There is no need to integrate over the entire solenoid. You can just integrate over the interval x=l+a to x=a.
  • #1
KracniyMyedved
1
0

Homework Statement


A solenoid of length L and radius R lies on the y-axis between y=0 and y=L and contains N closely spaced turns carrying a steady current I. Find the magnetic field at a point along the axis as a function of distance a from the end of the solenoid.


Homework Equations


Magnetic field on the axis (distance x from center) due to a current carrying ring : [itex]\vec{B}= \frac{\mu_o I R^2}{2(R^2 +x^2)^{3/2}}[/itex]
Biot-Savart Law: [itex]\vec{B}=\frac{\mu_o I}{4\pi} \int_A^B \frac{\vec{ds}\mathbf{x}\vec{r}}{r^3}[/itex]


The Attempt at a Solution


My best attempt is based around modelling the solenoid as a collection of current carrying rings covering a distance of L.

[itex]\vec{B}= \frac{\mu_o I R^2}{2(R^2 +x^2)^{3/2}}[/itex] for each ring of current

[itex]\vec{B}= \int_{l+a}^a \frac{\mu_o I R^2}{2(R^2 +x^2)^{3/2}}dx[/itex]

[itex]\vec{B}= \frac{\mu_o I R^2}{2} \int_{l+a}^a \frac{dx}{(R^2 +x^2)^{3/2}}[/itex]

Integrating via trig substitution, using x = Rtanθ, which implies [itex]dx=Rsec^2 θ dθ[/itex] and [itex](R^2 + x^2)^{3/2}=R^3 sec^3 θ[/itex]

[itex]\vec{B}= \frac{\mu_o I R^2}{2} \int_{x=l+a}^{x=a} \frac{Rsec^2 \theta d\theta}{R^3 sec^3 \theta}[/itex]

[itex]\vec{B}= \frac{\mu_o I}{2} \int_{x=l+a}^{x=a} \frac{d\theta}{sec\theta}[/itex]

[itex]\vec{B}= \frac{\mu_o I}{2} \int_{x=l+a}^{x=a} cos\theta d\theta[/itex]

[itex]\vec{B}= \frac{\mu_o I}{2} sin\theta [/itex] for x from x=l+a to x=a

Constructing a triangle from the substitution let's us find [itex]sin\theta[/itex] in terms of R and x:
[itex]sin\theta = \frac{x}{(x^2 + R^2)^{1/2}}[/itex]

Finally,

[itex]\vec{B}= \frac{\mu_o I}{2}(\frac{a}{(a^2+R^2)^{1/2}} - \frac{l+a}{((l+a)^2 +R^2)^{1/2}})[/itex]

I'm reasonably sure this is wrong because the next part of the question implies a dependence on N, as well I think my modelling the solenoid as I did was a bit shaky, but I cannot think of another way to go about this.
 
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  • #2
I'm reasonably sure this is wrong because the next part of the question implies a dependence on N, as well I think my modelling the solenoid as I did was a bit shaky, but I cannot think of another way to go about this.
Well you should expect a dependence on N wouldn;t you? After all, the field due to 2 loops must be different from the field due to 1 loop right?

This is a standard problem - so you could just look it up and see how other people approach it.
i.e. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html
 
  • #3
KracniyMyedved said:
My best attempt is based around modelling the solenoid as a collection of current carrying rings covering a distance of L.
[itex]\vec{B}= \frac{\mu_o I R^2}{2(R^2 +x^2)^{3/2}}[/itex] for each ring of current

[itex]\vec{B}= \int_{l+a}^a \frac{\mu_o I R^2}{2(R^2 +x^2)^{3/2}}dx[/itex]

What you are actually doing here is integrating the field due to small section of the solenoid of length dx (which you later substitute the limits for the length of the solenoid) which is not necessarily a single loop, but consists of ndx loops, n being the loops per unit length.
 

FAQ: How Do You Calculate the Magnetic Field at a Point Along the Axis of a Solenoid?

1. What is a solenoid?

A solenoid is a coil of wire that carries an electric current and produces a magnetic field. It is typically cylindrical in shape and can be made from various materials such as copper, aluminum, or steel.

2. How does a solenoid create a magnetic field?

When an electric current flows through a solenoid, it creates a magnetic field around the coil. The direction of the magnetic field can be determined using the right-hand rule, where your thumb points in the direction of the current and your fingers curl in the direction of the magnetic field.

3. What factors affect the strength of the magnetic field produced by a solenoid?

The strength of the magnetic field produced by a solenoid is affected by the number of turns in the coil, the current flowing through the coil, and the material of the core (if any) inside the coil. Increasing any of these factors will result in a stronger magnetic field.

4. How is the direction of the magnetic field inside a solenoid determined?

The direction of the magnetic field inside a solenoid is determined by the direction of the electric current flowing through the coil. The magnetic field lines inside the solenoid are parallel to each other and in the same direction as the current.

5. What are some practical applications of the magnetic field produced by a solenoid?

The magnetic field produced by a solenoid is used in a variety of applications, such as electromagnets for lifting or moving heavy objects, in electronic devices such as speakers and motors, and in medical devices such as MRI machines.

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