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KracniyMyedved

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## 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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