How can I solve the Magnetic Field Integral for a current loop?

[latentcorpse]

I need to do the following integral:

$B_z=\frac{\mu_0 I N a^2}{2} \int_{0}^{L} (a^2+(z-z')^2)^{-\frac{3}{2}} dz'$

so far i have set $x=z-z' \Rightarrow dx=-dz'$ and the limits also change to give

$B_z = - \frac{\mu_0 I N a^2}{2} \int_{z}^{z-L} \frac{dx}{(a^2+x^2)^{\frac{3}{2}}}=\frac{\mu_0 I N a^2}{2a^3} \int_{z-L}^{z} \frac{dx}{(1+(\frac{x}{a})^2)^{\frac{3}{2}}}$

im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?

 

[Hootenanny]

I need to do the following integral:

$B_z=\frac{\mu_0 I N a^2}{2} \int_{0}^{L} (a^2+(z-z')^2)^{-\frac{3}{2}} dz'$

so far i have set $x=z-z' \Rightarrow dx=-dz'$ and the limits also change to give

$B_z = - \frac{\mu_0 I N a^2}{2} \int_{z}^{z-L} \frac{dx}{(a^2+x^2)^{\frac{3}{2}}}=\frac{\mu_0 I N a^2}{2a^3} \int_{z-L}^{z} \frac{dx}{(1+(\frac{x}{a})^2)^{\frac{3}{2}}}$

im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?

You can evaluate this integral using exactly the same substitution. Try letting

$$\frac{x}{a} = \tan\theta$$

and see where it takes you.

 

[latentcorpse]

cheers. this gives

$\frac{\mu_0 I N}{2a} \int_{\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}} \frac{1}{\sec{\theta}} d \theta =\frac{\mu_0 I N}{2a} \int_{\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}} \cos{\theta} d \theta =\frac{\mu_0 I N}{2a} [\sin{\theta}]_{\theta=\arctan{(\frac{z-L}{a})}}^{\arctan{(\frac{z}{a})}}$

how do i evaluate sin theta at these points though?

 

[nicksauce]

Draw a right triangle, such than one of the angles, theta, has arctan z/a. Then find the sin of that angle.

 

[latentcorpse]

i can draw it if tan theta = z/a but what would it look like if arctan theta = z/a?

 

[xboy]

i can draw it if tan theta = z/a but what would it look like if arctan theta = z/a?

You have something like sin(arctan z/a). The term inside the bracket is an angle (let's call it alpha). As it is an angle, you can express it as arcsin something. So you need to know sine alpha, right? And you already know tan alpha.

 

[latentcorpse]

so now i get:

$B_z=\frac{\mu_0 I N}{2a} [\frac{z}{\sqrt{z^2+a^2}}-\frac{z-L}{\sqrt{(z-L)^2+a^2}}]$

is there anyway to further simplify this or am i done?

 

[gabbagabbahey]

so now i get:

$B_z=\frac{\mu_0 I N}{2a} [\frac{z}{\sqrt{z^2+a^2}}-\frac{z-L}{\sqrt{(z-L)^2+a^2}}]$

is there anyway to further simplify this or am i done?

You seem to have an extra factor of $$\frac{1}{a}$$...It probably arose when you did your substitution: $$dx=a\sec^2\theta d \theta$$

 

[latentcorpse]

cheers. but the rest of it is correct?

 

[gabbagabbahey]

Sure, but you don't really need me to tell you that: there are tools to help you check integrals