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

In summary, the conversation discusses a difficult integral and how to solve it using a substitution method. The final answer is given as B_z=\frac{\mu_0 I N}{2a} [\frac{z}{\sqrt{z^2+a^2}}-\frac{z-L}{\sqrt{(z-L)^2+a^2}}], but there may be a factor of \frac{1}{a} that needs to be accounted for.
  • #1
latentcorpse
1,444
0
I need to do the following integral:

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

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

[itex]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}}}[/itex]

im having trouble with the power of 3/2 as otherwise its just a standard integral to arctan. any advice?
 
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  • #2
latentcorpse said:
I need to do the following integral:

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

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

[itex]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}}}[/itex]

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

[tex]\frac{x}{a} = \tan\theta[/tex]

and see where it takes you.
 
  • #3
cheers. this gives

[itex]\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})}}[/itex]

how do i evaluate sin theta at these points though?
 
  • #4
Draw a right triangle, such than one of the angles, theta, has arctan z/a. Then find the sin of that angle.
 
  • #5
i can draw it if tan theta = z/a but what would it look like if arctan theta = z/a?
 
  • #6
latentcorpse said:
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.
 
  • #7
so now i get:

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

is there anyway to further simplify this or am i done?
 
  • #8
latentcorpse said:
so now i get:

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

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

You seem to have an extra factor of [tex]\frac{1}{a}[/tex]...It probably arose when you did your substitution: [tex]dx=a\sec^2\theta d \theta[/tex]
 
  • #9
cheers. but the rest of it is correct?
 

1. What is a Magnetic Field Integral?

A Magnetic Field Integral is a mathematical concept used to calculate the total magnetic field at a point in space. It takes into account the contributions of all magnetic sources in the vicinity of the point, including current-carrying wires, permanent magnets, and electric currents.

2. How is a Magnetic Field Integral calculated?

A Magnetic Field Integral is calculated by integrating the product of the magnetic field vector and the infinitesimal length vector along a chosen path. This path can be defined as a closed loop around the point of interest or a surface enclosing the point.

3. What is the significance of Magnetic Field Integral in physics?

Magnetic Field Integral plays a crucial role in many areas of physics, including electromagnetism, quantum mechanics, and astrophysics. It is used to study the behavior of magnetic fields in various systems and to make predictions about their effects on charged particles and other objects.

4. How is Magnetic Field Integral related to Ampere's law?

Magnetic Field Integral is closely related to Ampere's law, which states that the line integral of the magnetic field around a closed loop is equal to the current passing through the loop. In fact, the Magnetic Field Integral is often used to calculate the magnetic field in situations where the current is not constant or the loop is not closed.

5. Can Magnetic Field Integral be used to calculate the magnetic field inside a material?

Yes, Magnetic Field Integral can be used to calculate the magnetic field inside a material, as long as the material's magnetic properties are known. This is often used in the study of magnetic materials, such as ferromagnets, to understand their behavior and properties in different situations.

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