- #1
EmilyRuck
- 136
- 6
Suppose that a ferrite bead is put around a cable where a constant current [itex]I[/itex] flows, just like in this image.
The coordinate system has the [itex]z[/itex] axis along the cable. Let's evaluate the current through the [itex](x,y)[/itex] plane: according to the Ampère's law, the only magnetic field component generated by that current is
[itex]H_{\phi} = \displaystyle \frac{I}{2 \pi r}[/itex]
where [itex]r[/itex] is the distance from the origin.
In order to calculate the inductance of the wire with the ferrite bead, the magnetic flux through the ferrite should be calculated first.
The ferrite bead has [itex]a[/itex] as internal radius and [itex]b[/itex] as external radius, so it is present in the [itex](x,y)[/itex] plane only for [itex]a \leq r \leq b[/itex]. It surrounds the conductor (which has of course a section diameter less than [itex]2a[/itex]) and the flux of the magnetic field through the ferrite bead should be:
[itex]\Phi = \mu_0 \mu_r \displaystyle \frac{I}{2 \pi} \int_a^b \displaystyle \frac{1}{r} dr = \displaystyle \frac{\mu_0 \mu_r I}{2 \pi} \ln \left( \displaystyle \frac{b}{a} \right)[/itex]
Then the inductance is [itex]L = \Phi / I[/itex] as usual.
My question is: why is the flux calculated in such a way?? Should not be considered all the surface surrounded by the circuit or something similar? (I know, it is a linear cable and I didn't specify that area)
The coordinate system has the [itex]z[/itex] axis along the cable. Let's evaluate the current through the [itex](x,y)[/itex] plane: according to the Ampère's law, the only magnetic field component generated by that current is
[itex]H_{\phi} = \displaystyle \frac{I}{2 \pi r}[/itex]
where [itex]r[/itex] is the distance from the origin.
In order to calculate the inductance of the wire with the ferrite bead, the magnetic flux through the ferrite should be calculated first.
The ferrite bead has [itex]a[/itex] as internal radius and [itex]b[/itex] as external radius, so it is present in the [itex](x,y)[/itex] plane only for [itex]a \leq r \leq b[/itex]. It surrounds the conductor (which has of course a section diameter less than [itex]2a[/itex]) and the flux of the magnetic field through the ferrite bead should be:
[itex]\Phi = \mu_0 \mu_r \displaystyle \frac{I}{2 \pi} \int_a^b \displaystyle \frac{1}{r} dr = \displaystyle \frac{\mu_0 \mu_r I}{2 \pi} \ln \left( \displaystyle \frac{b}{a} \right)[/itex]
Then the inductance is [itex]L = \Phi / I[/itex] as usual.
My question is: why is the flux calculated in such a way?? Should not be considered all the surface surrounded by the circuit or something similar? (I know, it is a linear cable and I didn't specify that area)