Consider a straight vortex filament as shown below. At each point, there's a point vortex of strength $\it T$. Consider that a point $P$ is there on the outermost circle of flow induced by the vortex about the origin $O$. That is the point $P$ lies in the plane containing the flow induced by the vortex at $O$. So, the only point on the vortex filament that will induce a flow at point $P$ is $O$.

Therefore, the vortex at point $O^{'}$ will not induce any flow at point $P$. Vortex at point $O^{'}$ can only induce flow at those points that are in the plane containing the flow induced by the vortex at $O^{'}$.

Vortex Filament

So, my question is how Biot-Savart Law is applied for a straight filament vortex? How can the segment $dl$ induce a flow at a point $P$ that is not in the plane? Only the vortex centered at the point of intersection of line and perpendicular from point $P$ to the line can induce the flow at $P$. Please clear my query.

Biot-Savart Law


You need to think in 3D. A vortex will induce a flow speed everywhere.

That is the point 𝑃 lies in the plane containing the flow induced by the vortex at 𝑂

Why would you think that vorticity induces a speed only in one plane? You start from the wrong presupposition.

Now consider the rings around the vortex filament in your first drawing. All around those rings a flow speed is induced. Its magnitude drops with the inverse of the radius of one particular ring but is constant along the circumference of that ring. Actually, that ring is only a section through an infinitesimal tube, just as infinitesimal as that vortex filament.

Induced speed of a vortex

Irrotational vortex. © Silver Spoon, picture source.

Also, the symbol for vortex strength is a capital Greek letter Gamma ($\Gamma$), not $T$.

  • $\begingroup$ Thanks for the response! Now, I understood that I had a wrong idea about the 3D vortex filament. $\endgroup$
    – Pavan
    Aug 23 '20 at 16:30

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