Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Consider also a straight line through the center of the loop, perpendicular to the plane of the loop. Let A be the distance along the line, measured from the plane of the loop. Obtain an expression for the velocity at distance A on the line, as induced by the vortex filament.

Respuesta :

Answer:

[tex]\vec{V} = \frac{\Gamma}{2R}\vec{A}[/tex]

Explanation:

We define our values according to the text,

R= Radius

[tex]\vec{V} =[/tex]Velocity

[tex]\Gamma =[/tex]Strenght of the vortex filament

From this and in a vectorial way we express an elemental lenght of this filmaent as [tex]\vec{dl}[/tex]. So,

[tex]\vec{dl}x\vec{r} = R*dl*\vec{A}[/tex]

Where [tex]\vec{A}[/tex] imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

[tex]\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}[/tex]

Substituting the preoviusly equation obtained,

[tex]\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}[/tex]

[tex]\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}[/tex]

[tex]\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}[/tex]

So we can express the velocity induced is,

[tex]\vec{V} = \frac{\Gamma}{2R}\vec{A}[/tex]

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