Question

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.

Answer 1

Answer:

$\vec{V} = \frac{\Gamma}{2R}\vec{A}$

Explanation:

We define our values according to the text,

R= Radius

$\vec{V} =$Velocity

$\Gamma =$Strenght of the vortex filament

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

$\vec{dl}x\vec{r} = R*dl*\vec{A}$

Where $\vec{A}$ imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

$\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}$

Substituting the preoviusly equation obtained,

$\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}$

$\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}$

$\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}$

So we can express the velocity induced is,

$\vec{V} = \frac{\Gamma}{2R}\vec{A}$