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

`\vec{V} = (\Gamma)/(2R)\vec{A}`

Step-by-step 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} =(\Gamma)/(4\pi)\int\frac{\vec{dl}x\vec{r}}{r^3}`

Substituting the preoviusly equation obtained,

`\vec{V} = (\Gamma)/(4\pi)\int\frac{R*dl*\vec{A}}{R^3}`

`\vec{V} = (\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} = (\Gamma)/(2R)\vec{A}`