Question

2 Consider the same vortex filament as in Problem 5.1. Consider also a straight line through the center of the loop, perpendicular to the plane of the loop. Let A be the distance along this 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:

The expression for the velocity at distance A on the line is $V_A = \frac{\tau}{2} [\frac{R^2}{(A^2 + R^2 )^{\frac{3}{2} }} ]$

Explanation:

The free body diagram of the circular voltage filament is shown on the first uploaded image

Looking at the diagram we see that a straight pass through the center of the loop and this line is perpendicular to the plane of the loop

    R is the radius of this vortex filament ,  $\tau$ denoted the strength of the vortex filament , V is  the velocity that is been induce due to the distance A traveled, $\r {dl}$ is the elemental length of the vortex filament

   Now the velocity that is been induced perpendicular to the plane of the loop  According to Biot-Sarvart law is  mathematically represented as

             $\r dV = \frac{ \tau }{4 \pi} \frac{dl \ * \ r\ * \ sin \theta}{r^3}$

                  $= \frac{\tau }{4 \pi } \frac{r \ * dl \ * \ sin 90^o }{r^3}$

                 $= \frac{\tau }{4\pi} \frac{dl * 1}{r^2}$

Now the velocity induced at the distance A on the line is mathematically represented as

               $dV_A = dV\ cos \o$

              $V_A = [\int\limits^{2 \pi R}_0 {\frac{\tau}{4 \pi}\frac{dl}{r^2} } \, ] cos\o$

              $= \frac{\tau}{4 \pi}[\frac{1}{A^2 +R^2} ](2\pi R - 0 ) cos \o$

This is  because  $r^2 = A^2 + R^2$ from the diagram applying Pythagoras theorem  

             $= \frac{\tau}{2}[\frac{R}{A^2 +R^2} ][\frac{R}{\sqrt{A^2 + R^2} } ]$

This is  because  $cos \o = \frac{R}{\sqrt{A^2 +R^2} }$  from the diagram applying SOHCAHTOA            

             $= \frac{\tau}{2} [\frac{R^2}{(A^2 + R^2 )^{\frac{3}{2} }} ]$