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

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

Step-by-step 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 = ( \tau )/(4 \pi) (dl \ * \ r\ * \ sin \theta)/(r^3)`

`= (\tau )/(4 \pi ) (r \ * dl \ * \ sin 90^o )/(r^3)`

`= (\tau )/(4\pi) (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 {(\tau)/(4 \pi)(dl)/(r^2) } \, ] cos\o`

`= (\tau)/(4 \pi)[(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

`= (\tau)/(2)[(R)/(A^2 +R^2) ][(R)/(√(A^2 + R^2) ) ]`

This is because
`cos \o = (R)/(√(A^2 +R^2) )` from the diagram applying SOHCAHTOA

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