Chapter 5, Problem 5.1P

Consider a vortex filament of strength $\Gamma$ in the shape of a closed circular loop of radius R. Obtain an expression for the velocity induced at the center of the loop in terms of $\Gamma$ and R.

To determine

An expression for the velocity induced at the center of the loop.

Answer to Problem 5.1P

  $\text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}=\frac{\Gamma }{2R}\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

Explanation of Solution

Using Biot savrat law;

  $d\text{<mover accent="true"><mo>V</mo><mo>→</mo></mover> = }\frac{\Gamma }{4\pi }\frac{d\overrightarrow{l}\times \overrightarrow{r}}{{\left|\overrightarrow{r}\right|}^3}$

Now,

Velocity induced at the center;

  $\begin{array}{l}\text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}={\displaystyle \int \frac{\Gamma }{4\pi }\frac{d\overrightarrow{l}\times \overrightarrow{r}}{{\left|\overrightarrow{r}\right|}^3}}\\ \Rightarrow \text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}={\displaystyle \underset{0}{\overset{2\pi R}{\int }}\frac{\Gamma }{4\pi }\frac{\left(dl\times R\right)}{{\left|R\right|}^3}}\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ \Rightarrow \text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}=\frac{\Gamma }{4\pi }\frac{1}{R^2}{\displaystyle \underset{0}{\overset{2\pi R}{\int }}dl}\\ \Rightarrow \text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}=\frac{\Gamma }{4\pi }\times \frac{1}{R^2}\times 2\pi R\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ \text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}=\frac{\Gamma }{2R}\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Therefore, the obtained expression is $\text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}=\frac{\Gamma }{2R}\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .