MyOpenMath/Solutions/Electromotive force/Helicopter

The following problem is inspired by CHECK YOUR UNDERSTANDING 13.4 in University Physics Volume II Chapter 13:

File:Rescue Helicopter.svg
An approximation to ${\displaystyle \delta (x)}$

A metal helicopter blade will develop an emf between the central axis and the tip of the blade (the electric field inside the conducting blade is zero due to surface charges on the blade.) Calculate the emf between the blade tip and central axis if the tip is moving at $v-m/s and the entire blade describes a circle$d-meters in diameter.  Assume a magnetic field of \$B tesla oriented parallel to the axis of this circle.

Guy Vandegrift, myopenmath.com, Question ID 409154

If the moving blade develops a motional emf caused by the magnetic field, why doesn't a current flow? At one level, the answer is simple: Electrostatic charges build up to create an internal electric field to oppose the motional emf. But how do wires manage this? Consider an infinitely long thin wire and calculate how any electric field inside of that wire can be generated by a surface charge density. To keep things simple, consider only the electric field at the center of the wire. We begin with the well known formula for the electric field on the axis of a charged loop:[1] Integrate to find the electric field at z=0: ${\displaystyle E_{z}(0)=\int dE_{z}}$, where:

${\displaystyle dE_{z}={\frac {-1}{4\pi \varepsilon _{0}}}{\frac {z\lambda dz}{\left(z^{2}+R^{2}\right)^{3/2}}},}$

where ${\displaystyle dq=\lambda dz}$ is a charge distributed on the surface of the wire. We shall now show that an arbitrary electric field at the center of the wire can be created by a suitable choice of the line density, expressed as a function of z.

This expression for ${\displaystyle dE_{z}}$ diverges as ${\displaystyle R\rightarrow 0}$ when ${\displaystyle z=0}$. This suggests that a thin wire, perhaps the dominant contribution to the electric field at ${\displaystyle z=0}$ is due to charges located in the range ${\displaystyle -R\gtrsim z\lesssim R}$. To verify this suggestion, we use integration by parts to introduce the Dirac delta function:

${\displaystyle \int u\,dv\ =\ uv-\int vdu.}$
${\displaystyle u=\lambda (z)}$
${\displaystyle dv={\frac {-zdz}{\left(z^{2}+R^{2}\right)^{3/2}}}\Rightarrow v={\frac {1}{\sqrt {z^{2}+R^{2}}}}}$
${\displaystyle 4\pi \varepsilon _{0}E_{z}(0)=\left.{\frac {\lambda (z)}{\sqrt {z^{2}+R^{2}}}}\right|_{-a}^{b}+\int _{-a}^{b}{\frac {d\lambda }{dz}}{\frac {dz}{\sqrt {z^{2}+R^{2}}}}}$

We can discard the first term by integrating past the wire, allowing ${\displaystyle \lambda (z)}$ to be a Heavyside function