Motional EMF from Lorentz Force

What this figure shows. A straight conducting rod AB of length l lies in a uniform field $\vec B$ directed into the page, with the rod moving to the right with constant velocity $\vec v$ perpendicular to both its own length and the field. Panel (a) shows the rod at an instant during its motion, with the Lorentz force pushing free electrons from B towards A, so that end A accumulates negative charge and end B is left relatively positive. Panel (b) shows the resulting steady state: an electric field $\vec E$ has built up inside the rod pointing from B to A (from + to $-$), and a potential difference $\varepsilon$ (marked with $+$ and $-$ symbols) appears across the rod's two ends, with the magnitude of this motional emf given by $\varepsilon = Blv$ once the ma …

Worked out. A conducting rod of length 0.5 m falls freely from a height of 7.2 m at a place in Chennai where Earth's horizontal magnetic field is $4.04\times10^{-5}$ T, with the rod's length kept perpendicular to that horizontal field throughout the fall; the emf just before it touches the ground is required. Using $v^2 = u^2+2gh$ with $u=0$, $g=10\ \text{m/s}^2$ and $h=7.2$ m gives $v^2 = 2(10)(7.2)=144$, so $v=12$ m/s. Substituting into the motional emf formula, $\varepsilon = B_H\,l\,v = (4.04\times10^{-5})(0.5)(12) = 242.4\ \mu\text{V}$. The example shows that even ordinary, everyday falling objects have a (very tiny, but non-zero) emf induced across them by Earth's own magnetic f …

Worked out. A copper rod of length l rotates with angular velocity $\omega$ about one of its ends, in a plane perpendicular to a uniform field B, and the emf induced between its two ends is required. Because different points along a rotating rod move at different linear speeds, the rod is split into thin elements dx at distance x from the pivot, each moving with speed $v=x\omega$ and contributing an elemental emf $d\varepsilon = B(x\omega)\,dx$ by the motional-emf formula. Integrating over the whole rod, $\varepsilon = \displaystyle\int_0^l B\omega x\,dx = B\omega\left[\dfrac{x^2}{2}\right]_0^l = \dfrac{1}{2}B\omega l^2$. This result -- $\varepsilon=\tfrac12 B\omega l^2$ -- is the standard formula for the emf of a rotating conducting rod (as in a simple homopolar-type generator), obtained by integrating the elemental motional emf over …