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# A slender rod, $0.240 \mathrm{~m}$ long, rotates with an angular speed of $8.80 \mathrm{rad} / \mathrm{s}$ about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of $0.650 \mathrm{~T}$. Suppose instead the rod rotates at $8.80 \mathrm{rad} / \mathrm{s}$ about an axis through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?

Solution

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c)

This time the rotation axis is through the center of the rod, so the rotation of the rod results in a motional emf in both halves of the rod. The value of this emf is given by the integral of $d\varepsilon$ from the center of the rod $(r = 0)$ to either end of the rod $(r = L/2)$:

\begin{align*} \varepsilon &= \int_0^{L/2}\omega Br\;dr = \frac{1}{8}\omega BL^2\\ &= \frac{1}{8}(8.80\;\mathrm{rad/s})(0.650\;\mathrm{T})(0.240\;\mathrm{m})^2 = 41.2\;\mathrm{mV}. \end{align*}

By symmetry $|\varepsilon|$ is the potential difference between the rod's center and either end of the rod. The potential difference between the two ends of the rod is the algebraic sum of the individual potential differences between the rod's center and each end of the rod, which is zero since the polarities of the $\varepsilon$ induced in the two halves of the rod are reversed; $\varepsilon$ in each half tends to drive the current radially outward in opposite directions (the magnetic force $\vec{\pmb F}$, represented in Fig. 1b, points along each half of the rod in opposite directions). To summarize our results, the potential difference between the ends of the rod is zero and the potential difference between the center of the rod and one end is simply $41.2 \text{ mV}$.

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