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(D) The induced electric field $\vec{E}$ at a distance $r$ from the center of the circular region (where $r < R$) is given by $\oint \vec{E} \cdot d\v

Vedclass · Accepted Answer

$\frac{R^2 \alpha}{2}$

Vedclass · Answer

(D) The induced electric field $\vec{E}$ at a distance $r$ from the center of the circular region (where $r For a circular path of radius $r$,$E(2\pi r) = \pi r^2 \alpha$,which gives $E = \frac{r \alpha}{2}$.The direction of the induced electric field is tangential to the circular path.Consider the rod of length $2R$ placed tangent to the circle at its midpoint. Let the center of the circle be $O$ and the midpoint of the rod be $M$. The rod extends from $x = -R$ to $x = R$ along the tangent line.The induced emf $\varepsilon$ across the rod is $\int_{-R}^{R} \vec{E} \cdot d\vec{\ell}$.At a point on the rod at distance $x$ from the midpoint $M$,the distance from the center $O$ is $r = \sqrt{R^2 + x^2}$. The component of the electric field along the rod is $E_x = E \sin 	heta = E \frac{x}{r} = (\frac{r \alpha}{2}) \frac{x}{r} = \frac{\alpha x}{2}$.Thus,$\varepsilon = \int_{-R}^{R} \frac{\alpha x}{2} dx = \frac{\alpha}{2} [\frac{x^2}{2}]_{-R}^{R} = 0$.However,the question likely implies the magnitude of the potential difference between the center and the ends,or the emf induced in a specific configuration. Re-evaluating the standard problem for a rod tangent to the field region: the emf induced between the center and the point of tangency is $\int_0^R E dr = \int_0^R \frac{r \alpha}{2} dr = \frac{R^2 \alpha}{4}$. Since the rod is symmetric,the potential difference between the center and either end is $\frac{R^2 \alpha}{4}$. The magnitude of the induced emf across the entire rod is $0$ due to symmetry,but typically these problems ask for the potential difference between the midpoint and an end,which is $\frac{R^2 \alpha}{4}$.

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