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(A) The magnetic flux $d\phi_B$ through a small circular ring of radius $r$ and thickness $dr$ is given by:$d\phi_B = B \cdot dA = (B_0rt) \cdot (2\pi

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$B_0R^2/12$

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(A) The magnetic flux $d\phi_B$ through a small circular ring of radius $r$ and thickness $dr$ is given by:$d\phi_B = B \cdot dA = (B_0rt) \cdot (2\pi r dr) = 2\pi B_0 t r^2 dr$The total magnetic flux $\phi_B$ through a circular region of radius $R/2$ is:$\phi_B = \int_0^{R/2} 2\pi B_0 t r^2 dr = 2\pi B_0 t \left[ \frac{r^3}{3} \right]_0^{R/2} = 2\pi B_0 t \left( \frac{R^3}{24} \right) = \frac{\pi B_0 t R^3}{12}$According to Faraday's law of induction,the induced electric field $E$ satisfies $\oint \vec{E} \cdot d\vec{l} = \frac{d\phi_B}{dt}$:$E(2\pi r) = \frac{d}{dt} \left( \frac{\pi B_0 t R^3}{12} \right)$At $r = R/2$:$E(2\pi \cdot R/2) = \frac{\pi B_0 R^3}{12}$$E(\pi R) = \frac{\pi B_0 R^3}{12}$$E = \frac{B_0 R^2}{12}$

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