Q: A magnetic field directed into the page changes with time according to the expression B = (0.03 t^{2} + 1.4)T, where t is in seconds. The field has a circular cross – section of radius R- 2.5 cm. What is the magnitude and direction of electric field at p, when t = 3.0 s and r = 0.02 m.

Sol: $\large e = \oint \vec{E}.\vec{dl} = \frac{d\phi}{dt}$

$\large E(2\pi r) = A \frac{dB}{dt}$

$\large E(2\pi r) = \pi r^2 \frac{d}{dt}(0.03 t^2 + 1.4 t) $

$\large E = \frac{\pi r^2}{2\pi r} (0.06 t) = \frac{r}{2} (0.06 t) $

$\large = \frac{0.02}{2} (0.06 \times 3)$

= 18 × 10^{-4} N/C