Question

Inside a long cylinder of radius $R=10cm$, there is a smaller cylinder of radius $R/2$. These are arranged in such a way that the two cylinder's touch each other as shown in diagram. There is no magnetic field inside the small cylinder, and in the remaining part of the big cylinder there is homogenous magnetic field which is changing uniformly with time. The rate of change of the magnetic field is $△B/△t=80V/m^2$, and the magnetic field is parallel to the axis of the cylinder. Determine the induced electric field inside the small cylinder is $V/m$: ($O_1$ and $O_2$ are centres of the circular cross-sections shown).

Answer 1

As given in diagram,area where magnetic field is =$\pi R^2-\pi {\left(\frac{R}{2}\right)}^2$

$A=\frac{3\pi R^2}{4}$

Flux passing through this are=$B.A$

$\varphi =\overrightarrow{B}\left(\frac{3}{4}\pi R^2\right)$

emf induced $=-\frac{dQ}{dt}$

$V=\frac{3\pi R^2}{4}\frac{d\overrightarrow{B}}{dt}$

$=-\frac{3\pi R^2}{4}\times 80$

$=-60\pi R^2$

As we know , $V=\int E.dl$ ($dl$ for small cylinder)

$\left|V\right|=|E.\frac{2\pi R}{2}|$

$E=\frac{60\pi R^2}{2\pi R}=60R$

$E=60R$ at point $O_2$.