Question

A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases at a constant rate $\frac{dB}{dt}$. Consider a circle of radius r coaxial with the cylindrical region. (a) Find the magnitude of the electric field E at a point on the circumference of the circle. (b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude found in part (a)

Answer 1

(a) The emf induced in the circle is given by
$$\begin{gathered} e=\frac{d\varphi }{dt}=\frac{d(B.A)}{dt} \\ =A\frac{dB}{dt} \end{gathered}$$
The emf induced can also be expressed in terms of the electric field as:
E.dl = e
For the circular loop, $A=\pi r^2$
$\Rightarrow E2\mathrm{\pi }r=\mathrm{\pi }{r}^2\frac{dB}{dt}$
Thus, the electric field can be written as:
$E=\frac{\mathrm{\pi }{r}^2}{2\mathrm{\pi }r}\frac{\mathrm{d}B}{\mathrm{d}t}=\frac{r}{2}\frac{\mathrm{d}B}{\mathrm{d}t}$
(b) When the square is considered:
E.dl = e
For the square loop, $A={\left(2r\right)}^2$
$$\begin{gathered} \Rightarrow E\times 2r\times 4=\frac{dB}{dt}(2r{)}^2 \\ \Rightarrow E=\frac{dB}{dt}\frac{4r^2}{8r} \\ \Rightarrow E=\frac{r}{2}\frac{dB}{dt} \end{gathered}$$
The electric field at the given point has the value same as that in the above case.