Question

A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases at a constant rate `(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

\[e = \frac{d\phi}{dt} = \frac{d(B . A)}{dt}\]

\[ = A\frac{dB}{dt}\]

The emf induced can also be expressed in terms of the electric field as:-

E.dl = e

For the circular loop,

A = πr²

\[\Rightarrow E2\pi r = \pi r^2 \frac{dB}{dt}\]

Thus, the electric field can be written as:-

\[E = \frac{\pi r^2}{2\pi r}\frac{dB}{dt} = \frac{r}{2}\frac{dB}{dt}\]

(b) When the square is considered:-

E.dl = e

For the square loop,

A = (2r)²

\[\Rightarrow E \times 2r \times 4 = \frac{dB}{dt}(2r )^2 \]

\[ \Rightarrow E = \frac{dB}{dt}\frac{4 r^2}{8r}\]

\[ \Rightarrow E = \frac{r}{2}\frac{dB}{dt}\]

The electric field at the given point has the value same as that in the above case.