4.2 Induced EMF in a rotating rod:

4.1 Motional EMF
4.2 Induced EMF in a rotating rod:

7.1 Inductors connected in series:
7.2 Inductors connected in parallel:
7.3 Mutually coupled inductors in parallel:

Consider a metallic rod of length $L$ is rotating with angular velocity $\omega $ around point O. There is no magnetic field present so there will not be any magnetic force on electrons. We know that if a particle of mass $M$ is rotating around any point, centrifugal force will try to push it in outward direction. Now we can say that there will be centrifugal force on electrons which will push them in direction of free end $P$. So an electric field is induced from $O$ to $P$.

Consider an electron at a distance of $x$ from point $P$. There are two forces on this electron, electric force and magnetic force which are in opposite indirections. So after some electron flow will stop.

Let $E$ is the magnitude of electric field at $x$ distance from $O$, so situation at which electron flow will stop,

$$eE = {M_e}{\omega ^2}r$$

$$E = \frac{{{M_e}{\omega ^2}r}}{e}$$

$$\int {E.dx} = \int\limits_0^l {\frac{{{M_e}{\omega ^2}x}}{e}} dx$$

$${V_ + } - {V_ - } = \frac{{{M_e}{\omega ^2}{r^2}}}{{2e}}$$

We have same situation as was in previous but in this case rod is moving in a magnetic field of magnitude $B$, so there will be one more force on electrons. In this case, Centrifugal force will be much less with compare to magnetic force because mass of electrons is very small(in the factor of 10-31) .

Let $E$ is the magnitude of electric field at $R$ so situation at which electron flow will stop,

$$eE = {M_e}{\omega ^2}r + ewrB\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because (v = rw)$$

${M_e}$ is very small so ${M_e}{\omega ^2}r \ll ewrB$, we can ignore centrifugal force.

$$eE = e\omega rB$$

$$E = \omega rB$$

On intregating both side from $0$ to $l$,

$${V_ + } - {V_ - } = \frac{{Bw{l^2}}}{2}$$

$$\varepsilon = \frac{{Bw{l^2}}}{2}$$