A metallic rod of length l is rotated in a horizontal plane with an angular velocity $\omega$ where no magnetic field exists. If emf induced in the rod, (charge and mass of electron is e and m respectively) is $\frac{m{\omega }^2{l}^2}{Ne}$, then value of N is _________.

 

Given a metallic rod of length l, rotating in a horizontal plane with an angular velocity ω, and no external magnetic field, we are tasked with finding the induced EMF in the rod. We are also provided that the induced EMF is given by the expression mω²l² / eN, where e and m represent the charge and mass of an electron, respectively.

Step-by-Step Derivation

Consider a metallic rod of length l rotating about a fixed point O with an angular velocity ω. We will analyze the induced EMF generated in the rod.

Let us take an elemental charge at a distance r from the center of the rod. The velocity v of this element is given by:

v = rω

The electric field E experienced by the charge at a distance r is given by the relation:

E = mω2r / e

This electric field is responsible for providing the necessary centripetal force to the moving charge. The electric field creates a potential difference across the length of the rod.

To find the induced EMF across the length of the metallic rod, we need to integrate the electric field over the length of the rod. The total induced EMF ε is given by:

ε = ∫0l E(r) dr

Substituting the expression for E, we get:

ε = ∫0l mω2r / e dr

Now, integrating the above expression:

ε = mω2 / e ∫0l r dr

The integral of r with respect to r is:

∫ r dr = (r2) / 2

Thus, we get:

ε = mω2 / e [ (l2) / 2 ]

Finally, we arrive at the expression for the induced EMF:

ε = mω2l2 / 2e

Comparing this with the given expression for the induced EMF, we conclude that the value of N is:

N = 2

Hence, the induced EMF in a metallic rod of length l rotating with angular velocity ω in the absence of any magnetic field is mω²l² / 2e, and the value of N is 2.