# E.m.f across the ends of a rod rotating in a magnetic field

1. May 22, 2008

### rohanprabhu

I am trying to derive the formula for E.m.f across the ends of a rod rotating in a magnetic field when the field is perpendicular to the plane of rotation:

$$|\epsilon| = B \omega r^2$$

using the Lorentz Formula only. Suppose, there is a rod of length 'R'. Let, 'n' be the no. of electrons per unit length. Let, at a distance 'r' from the center of rotation, there be a small length element dr. The no. of electrons in this length element is $ndr$. Also, the velocity, $v = r\omega$ is perpendicular to the field always as it lies in the plane of rotation of the rod. Hence, the Lorentz force here is given as:

$$dF = en r\omega B dr$$

What do i do after this? Do integrate it? This integral only talks about a 'net force' on all the electrons. How do I use it to compute the e.m.f across the two ends? Do I first compute work using $\int W = \int F.dr$ and the differentiate w.r.t charge. But, what charge is the Work a function of?

2. May 29, 2008

### mukundpa

I think the EMF will be $$\frac{1}{2}$$ B $$\omega$$ r$$^{2}$$

3. May 29, 2008

### mukundpa

The magnetic force on every electron is balanced by the force due to electric field. Thus

eE = Bev = Be $$\omega$$ r

and d$$\xi$$ = Edr