- #1
rohanprabhu
- 410
- 2
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:
<br /> |\epsilon| = B \omega r^2<br />
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:
<br /> dF = en r\omega B dr<br />
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?
<br /> |\epsilon| = B \omega r^2<br />
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:
<br /> dF = en r\omega B dr<br />
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?
