$\begin{array}{1 1} 157\;V \\ 162\;V \\ 175\;V \\ 135\;V \end{array}$

Free electrons in the rod move towards the outer end due to Lorentz force when the rod is rotated.

As they get distributed over the ring, the resulting separation of charges produces an emf across the ends of the rod.

At a certain value of emf, there is no more flow of electrons and a steady state is reached.

The magnitude of the emf generated across the rod as it moves at right angles to the magnetic field is given by: $d\varepsilon = Bv \;dr$

Hence $\varepsilon = \int d\varepsilon = \int_{R}^{0} Bv\; dr = \int_{R}^{0} B \omega dr = \large\frac{B\omega R^2}{2}$

Given that $v = \omega r$, we get:

$\varepsilon = \large\frac{1}{2}$$ \times 1.0 \times 2\;\pi \times 50 \times 1^2$

$\varepsilon = 157 V$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...