1. Potential Difference in Rotating Rod
Part 1: Potential Difference due to Rotation
When the rod rotates, the free electrons experience a centrifugal force pushing them outwards. This creates a charge separation and thus an electric field $E$. In equilibrium, the electric force balances the centrifugal force.
Consider an electron at a distance $x$ from the axis of rotation:
$$ eE = m\omega^2 x \implies E = \frac{m\omega^2 x}{e} $$The potential difference $V_1$ is the integral of this field along the length of the rod (from $x=l$ to $x=l+L$):
$$ V_1 = \int_{l}^{l+L} E \, dx = \int_{l}^{l+L} \frac{m\omega^2 x}{e} \, dx $$ $$ V_1 = \frac{m\omega^2}{e} \left[ \frac{x^2}{2} \right]_{l}^{l+L} = \frac{m\omega^2}{2e} \left[ (l+L)^2 – l^2 \right] $$ $$ V_1 = \frac{m\omega^2}{2e} (L^2 + 2lL) = \frac{m\omega^2 L(L+2l)}{2e} $$Part 2: Potential Difference with Magnetic Field
A vertical magnetic field $B$ induces a motional EMF. The electrons move with velocity $v = \omega x$ perpendicular to $B$. The magnetic force is $F_m = evB = e(\omega x)B$.
The electric field equivalent is $E_B = \omega B x$. The potential difference induced by the magnetic field is:
$$ V_B = \int_{l}^{l+L} \omega B x \, dx = \omega B \left[ \frac{x^2}{2} \right]_{l}^{l+L} = \frac{\omega B L(L+2l)}{2} $$The problem states the total potential difference becomes double the original. Thus, the magnetic contribution must equal the centrifugal contribution ($V_B = V_1$).
$$ \frac{\omega B L(L+2l)}{2} = \frac{m\omega^2 L(L+2l)}{2e} $$Solving for $B$:
$$ B = \frac{m\omega}{e} $$