Question

A copper rod AB of length l is rotated about end A with a constant angular velocity ω. The electric field at a distance x from the axis of rotation is

• A. \(\frac{m\omega^2x}{e}\)
• B. \(\frac{m\omega x}{el}\)
• C. \(\frac{mx}{\omega^2l}\)
• D. \(\frac{me}{\omega^2x}\)

Answer 1

The Correct Option is A

A copper rod of length $l$ is rotated about end A with constant angular velocity $\omega$. We want to determine the electric field at a distance $x$ from the axis of rotation.

Consider a small element of length $dr$ at a distance $r$ from the axis of rotation. The centripetal force acting on this element is $dm \cdot \omega^2r$. The copper rod is electrically neutral, and the rotation is causing a radial electric field E due to the redistribution of electrons. This creates a force equal to the centripetal force for the element to be in equilibrium.

The mass of the small element is given by $dm = \frac{m}{l} dr$ where $m$ is the total mass. The charge of this small element is $dq$. The force due to the electric field is $E \cdot dq$. In equilibrium:

$E dq = dm \omega^2 r = \frac{m}{l} \omega^2 r dr$.

The small charge $dq$ can be expressed using linear charge density $\lambda$ as $dq = \lambda dr$, but since the rod is neutral and $\lambda$ would be 0, we write the charge in terms of the free electron density. $n$: the number of free electrons per unit volume. $e$: charge of an electron. $A$: cross-sectional area. $dq = n e A dr$

$E \cdot n e A dr = \frac{m}{l} \omega^2 r dr$

Integrating both sides from $0$ to $x$:

$\int_0^x E \cdot neA dr = \int_0^x \frac{m}{l} \omega^2 r dr$

$E \cdot neA \cdot x = \frac{m \omega^2}{l} \frac{x^2}{2}$

$E = \frac{m\omega^2 x}{2nelA}$

If we consider $m$ to be the mass of the free electrons and $e$ the charge of a single electron, then $neA l = total \ charge \ of\ free \ electrons = q$, and $m=n_{electrons}\cdot m_e$ is the total mass of the free electrons, then, if we have one electron per atom then $q=e$ and $m=m_e$.

$E= \frac{m_e \omega^2x}{2e} $

Therefore the closest answer is (A) $\frac{m \omega^2 x}{e}$