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A copper rod $A B$ of length $l$ is rotated about end $A$ with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is
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Verified Answer
The correct answer is:
$\frac{m \omega^{2} x}{e}$
In circular motion, net force on the particle is given as
$F_{c}=\frac{m v^{2}}{r}=m \omega^{2} r$
where, $\omega$ is angular speed.
When the rod rotates, electrons in it also rotates which produce electric field $E$ at a distance $x$.
So, force on electron,
$F_{e}=e E$
This force provide the centripetal force,
i.e., $\quad F_{e}=F_{c}$
or
$e E=m \omega^{2} x$
or
$E=\frac{m \omega^{2} x}{e}$
$F_{c}=\frac{m v^{2}}{r}=m \omega^{2} r$
where, $\omega$ is angular speed.
When the rod rotates, electrons in it also rotates which produce electric field $E$ at a distance $x$.
So, force on electron,
$F_{e}=e E$
This force provide the centripetal force,
i.e., $\quad F_{e}=F_{c}$
or
$e E=m \omega^{2} x$
or
$E=\frac{m \omega^{2} x}{e}$
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