A metallic rod of length L and mass M is movi | vedclass

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(B) The acceleration $a$ of the rod is given by $a = (F_1 - F_2) / M$.In the steady state,the electrons inside the rod experience a pseudo-force $F_p

Vedclass · Accepted Answer

$(F_1 - F_2) mL / eM$

Vedclass · Answer

(B) The acceleration $a$ of the rod is given by $a = (F_1 - F_2) / M$.In the steady state,the electrons inside the rod experience a pseudo-force $F_p = m a$ in the direction opposite to the acceleration,which causes charge separation and creates an internal electric field $E$.For an electron of mass $m$ and charge $-e$,the force due to the electric field is $F_e = e E$.In steady state,the net force on the electrons is zero,so $e E = m a$.Substituting $a = (F_1 - F_2) / M$,we get $e E = m (F_1 - F_2) / M$,which implies $E = m (F_1 - F_2) / (M e)$.The potential difference $V$ between the ends of the rod is $V = E L = \frac{m (F_1 - F_2) L}{M e}$.

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