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A metallic rod of length L and mass M is pushed along its own length by two unequal, oppositely directed forces F1 and F2 applied at its two ends. Gravity and external magnetic fields are absent. If the specific charge of an electron is e/m, what is the steady-state potential difference between the ends of the rod?

1. |F1 - F2|*m*L/(e*M)
2. (F1 + F2)*m*L/(e*M)
3. (m*L/(e*M))*ln(F1/F2)
4. None

Correct answer: |F1 - F2|*m*L/(e*M)

Solution

The rod has a net force |F1 - F2| and accelerates with a = |F1 - F2|/M. The free electrons inside, of mass m, require force m*a to accelerate with the rod. This force is supplied by an internal electric field E set up by charge separation: eE = m*a, so E = m*a/e. The potential difference across length L is V = E*L = m*a*L/e = m*L*|F1 - F2|/(e*M).

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