Consider the rod in a circuit with current $i$, then the force is $$F=BiL$$ the mass is $m$ and the velocity is $v$, then $$F=m \dot{v}$$ $$E=BLv$$ then $$i=\frac{m}{B^2L^2}\dot{E}$$ This seems like the equation of a circuit component, like a capacitor of $C=\frac{m}{B^2L^2}$. Is this approach valid?
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$\begingroup$ I don't see why not. The key assumptions are (1) other forces acting on the rod are negligible, including by the external circuit it is connected to, and (2) the rod has negligible resistance. $\endgroup$– PukCommented Jun 8 at 6:42
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$\begingroup$ In practice one must also take into account the magnetic field generated by the external circuit, which is what makes a railgun work. $\endgroup$– PukCommented Jun 8 at 6:52
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