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Due to the current flow, supplied by the battery, and the production of the magnetic field($-B\hat{k})$, A Lorentz force($f_L = IL\times B$) will accelerate the rod.

My issue with this, is classical problems would have $B$-uniform and constant, but in reality it's neither. $B$ changes over time due to the addition of more current carrying elements $dl$ with the changes of $x$, leading to a sum $B$ that's greater at it's final position.

However, these problems it's always assumed that the induced $\varepsilon$ is due to the separation of charges caused by the magnetic force acting on them($vBL$), but isn't the an induced electric force due to the changed of B?

Instead of:

$$ \varepsilon = - \frac{\delta\Phi}{\delta t} = - vBh$$

Wouldn't it be:

$$ \varepsilon = - (vBh + \int \frac{\delta B}{\delta t} \cdot da)$$


1 Answer 1


You are right. Rigorously, one should write ${{e}_{ind}}=-\frac{d\Phi }{dt}$ with $\Phi =B(t)lx$. It would give ${{e}_{ind}}=-lx\frac{dB}{dt}-Blv=-S\frac{dB}{dt}-Blv$

In practice, (if these rail's experiment have any practice....) the "self field" is related to the self-inductance coefficient which is very low for a filiform circuit.

Sorry for my poor english !


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