0
$\begingroup$

enter image description here

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:

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

$\endgroup$

1 Answer 1

1
$\begingroup$

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 !

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.