Electric field in a sliding bar along frictionless conducting rails

When discussing a conducting bar sliding frictionless over two parallel conducting rails in the context of motional emf as in the picture below, Chabay and Sherwood write in their book Matter and Interactions (4th edition, page 821)

The electron current through the resistor is continually depleting the charges on the ends of the moving bar with the result that the electric field inside the bar is always slightly less than is needed to balance the magnetic force. The downward electric force $$eE$$ is slightly less than the upward magnetic force $$evB$$, so the electrons move toward the negative end of the bar.

How much is slightly less? How can it be quantified depending on $$R$$ and $$v_{\text{bar}}$$ or maybe other parameters of the system?

Conductors have always an intrinsic resistance, let's say $$r$$ the resistance of the bar and $$r'(t)$$ the resistances of the parallel rails (increasing with time as well as the length of the circuit), we consider $$r< and $$r'(t)<. Let's call $$I_{ideal}$$ the current needed to equalize the magnetic field effect and $$I_{real}$$ the current $$I$$ that flows into $$R$$ due to the EMF $$\varepsilon$$. Surely it will be $$I_{real} since the resistance in the circuit is slightly bigger than the ideal case, moreover the real current will depend on time. We can calculate $$I_{real}$$ by firstly obtaining the velocity of the bar $$v_{bar}(t)$$ in this way:

$$m\frac{dv_{bar}(t)}{dt}=\frac{v_{bar}(t)l^2B^2}{R+r+2r'(t)}$$

where $$m$$ is the mass of the bar and $$l$$ is the vertical length of the circuit. When we have the solution $$v_{bar}(t)$$ we can calculate $$I_{real}(t)$$ with this formula:

$$I_{real}(t)=\frac{v_{bar}(t)lB}{R+r+2r'(t)}$$

For simplicity we can neglect the term $$2r'(t)$$ since the dependence on time leads to a complicated calculus, this assumption gives rise to an exponential law:

$$I_{real}(t)=\frac{v_{0}lB}{R+r}\exp\left(-\frac{l^2B^2}{m(R+r)}t\right)$$

where $$v_{0}$$ is the initial velocity of the bar. Note that also velocity will decrease exponentially with time. The variations of current on time imply a variation of the flux of the magnetic field that generate EMF' different from the one associated with the field B. This new EMF' always opposes to the variation of current and it is given by

$$\varepsilon'(t)=-L(t)\frac{dI_{real}}{dt}$$

where $$L(t)$$ is the inductance of the circuit, it is a time dependent quantity since the circuit's area is changing on time. Now, the electric field can be calculated using the following formula:

$$\int \vec{E}(t)\cdot d\vec{l}=\varepsilon(t)+\varepsilon'(t)$$

Since the $$\varepsilon'(t)$$ has a negative sign, we have (at any time $$t$$):

$$eE(t) < ev_{bar}(t)B$$

Having an idea of the quantity "slightly less" implies facing a problem that can be solved computationally, I'll try to give an example: you can choose an interval of time $$[0, t_{final}]$$ and divide it into many steps $$\Delta t$$. Given a step $$t_{i}$$ you can calculate $$L(t_{i})$$ (since you know $$v_{bar}(t_{i}$$) you know how the area changes) and multiply it by the derivative $$\frac{dI_{real}}{dt_{i}}$$ to obtain $$\varepsilon'(t_{i})$$. Then you can easily obtain $$\varepsilon(t_{i})$$ and $$E(t_{i})$$ by simply

$$E(t_{i})=\frac{\varepsilon(t_{i})+\varepsilon'(t_{i})}{P(t_{i})}$$

where $$P(t_{i})$$ is the perimeter of the circuit. At this point, you can see how $$eE(t_{i})$$ differs from $$ev_{bar}(t_{i})B$$ and you can also see what happens when you increase the time by calculating $$eE(t_{i}+\Delta t)$$ and $$ev_{bar}(t_{i}+\Delta t)B$$.

• You may assume that $v_{\mathrm{bar}}$ is constant. Does this allow to quantify the "slightly" less analytically? Apr 24, 2020 at 21:05
• I don't think there's a way to quantify that "slightly" less with a formula, however, if $v_{bar}$ is constant you can say that $L$ will depend linearly on time $L\propto t$ and. You can take into account the resistances of the parallel rails $2r'(t)$ so that current will depend on time, let's say that $I_{real} \propto 1/(c+t)$ where $c$ is some kind of constant, then we'll have $\varepsilon'=-L(t)\frac{dI_{real}}{dt}\propto -1/(k+t)$ where $k$ is another kind of constant. At this point you can try with the method I proposed above. May 1, 2020 at 21:45

Let's call the $$+x$$ direction to the left, the direction the bar is moving. The current through the bar is up. A current up in a field out of the page gives a force to the right:

$$F_x = ma = -ILB \Rightarrow\\ m \frac{dv}{dt} = -ILB.$$

The current is given by Ohm's Law, where the emf is the motional emf: $$I = \frac{ε}{R} = \frac{vBL}{R}$$ Plugging this into our force expression gives: $$m \frac{dv}{dt} = -\frac{vB^2L^2}{R}$$ The solution to this is the function that is basically the negative derivative of itself.

We need a negative exponential: $$v(t) = v_i e^{-t/τ}$$ The time constant here is $$τ = mR/(B^2L^2)$$

• Please use MathJax for typesetting mathematical expressions.
– user258881
Apr 23, 2020 at 16:27
• ok, I will make sure from the next time, thanks for your concern @FakeMod Apr 23, 2020 at 18:24