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The circuit is concatenated with a constant flux $B$ and the rod slides with constant velocity $v$. Find the current through the rod.

Before I always solved the problem without the resistor $R_2$ by placing a generator $V_i=Blv$ on the rod in series with the resistor $R$ but this time I'm not sure about doing the same thing with this problem.

I solved it this way: the moving rod changes the flux of $B$ both on its right and on its left and thus induces an emf $V_i=Blv$ on both sides. Then I have two loops and I would apply the following equations:

$R_1 i_1 + R i_1 = V_i$

$R_2 i_2 + R i_2 = V_i$

With $i_1$ clockwise in the left-hand side loop and $i_2$ anti-clockwise in the right-hand side loop and the total current through the rod is:

$i=i_1+i_2$ from top to bottom.

Is this approach correct or I have to place the generator on the rod?

I understand that in a moving rod a electric field $E=v\times B$ is generated and this would support the approach with the generator but Faraday-Maxwell laws says $\nabla\times E=-\partial_tB$ and in integral form $\oint E \cdot dl = - \partial_t \Phi_B$ that I assumed equivalent to my application of KVL.

  • $\begingroup$ is this a home-work problem? $\endgroup$
    – hyportnex
    Commented Jan 12, 2019 at 21:29
  • $\begingroup$ No, it was in my exam. $\endgroup$ Commented Jan 12, 2019 at 23:14
  • $\begingroup$ Before someone steps in to say that, even if not a homework question, this is a homework $type$ question, may I point out that there are real conceptual issues, especially in the last paragraph? $\endgroup$ Commented Jan 12, 2019 at 23:35

1 Answer 1


(a) "Faraday-Maxwell laws says ∇×E=−∂tB" The equation is not applicable to this set-up as $\vec{B}$ is not changing with time.

(b) "in a moving rod a electric field $\vec{E}=\vec{v}\times \vec{B}$ is generated." Indirectly, perhaps, but the component along the moving wire of the Lorentz force on a free electron in the wire is$$\vec{F}=-e\vec{v}\times \vec{B}$$and this is not an electric field force, but a magnetic force.

Thus (if $\vec{B}$ is normal to plane of diagram and the rod is moving in the direction shown) the emf in the moving rod is $$\mathscr{E}=\frac{1}{-e}\vec{F}.\vec{L}= BLv.$$

(c ) Your Kirchhoff II equations are correct, except that the pd across $R$ is $R(i_1+i_2).$ You get exactly the same equations whether you consider the seat of the emf to be the moving wire (as I prefer) or rate of change of flux linked with each loop.

  • $\begingroup$ But the concatenated flux of B is changing. Is adding the generator on the rod the correct way of solving the problem? $\endgroup$ Commented Jan 12, 2019 at 23:16
  • 1
    $\begingroup$ there are also "static" induced charges on the surface on the conductor and their field is $\mathscr{E}$ and can be regarded as being the mover of the current. $\endgroup$
    – hyportnex
    Commented Jan 12, 2019 at 23:25
  • $\begingroup$ In moving conductor set-ups the fundamental seat of the emf is the magnetic Lorentz force, as I've tried to explain. Perhaps this is what you mean by "adding the generator on the rod". You $can$ solve the problem by using rate of change of concatenated flux in the two loops, but it's less fundamental, in my opinion, and more likely to cause confusion. [But it's the method you'd $have to$ use if the emfs were due to $\vec{B}$ changing with time.] $\endgroup$ Commented Jan 12, 2019 at 23:32
  • $\begingroup$ With a single emf (in the moving rod) there's actually no need to use Kirchhoff's second law in two loops. You can find the currents in a couple of lines of algebra by using series and parallel resistances. I leave the details to you. $\endgroup$ Commented Jan 13, 2019 at 0:03
  • $\begingroup$ Yes I know it's a simple circuit with a parallel but I'm not interested in the actual result, I'm interested in the approach this kind of problems. The fact that I haven't written the pd across the rod as caused by both the currents 1 and 2 is because I interpreted the effects caused by the two fields on the right and on the left separately and then added the two contributions to the current. How would the approach change if the right loop has no concatenated flux? It stays on a plane parallel to the B field for example. Thank you already for your time! $\endgroup$ Commented Jan 13, 2019 at 0:32

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