A moving rod placed in a stationary U-shaped frame.

When a conductor (the moving rod in this case) moves perpendicular to a magnetic field, an electromotive force (EMF) is induced according to Faraday's law of electromagnetic induction. The rate of change of magnetic flux through the loop formed by the moving rod and the stationary U-shaped rods is given by = B⋅L⋅V where B is the magnetic field strength,L is the length of the rod, and v is its velocity.

Now, the negative of rate of change of magnetic flux = work done by electromagnetic force throughout the loop, which is BLV+IR, i=current , r=resistance

But this leads to that,BLV=BLV+I--->>IR=0 , means electric current through loop is 0, So, Just Please Clarify My Confusion


1 Answer 1


The emf induced in a rod of length $\ell$ travelling with speed $v$ in a plane perpendicular to a magnetic field $B$ is $B\ell v$.
If the resistance of the complete circuit is $R$ then the power dissipated in the circuit is $\dfrac{(B\ell v)^2}{R}= \dfrac{B^2\ell^2 v^2}{R}$.

A current carrying conductor moving in a magnetic field experiences a force acting on it $B I\ell$ where $I$ is the current in the circuit, $\dfrac{B\ell v}{R}$.

For the rod to move at constant velocity, $v$, an external force, $\dfrac{B\ell v}{R}$ must be applied to the rod and the work done per second by the external force is $B I\ell\cdot v = B\cdot \dfrac{B\ell v}{R}\cdot \ell\cdot v = \dfrac{B^2\ell^2 v^2}{R}$ which is exactly equal to the power dissipated in the circuit.

  • $\begingroup$ That's not what i asked. $\endgroup$ Nov 1, 2023 at 7:48
  • $\begingroup$ The statement the negative of rate of change of magnetic flux = work done by electromagnetic force throughout the loop, which is BLV+IR is not correct. $B\ell v$ is the induced emf and $IR$ is the potential difference across the resistance $R$. These two quantities are equal in magnitude which means that the current in the circuit is $B\ell v/R$ $\endgroup$
    – Farcher
    Nov 1, 2023 at 8:30
  • $\begingroup$ MultiversalExplorers it is true that this answer is not what you asked for specifically, in fact your post had no question in it. Instead it had a request for clarification and if it is not clear yet then you should read Farcher's post more carefully a few mote times and next time ask a clear question. $\endgroup$
    – hyportnex
    Nov 1, 2023 at 12:57
  • $\begingroup$ If you equate line integral of lorentz force per coulomb of charge to negative rate of change of magnetic flux, then you will get, BLV+IR=BLV. $\endgroup$ Nov 2, 2023 at 0:39
  • $\begingroup$ @MultiversalExplorers - You either treat the $B\ell v$ as an emf on the right hand side of the equation or the $B\ell v$ as a potential difference on the left hand side of the equation. That term cannot be on both sides of the equation. $\endgroup$
    – Farcher
    Nov 2, 2023 at 8:26

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