In school we are taught about the motor effect and the dynamo effect. We are also taught Fleming's left-hand rule and Fleming's right-hand rule to describe the motor effect and dynamo effect (respectively).

We are taught that Fleming's left-hand rule describes the force that acts on a wire (technically the charge-carrying particles) that has a current which passes through a magnetic field.

We are taught that Fleming's right-hand rule describes the direction of a induced emf (or current if the circuit is complete) when a conductor is passed through a magnetic field.

The only difference between these that I can see is the difference between whether the force is inducing the current or if the current is inducing the force. To me, it would make sense that in either scenario the current would flow in the same direction, regardless of whether it is doing work on the system or if the system is doing work to produce it. However that's not how it works, if the current induces the force, the current is flowing in the opposite direction to which it would if the force induced the current.

I'm clearly misunderstanding something here, any help would be appreciated.

  • $\begingroup$ When the force produces the current, isn't it supposed to be a forcefield that is generated by the exterior of the system (not by the system)? $\endgroup$
    – Emil
    Commented Feb 28, 2019 at 9:24

1 Answer 1


Why is the direction of current in a wire different depending on if a force induced a current or a current induced a force?

It is because in one case you have a dynamo - force induced a current and in the other a motor - current induced a force.

Imagine that you have a rechargeable battery of emf $\mathcal E_{\rm battery}$ which is connected to a simple dc motor and the total resistance of the circuit is $R$.

As the motor coil rotates in the magnetic field a back emf $\mathcal E_{\rm back}$ is produced which will depend on the speed of revolution of the coil.

When the coil is rotating at a constant speed $\mathcal E_{\rm battery}-\mathcal E_{\rm back}=IR$ where $I$ is the current flowing in the circuit.

Multiplying by the current and rearranging slightly gives $\mathcal E_{\rm battery}I=I^2R+\mathcal E_{\rm back}I$

The term $\mathcal E_{\rm battery}I$ is the power supplied by the battery and $I^2R$ is the power dissipated as heat in the part of the circuit which have resistance.
$\mathcal E_{\rm back}I$ is the mechanical work done by the motor.

Now suppose that in some way the speed of the coil in the motor is increased to such a degree that $\mathcal E_{\rm back}>\mathcal E_{\rm battery}$.

Again $\mathcal E_{\rm battery}-\mathcal E_{\rm back}=IR$ but now because $\mathcal E_{\rm back}>\mathcal E_{\rm battery}$ the current must be in opposite direction to what it was before and what was the motor (electrical energy converted to mechanical energy) is now acting as a dynamo (mechanical energy - something had to make the coil rotate faster - converted to electrical energy - recharging the battery.

$\mathcal E_{\rm back}(-I) = I^2R + \mathcal E_{\rm battery}(-I)$ remembering that the current is travelling in the opposite direction ie it has a negative value.

The left hand side if the rate at which mechanical work is being done to rotate the coil and then on the right hand side you have the rate at which heat is dissipated and the rate at which electrical energy is converted into chemical energy in the rechargeable battery.


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