0
$\begingroup$

In Lenz's law we state that direction of the induced current is such that it opposes the cause that produces it.

But can we state the direction of induced emf instead of current?

If by direction of emf we mean higher to lower potential, then is the above statement correct? If wrong why is it wrong? Please help me understand. Thank you.

$\endgroup$
2
  • $\begingroup$ I think it is correct... but please clarify. $\endgroup$
    – Suprabha
    Apr 21, 2018 at 7:47
  • $\begingroup$ I said what i meant by direction of emf. By Faraday's law we get the term induced emf... i used it. $\endgroup$
    – Suprabha
    Apr 21, 2018 at 9:11

2 Answers 2

1
$\begingroup$

The EMF is generated by the change of the magnetic flux.

So, your proposed statement comes down to the following: the direction of the induced EMF is such that it opposes the cause that produces it, i.e. the change of the magnetic flux.

Although this interpretation is reasonable in spirit, it is not technically accurate, since you can not directly compare EMF and magnetic flux - they have different units, so to speak. On the other hand, you can compare the magnetic flux of the field generating an EMF with the magnetic flux of the current caused by that EMF.

$\endgroup$
1
$\begingroup$

Yes and no. Lenz's law is more naturally defined in terms of current direction.

The problem with your statement is that you assume there is a place of low EMF, and a place of high EMF, sort of like in a simple electric circuit hooked up to a battery.

Let's look at a simple example and see why this gets troublesome. Imagine a loop of perfectly conducting wire with a 1 Ohm resistor, with an ever increasing magnetic field perpendicular to the loop. a loop with a resistor, and a perpendicular magnetic field The increase in magnetic field will generate a counter clockwise current, let's say 1 A. The current direction is well defined.

But where is the high potential in the loop? Using electronics formulas we know and love, we would say that there is a voltage drop V = IR of 1 Volt over the resistor. Assume we are at 10 volts on the right, we would be at 9 Volts after the resistor. But wait a minute! If you go back to the right side, there is no drop since there is no resistor. So the voltage must be 9 on the right. Go across the resistor again, you get a voltage drop of 1 Volt, so it must now be 8 Volts on the left of the resistor, and so on. Clearly, this does not work.

This simple example shows that the generated EMF does not work as a simple circuit with a high and low voltage point. Rather, the EMF that drives the current is generated locally throughout the circuit.

So we can't easily define the direction of the EMF, since there is not a point of high voltage and a point of low voltage. What we do for the direction of the EMF, is we look at the current, and we say that the direction of the EMF is such that it generates a current in that direction.

So I suppose you could define Lenz's law in terms of EMF direction, and then define EMF direction in terms of current direction. However, that is not as neat as the original, since current direction really is the underlying, easily defined measurable.

(Also see this post for more info.)

$\endgroup$

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.