0
\$\begingroup\$

I am trying to understand something about Lenz law. First, let me show an example of what I do understand. Lenz Law

In this image, I understand

  • a) As the loop moves closer to the current the field increases. So therefore the current must go in the opposite direction, thus counter clockwise.

  • b) the loop moves farther away, the field decreases, so the current moves clockwise around the loop.

  • c) the field doesn't change, so there is no current change.

  • d) the current increases, which increases the field, and that has to be negated, so counter clockwise.

Those are the correct answers.

However, when looking at another example, I am having some troubles.

Lenz Law

With this I understand b) and c).

  • b) The current is constant, no change.

  • c) Again, there current is constant, and the loop is not moving in a way to change the field.

But for a) and d). I don't understand.

  • a) is clockwise, so this is acting a decrease.
  • d) is counter clockwise, so this is acting as an increase.

But I don't understand how the sinusoidal are interacting differently. Can someone please explain how the sin and cos functions act on the loop?

Thanks!

\$\endgroup\$
0
1
\$\begingroup\$

The current in the loop being sinusoidal creates a magnetic field that is in-phase with current.

Using V = N\$\dfrac{d\Phi}{dt}\$ we can predict what the phase angle is of the driving voltage that is induced into the loop by the changing magnetic field.

This means that \$V_{INDUCED}\$ is leading the sinewave current by 90 degrees.

However, the current produced in the loop is dependant on this voltage and the load resistors plus one other factor that your question omits. This other factor is the loop inductance of the secondary. It will form a complex impedance with the two resistors. Please note that I am now going to ignore the loop inductance.

So, the induced voltage generated by the sinusoidal current leads by 90 degrees and this causes a current to flow that is in phase with the induced voltage (ignoring loop inductance).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.