Electronic – Polarity of the induced voltage on inductor for an applied magnetic flux according to the Lenz’s Law

fluxinductioninductormagnetics

I'm trying to visualize the Lenz's Law on a practical inductor.

The Lenz's Law says that

$$ \mathcal{E} = -\dfrac{\partial \Phi}{\partial t} . $$

Now, assume that there is a magnetic flux inside the coil is in the given direction in the image below, and magnitude of this flux is increasing in time (i.e.; \$\frac{\partial \Phi}{\partial t}>0\$).

My two questions are:

  1. What will be the polarity of induced voltage on the resistor? Please give an answer like "'A' will be more positive than 'B'.".

  2. What will be the direction of current in the system. Please give an answer like "It will be in the 'R' direction.".

enter image description here

Best Answer

If the changing magnetic field totally surrounds the coil, inside and out then there will be no net voltage or current but, i think you are probably referring to the flux only being concentrated through the centre of the coil in which case you use the right hand rule.

The thumb points in the direction of the flux that will counter the original flux ie it points to the left of your picture. The fingers of your right hand represent the direction that current flows.

This means that point A is more negative than point B and current flows, using your terminology, will be in the R direction.

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