# Lenz Law

## Main Question or Discussion Point

When a magnet is moved towards a coil, it produces a current in the coil. The current flows in the coil in the direction which produces an "induced" magnetic field that pushes away the motion of the approaching magnet. Faster the magnet approaches, bigger the current and the larger the induced magnetic field.

Suppose Lenz Law was exactly the opposite: When you move a magnet towards a coil, the induced magnetic field pulls the magnet closer. What do you think would happen?"

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You would have the current flowing in the opposite direction from the real Lenz Law scenario. Although, if Lenz law were the opposite of what it is, I believe you would get a runaway feedback process that would eventually give the currents infinite speed, with the energy coming out of nowhere, thus breaking the law of conservation of energy.

Will the induced magnetic field pull in the magnet instead of opposing it?

Minor comment: The magnet induces a voltage in the coil, and the instantaneous current is the instantaneous voltage divided by the coil resistance.

Minor comment: The magnet induces a voltage in the coil, and the instantaneous current is the instantaneous voltage divided by the coil resistance.

The magnet induces current & voltage in unison. Is it constant voltage, with current determined as V/R? Or is it constant current, with V determined as I*R?

WIth high R values, constant voltage is induced, I = V/R. But with low R values, I is constant, V = I*R. For example, if the loop is 1 ohm, changing it to 2 ohms results in very nearly the same current with twice the voltage.

If the loop is 1.0 megohm, raising it to 2.0 megohm results in the same voltage with half the current. The R value is critical. Refer to the above link for the equations.

The induced voltage and current both depend upon R, as well as L, and frequency. If R >> X (X = L*omega), we get constant voltage variable current. If R << X, we get constant current variable voltage.

It has to be this way re conservation of energy law - CEL. The magnetic field has a limited power. The induced current-voltage product is power. This power cannot exceed the field power. At low R values, Mother Nature exhibits current source behavior. At high R values, she axhibits voltage source behavior. No surprise at all.

Mother Nature is always conserving energy/power. A low resistance dissipates low power when current source driven, high power when driven from a voltage source, and vice-versa for high R values.

Please examine the equations. It's all there. I'll accept correction, but these equations were posted a couple years ago, & nobody has found an error. Best regards.

Claude

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The basic equation is Faraday's Law:
$$\oint E\space d\ell=V = - N \int_{A}^{} n \cdot\frac{dB}{dt} \space dA$$
The moving magnet induces a voltage V in an N-turn coil. The "induced current" is the induced voltage V divided by the circuit resistance (plus inductance term). The minus sign represents negative feedback producing a magnetic field opposing the magnet, i.e., Lenz Law. If the coil were replaced with a superconducting sheet (similar to a shorted zero resistance coil), the magnet would float above the sheet. See

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The basic equation is Faraday's Law:
$$\oint E\space d\ell=V = - N \int_{A}^{} n \cdot\frac{dB}{dt} \space dA$$
The moving magnet induces a voltage V in an N-turn coil. The "induced current" is the induced voltage V divided by the circuit resistance (plus inductance term). The minus sign represents negative feedback producing a magnetic field opposing the magnet, i.e., Lenz Law. If the coil were replaced with a superconducting sheet (similar to a shorted zero resistance coil), the magnet would float above the sheet. See

You're assuming that phi and B are entirely due to the external flux. The "B" in your equation includes the contribution due to the induced current. This internal flux opposes the external per law of Lenz. Your mistake is a very common one. You must also remember that E in the left side integral does indeed depend upon R & X.

Please examine my equations per attached link. When both internal and external flux are included, the induction results in constant current at low R values, constant voltage at high R values. Your presentation regards the external flux only. The B in your equation consists of intetrnal Bi plus external Be.

Claude

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Redbelly98
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