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Current induced in a loop

Physics Asked on August 14, 2021

Change in magnetic flux through a area enclosed by a loop induces current in it.
How then can the current flow as no potential difference is developed in the wire?

2 Answers

A magnetically induced emf drives current around the loop, but there is no variation in charge density which would produce a voltage difference.

Correct answer by R.W. Bird on August 14, 2021

How then can the current flow as no potential difference is developed in the wire?

To answer that question let's begin with another circuit:

enter image description here

In this schematic, the resistance and voltage source R1 and V1 are meant to belong to the same physical component. Similarly for R2 and V2, and R3 and V3.

The total emf around the circuit is 3V. By Ohm's law and KVL, the total current in the circuit is 3V/3$Omega$ = 1A.

However the voltage drop across each component will be

  • the 1V voltage source minus
  • a 1V voltage drop across the associated resistor (1A x 1$Omega$)

or 0 volts!

Now, imagine that instead of three components, there were an infinite number of components, each infinitely small. Again, there would be zero voltage drop across each of the infinitely small components and yet a current would flow.

This leads us to the necessity of distinguishing between emf and voltage drop. It also shows that a voltage drop between nodes in a circuit is not necessary for current to flow through the circuit.

In the original question, the emf is provided by a time-varying magnetic field (which we designate as $vec{B}$). There is an electric field $vec{E}$ associated with any time-varying magnetic field.

$$nabla times vec{E} = -frac{partialvec{B}}{partial t}$$

There are many solutions (possible $vec{E}$ fields) to this equation for a given $vec{B}$ field. We are interested in the one where the divergence is 0, and we will call that solution $vec{E_{rot}}$ ("rot" because it is purely "rotational").

$$nabla cdot vec{E_{rot}}=0$$

We do this because the electrons in the wire will also create a $vec{E}$ field. However, we want to distinguish between the field that is induced by the time-varying magnetic field, from the total electric field which also contains the contribution of the electrons in the wire.

The path integral of this $vec{E_{rot}}$ field, along the wire is the EMF induced by the time-varying magnetic field.

$$ mathscr E_{induced} = int_C vec{E_{rot}} cdot dvec{ell}$$

If we take the total EMF around the loop, and divide it by the resistance around the loop, we get the current around the loop. The voltage drop along a section of the loop will be the EMF along that section, minus the $IR$, where R is the resistance of that section of the loop.

It should be clear that

$$int_C vec{E_{rot}} cdot dvec{ell} - I int_C R' d{ell}$$

can be zero, even if

$$int_C vec{E_{rot}} cdot dvec{ell}$$

is not.

That is, the EMF induced in any section of a ring excited by a time-varying magnetic field may be non-zero, even though the "voltage drop" across that section of the ring might be zero.

Answered by Math Keeps Me Busy on August 14, 2021

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