How do we calculate the emf and the current induced in two loops of wire, that have a portion in common between them? The specific example we had solved in class is:

Two simple rectangular loops conjoined. Even loops are born conjoined, dude.

The left side loop is a square and the right side loop has half the width. The magnetic field as a function of time is known, and the resistance per unit length of the wire is known.

Does a loop's being conjoined with another affect the emf induced across it? I do not understand how i can calculate the current in each wire, because there is no particular point from where the emf is induced (unlike a typical circuit with batteries). Our teacher and textbook pulled it off by assuming the induced emf in each loop to be supplied by two equivalent batteries, and using Kirchhoff's law to find the currents in each portion. Is this method correct? If not, how do we calculate the current induced in each loop?

More clarification: Circuit diagram: [Circuit diagram]

ABEF is a square with side length $\ell$, BC = ED = $\ell/2$. The part AB has a resistance $R$, and all wires have same resistivity and cross sectional area. Now, my doubt is, will we write Kirchhoff's Loop Law equations as: $$V_1 = i_1(3R) + (i_1 - i_2)(R) \\ V_2 = i_2(2R) - (i_1 - i_2)(R) $$ or, $$\begin{align} & V_1 - V_2 = i_1(3R) + (i_1 - i_2)(R) \\ & V_2 - V_1 = i_2(2R) - (i_1 - i_2)(R) \quad\quad\text{?} \end{align} $$

Thank you!


1 Answer 1


When you write Kirchoff's law going around a loop, there's nothing in the equation itself that ties a voltage (or EMF) source to a particular point. The only reason you need to know the location of a battery is so that you can tell whether to include the battery's voltage in a given loop. But there's no such requirement with a magnetic EMF. You can calculate the EMF around the loop without having to place that EMF at a single location. The EMF is actually associated with the loop as a whole, not with any specific point on it.

In this way, you can include an EMF term in each loop law you write. Just use Kirchoff's laws as normal, and make sure to include the term corresponding to the EMF for each loop you use. With the specific example you give, when you write the loop law for the left loop, you use the EMF associated with the left loop - I suppose that would be $V_1$. The EMF associated with the right loop ($V_2$) is entirely separate, and should not appear in the left loop equation at all.

Note that you can check this procedure by writing Kirchoff's loop law for the outer loop, which encloses both the left loop and the right loop. The equation you get should be the sum of the left loop equation and the right loop equation. You'll need to use the fact that the EMF associated with the outer loop is the sum of the EMFs associated with the left loop and the right loop (as you could derive from Faraday's law).

  • $\begingroup$ Thank you for the answer! So considering the left loop only, there is the emf $V_1$ for the left loop in it, and the right loop only, there is emf $V_2$ only? So the circuit would be something like $V_1$ and $V_2$ in parallel? Or would i put both the batteries in the common arm (with polarity opposite to each other)? $\endgroup$ May 17, 2016 at 15:32
  • $\begingroup$ What are $V_1$ and $V_2$? In any case, my point is really that you cannot think of these EMFs as batteries. They are not batteries. You cannot put them anywhere; you have to dispense with concepts of parallel and series. $\endgroup$
    – David Z
    May 17, 2016 at 15:33
  • $\begingroup$ $V_1$ is the emf calculated for the left loop independently, and $V_2$ for the right loop similarly. My point was, do we include both these emfs while writing the Kirchhoff's Law equations for each of the loops? I think that we do.... $\endgroup$ May 17, 2016 at 15:36
  • $\begingroup$ Yeah. You include $V_1$ for the left loop, and $V_2$ for the right loop. $\endgroup$
    – David Z
    May 17, 2016 at 16:32
  • $\begingroup$ Please see my answer for more doubts....it is a little long for a comment! $\endgroup$ May 18, 2016 at 4:15

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