0
$\begingroup$

Suppose there are 2 square loops of wires placed in the same plane symmetrically(having the same centre and orientation). The smaller one is ABCD with a side 10cm and the bigger one is EFGH with a side 20cm. Now suppose a time-varying magnetic field going into the plane is introduced. The rate of change of magnetic field with time is K. Current should flow through the wire loops as EMF is induced. If the 2 loops are connected by wires PQ and RS, then what would be the direction of current in PQ and RS?.

circuit diagram

My logic:

As flux through a bigger loop is more(more area), induced EMF in EFGH should be greater than ABCD. Hence current should flow from P to Q and R to S. But here's a problem. If PQ , RS weren't there the current in the loop would have been anticlockwise. Suppose current in FE is i1 in HG is i2 in BA is i$_3$ and in DC is i$_4$.

If current were to flow from R to S and P to Q then it would mean i$_1$>i$_2$ but that doesn't make sense... because circuit is symmetric so equal current should flow through both upper and lower parts.

current in loops as per me

$\endgroup$

3 Answers 3

0
$\begingroup$

As flux through a bigger loop is more(more area), induced EMF in EFGH should be greater than ABCD

Correct.

Hence current should flow from P to Q and R to S.

We can't conclude that, because we do not know details of the induced electric field, which depends on details of the magnetic field (position and orientation of currents/magnets that generate both fields). Magnetic field can't be uniform everywhere, it is stronger at some region of space and weaker at others, and it decays to zero at large distances. The field is usually generated by some electromagnet or moving magnet with some position with respect to the circuit, so all these details define magnetic and induced electric field in that space.

Depending on where the electromagnet/magnet exactly is, the current in $R \to S$ can have any positive or negative value. It obeys the generalized Ohm law

$$I = \mathscr{E}_{R\to S}/R_{R\to S}$$

where $\mathscr{E}_{R\to S}$ is induced emf in the conductive path $\gamma_{R\to S}$ joining $R$ with $S$:

$$ \mathscr{E}_{R\to S} = \int_{\gamma_{R\to S}} \mathbf E_i\cdot d\mathbf s. $$ Here integration path is the conductive path $\gamma_{R\to S}$, $\mathbf E_i$ is induced electric field, and $R_{R\to S}$ is ohmic resistance of the element.

If the magnetic field is cylindrically symmetric and perfectly concentric with the center of the square symmetrical circuit, then $\mathscr{E}_{R\to S}$ should vanish and thus current should be zero.

$\endgroup$
1
  • $\begingroup$ Thanks for ur time and trouble $\endgroup$ Aug 18, 2023 at 19:15
0
$\begingroup$

The emf generated in EFGH (or ABCD) is generated around the loop, not between the loops. PQ and RS are in the loops EFPQBASR and HGPQCDSR. We can see that the directions of the emfs in PQ produced by these loops are opposed to each other, and symmetry would suggest that they would be equal in magnitude, so the result would be zero. A similar consideration would apply for RS.

$\endgroup$
2
  • $\begingroup$ Your logic seems right.... But in case as I mentioned if we consider the RS and PQ as connecting wires between 2square loops won't it mean that inspire of R being at higher potential than S current flows in both directions? In other words how do u decide that we shall considee loops EFPQBASR and HGPQCDSR not the square loops for solving this $\endgroup$ Aug 18, 2023 at 5:58
  • $\begingroup$ Also can u pls explain what would happen if PQ and RS were not symmetrically placed suppose R were connected to A and P to B then what? Shall current still be zero? Assume (if needed) that resistance is proportional to length all other factors are same in all wires $\endgroup$ Aug 18, 2023 at 5:59
0
$\begingroup$

The difficulty you face is that the induced emfs are not localised and relates to a complete loop and this is unlike a battery where you can say for certain that the seat of emf is between the two terminals of the battery.

You can investigate the circuit by applying the following equation for a complete loop,

$\sum\limits_{\rm loop}\, (\text{current} \times \text{resistance})= \sum\limits_{\rm loop}\,\rm emf$

To simplify matters assume a slightly modified circuit, the changing magnetic field is uniform at any instant of time, the resistances between nodes are $A,\,B,\,C$ and $D$, the corresponding currents are $a,\,b,\,c$ and $d$ as shown in the circuit diagram below and the emf generated around each small loop is $1\,\rm V$.

enter image description here

Loop $FAEBF \quad a\,A - b\,B = 3$
Loop $FCEDF\quad c\,C - d\, D =1$
Loop $FAECF\quad a\, A- c\, C =1$

Summing the currents entering node $E$ gives $a+b+c+d=0$

If the resistance values are given the there are four equations to solve for four unknowns.

If $A=B=C=D = 1\,\Omega$, then using WolframAlpha gives the solution $a=3/2,\,b=-3/2,\,c=1/2,\,d=-1/2$.
Note the currents entering node $E$ from the right add up to $1/2+(-1/2) = 0$.

However, if $A=1,\,B=2,\,C=3,\,D = 4\,\Omega$, then using WolframAlpha gives the solution $a=28/25\,{\rm A},\,b=-47/50\,{\rm A},\,c=1/25\,{\rm A},\,d=-11/50\,{\rm A}$. Note the currents entering node $E$ from the right add up to $1/25+(-11/50) = -9/50\,\rm A$.

I have avoided the use of potential, potential difference and voltage because in the context of this example with a changing magnetic field present the work done is moving charge from one position to another is path dependent.

Suppose that $A,\,B,\,C,$ and $D$ were voltmeters then their respective readings across node $FE$ would be,

$1\times 28/25 = 28/25\,{\rm V},\,2\times (-47/50)=-47/25\,{\rm V},\,3\times 1/25=3/25\,{\rm V},\,4\times (-11/50)=-22/25\,{\rm V}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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