The induced emf is given by the formula $$\varepsilon = -\frac{d\Phi}{dt}\;.$$

Lenz's Law tells that

The direction of the induced current be such that it opposes the change that has induced it.

So, that means the current would be induced such that it counteracts the change in magnetic flux through the loop caused due to the approaching of the magnet.

If the flux is increasing when the north pole of the magnet approaches the loop, there would be counter-clockwise current so that it counteracts the change of magnetic flux of the moving magnet towards the loop.

But... I'm not getting one thing.

Does counteracting the flux mean at every moment the flux produced by the induced current in the loop nullifies the flux through the loop by the magnetic field of the moving magnet?

The equation only says the emf is the negative of the time rate change of flux through the loop by the moving magnet.

But as Lenz's law says the induced current counteracts the changing flux of the moving magnet, does that mean it nullifies the flux through the loop of the moving magnet? Is the total magnetic flux - sum of the magnetic flux due to the induced current & the magnetic flux through the loop is always constant- zero when the north pole of the magnet approaches & a non-zero constant value when the north pole of the magnet moves away?

Also, is the rate of change of flux through the loop due to the induced current equal to the rate of change of flux through the loop by the moving magnet that is $$\frac{d\Phi_\text{due to induced current}}{dt}= \frac{d\Phi_\text{due to the moving magnet}}{dt}\;?$$

  • $\begingroup$ That is a very good question. I don't think your argument has any flaws. You are correct. $\endgroup$
    – ShankRam
    Dec 13, 2015 at 12:38
  • 1
    $\begingroup$ Short answer is yes if the loop is made from perfectly conducting wire since, in that case, there can be no emf. See this related question $\endgroup$ Dec 13, 2015 at 13:51

1 Answer 1


Generally, it isn't the case that the (net, total) magnetic flux threading the conducing loop is constant.

Note that the quote uses the word opposes which you have evidently taken to mean nullify. But in fact, if the solution requires an emf, then there must be a non-constant flux threading the loop.

When the conductive material forming the loop has non-zero resistivity, the solution must satisfy

$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt}$$


$$\mathscr E = R i$$

where $R$ is the resistance 'round the loop and $i$ is the current.


$$\Phi = \Phi_\textrm{ext} + \Phi_i = \Phi_\textrm{ext} + L i$$

where $L$ is the inductance of the conducting loop.

Combining the above yields

$$R i + L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$

For example, consider the case that $\Phi_\textrm{ext} = \Phi_0$ is constant then the solution is

$$i(t) = I_0e^{-\frac{R}{L}t}$$

$$\Phi(t) = LI_0e^{-\frac{R}{L}t} + \Phi_0$$

$$\mathscr E(t) = -\frac{\mathrm d\Phi}{\mathrm dt} = -\left(-\frac{R}{L} \right)LI_0e^{-\frac{R}{L}t} =Ri(t)$$

So, this is all consistent and notice that the flux is not generally constant though the external flux is.

Now, consider the case that $\Phi_\textrm{ext}$ is increasing linearly with time

$$\Phi_\textrm{ext} = \Phi_0 + \phi t$$

then the particular solution is

$$i(t) = -\frac{\phi}{R}$$ $$\Phi(t) = \left(\Phi_0 -\frac{L}{R}\phi\right) + \phi t$$ $$\mathscr E = -\phi = Ri $$

Again, this is consistent and the flux is not generally constant. However, note that this result depends on $R$ being non-zero.

For the $R = 0$ case, we see that

$$L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$


$$\Phi_i = - \Phi_\textrm{ext} + \mathrm{constant}$$

$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt} = 0$$

thus we conclude that, for a perfectly conducting loop, the magnetic flux threading the loop is constant.

  • $\begingroup$ Oh! So, that means, the current induced in the loop is not only due to the flux by the moving magnet but also due to self-induction. But, can you tell why Purcell in his book deduce the Faraday's law only for the change in flux of the external magnetic field & not took the flux change due to self-induction? $\endgroup$
    – user36790
    Dec 13, 2015 at 20:31
  • $\begingroup$ Here is the Google Book. Isn't it wrong not to take the self-induction part of flux in the derivation of $$\varepsilon= -\frac{d\Phi}{dt}\;?$$ $\endgroup$
    – user36790
    Dec 13, 2015 at 20:32
  • $\begingroup$ @user36790, Re your first comment which isn't quite correct. If you look closely at the equations in my answer, you'll see that the loop current must satisfy a differential equation for which the forcing or driving term is the (negative) time rate of change of external magnetic flux. $\endgroup$ Dec 13, 2015 at 22:38
  • $\begingroup$ @user36790, Re Purcell, if the self-inductance term is insignificant (as is evidently the case in the Purcell example), the emf is effectively just the emf due to the $-\frac{d\Phi_{ext}}{t}$ term. $\endgroup$ Dec 13, 2015 at 22:46
  • $\begingroup$ Why is the self-inductance term insignificant in the examples given by Purcell? $\endgroup$
    – user36790
    Dec 14, 2015 at 1:17

Your Answer

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