# Analytical expression for the EMF induced in a solenoid due to motion of a permanent magnet

A permanent magnet (cylindrical) of length (l) and radius(r) with velocity v(t) along z-axis as shown in the figure. I am looking for analytical expression for the induced emf in a coil with N turns, due to the motion of a permanent magnet. • Do you have an analytical expression for the field of your permanent magnet? – probably_someone Oct 16 '19 at 8:57
• @probably_someone Not really. But I'm thinking of treating permanent magnet as a series of dipoles (not sure how valid this is ) With that assumption, yes, the field at any point can be calculated using the standard expression $B(r) = \mu_0 / 4* \pi (3 r ( m.r) - m) / r^3$ Where m is dipole magnetic strength. – Kumara Oct 16 '19 at 9:33

Suppose your coil has $$N$$ turns, spaced a distance $$d$$ apart from each other. It's simpler to approximate a helical coil as $$N$$ loops spaced a distance $$d$$ apart, so that we have a bunch of discrete shapes which have an easy-to-identify area vector.

If you have an analytical expression for the magnetic field $$\mathbf{B}(\mathbf{r})$$, then you can compute the flux through one loop by the following integral:

$$\Phi=\iint \mathbf{B}(\mathbf{r})\cdot\mathbf{dA}$$

This integral will obviously depend on the distance of the loop from the magnet, so the flux through each loop will be different. So what you need to compute, in the end, are the fluxes $$\Phi_1,...,\Phi_N$$ through each of the $$N$$ loops, as a function of the distance $$x(t)$$ from the first loop to the magnet (which, since the loops are all fixed relative to each other, determines the distance from all of the other loops to the magnet). Once you have $$\Phi_1(x),...,\Phi_N(x)$$, then you can compute the total EMF induced $$\mathcal{E}$$ using:

$$\mathcal{E}=-\sum_{i=1}^{N}\frac{d\Phi_i(x(t))}{dt}=-\sum_{i=1}^{N}\frac{d\Phi_i}{dx}v(t)$$

where $$v(t)$$ is the velocity of the magnet.

At this point, your problem reduces to a sum of surface integrals. Depending on what $$\mathbf{B}(\mathbf{r})$$ is, there may or may not be an analytical solution to these surface integrals.

One thing we can say immediately, though, just by looking at the general formula, is the following: the EMF generated at any point in time is a linear function of the velocity of the magnet. If you double the velocity, you double the EMF produced at that point in time.