I am an undergrad intern at a national lab currently working with a basic proton NMR device. The device consists of two big coils which provide the static magnetic field, and a smaller coil, which sends both the "excitation" signal and receives the NMR signal. A week or so ago, my supervisor asked me to calculate the magnitude of the magnetic field of the inner coil.

Since I knew the maxiumum voltage of my signal, I chose to use Faraday's law of induction for a tightly wound coil of wire: $\mathcal{E}=-N\frac{d\phi_B}{dt}$. Knowing that the excited net magnetization vector $\vec{M}$ has a torque exerted on it (thanks to the static field), I reasoned that the magnetic flux through the smaller coil would be: $\phi_{B}=BA\cos\omega t$. Taking the derivative of this, I reasoned that the maximum voltage would be equivalent to the maximum of $-N\frac{d\phi_B}{dt}$:


I figured that from this equation; knowing the area, precession frequency, and number of coils; I would be able to find $B$ pretty easily.

But we measured B another way, using $\theta=\gamma B_1t_p$, and got a result that was half the size of the "Faraday way". At first, even my supervisor was confused, but then he quickly remembered that we forgot to remember that we were working in a "rotating frame".

For this reason, our supervisor said that the correct relation between max voltage and max flux was really $\frac{V_{max}}{2}=NBA\omega$.

Well, this baffled me. It still baffles me. He tried to explain to me that we can think of our signal as two arrows which oscillate in opposite directions in a circle, each arrow having a magnitude half the size of the actual thing. I know I'm using vague language right now, but that's because I don't get it. Why do I cut my max voltage in half. Even if we did use this rotating frame, wouldn't the max voltage be when the arrows are in phase, and add up together.

If you guys could help me out, that would be awesome. Let me know if you need any clarification on the setup of the experiment; I'll be happy to elaborate.

  • $\begingroup$ By the way, NMR jocks usually call it 'the rotating frame', not 'the revolving frame.' Laser people often speak of the 'rotating wave approximation.' $\endgroup$ Jun 1, 2014 at 23:45

1 Answer 1


Think of it this way.

First of all, it is perfectly reasonable, and not at all vague, to think of a linearly polarized electromagnetic field as being a superposition of two circularly polarized fields, of opposite polarity, and half magnitude. You should just work this out with some simple algebra; you will not find it difficult.

Second, when a nuclear spin 1/2 is polarized in a static magnetic field, the transition from lower energy to higher energy state (or vice versa) has a selection rule for polarization-- only circular polarization is allowed, and only a particular sense of circular polarization, corresponding to the to the sense of Larmor precession dictated by the magnetic moment of the spin. There is a good discussion of the Larmor precession by a 'classical' magnetic moment in Goldstein's book Classical Mechanics (first edition for sure--don't know about later editions.)

You can think of the selection rule quantum mechanically in this way. The transition (spin flip + emission of photon, or spin flop + absorption of photon-- it can be either) must conserve each component of angular momentum. The spin pointing along the magnetic field (conventionally the 'z' direction) has e.g. angular momentum 1/2. After flipping, the spin points against the magnetic field and has z component angular momentum -1/2. (Actually the exct signs here depend on the sign of the spin gyromagnetic ratio; but whether there's an up-down or a down-up transition doesn't really matter here.) The delta is 1, which corresponds to the angular momentum of a circularly polarized photon. A linearly polarized photon, being a superposition of positive and negative spirals, has zero angular momentum about the propagation direction. Therefore, if excitation is made with linearly polarization, only that component will be active which sums to zero with the existing delta. That is, for delta = 1 (as above) the photon must have angular momentum -1 (negative polarization in this arbitrary example.) But this corresponds (by superposition) to just one half of the incident field amplitude.

This is a bit long-winded; I hope it helps.

  • $\begingroup$ I think I understand how a linearly polarized field can be thought of as two circularly polarized fields. What I don't understand is why the maximum voltage divided by two is equal to the maximum change in magnetic flux, rather than just the maximum voltage. Does that make sense? Also, your last paragraph was somewhat hard to follow, probably because I'm only a junior undergraduate. I was following it up to the photon part. $\endgroup$ Jun 6, 2014 at 3:49

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