# Induced current in NMR

1. Apr 7, 2012

### simorgh

Hi,
Question:
we have a magnetization $M_x$ that varying as $M_x(t)=M_x e^{-t}$.Threre is a coil on the x axe. The magnetization will induce a current in the coil. How can get value of this current.
This situation happen in NMR device.

I try to caculate magnetic flux and then voltage ($v=-N \frac {d \Phi} {dt}$) to find induced current. But to calculate magnetic flux we need to have magnetic field (B). How can get magnetic field from magnetization?

thanks all to help me.

2. Apr 7, 2012

### Hassan2

If the body has an infinite dimension along x axis but finite dimension along the other two axis , then we have the following relation:

$\vec{B}=\mu_{0}(\vec{M}+\vec{H})=\mu_{0}(\vec{M} +\frac{\vec{M} }{\chi})=\mu_{0}(\vec{M}+\frac{1}{\mu_{r}-1}\vec{M})=\frac{\mu_{0}\mu_{r}}{\mu_{r}-1}\vec{M}$

3. Apr 7, 2012

### QuantumBunnii

I've personally never heard of this relation, but another method (I think) might help cast some light on the situation is through the direct calculation of the induced bound currents:

Jb = $\nabla$ χ M

and

Kb = M χ $\hat{n}$

where Jb is the induced volume current density (I/A), and Kb is the induced surface current density (I/L). From these, you can directly determine the total induced current.

Hope this helped. :3

P.S., Hassan, do you know what that relation is called? I thought the H field would be 0 in this case, since there is no free current and the divH = 0? I'm rather interested.

4. Apr 7, 2012

### Hassan2

In my opinion, the above relations don't give the induced current. They give the equivalent surface and volume current densities which are sometimes called material current. They are totally different from the induced current which depends on rate of change and the conductivity. For example when M is time-invariant, we don't have an induced current while your equation could give non-zero currents.

About the relation, it comes from the definitions of $\vec{M}$ , $\vec{H}$ and $\mu_{r}$:

Maxwell equation for $\vec{B}$ is

$\frac{1}{\mu_{0} }\nabla \times\vec{B}=J=(J_{f}+\nabla \times \vec{M})$

or
$\nabla \times(\frac{\vec{B}}{\mu_{0}}-\vec{M})=J_{f}$

then $\vec{H}$ is defined as:

$\vec{H}=\frac{\vec{B}}{\mu_{0}}-\vec{M}$ or equivalently $\vec{B}=\mu_{0}(\vec{M}+\vec{H})$
so that

$\nabla \times \vec{H}=J_{f}$

Then we have the experimental relation between an applied field and induced magnetization:

$\vec{M}=\chi \vec{H_{app}}$

Note that $\vec{H}$ in Maxwell equation is the sum of the applied field and the field rising from the magnetization itself, called demagnetizing field. Under the condition I mentioned, the demagnetizing field become zero, and you get the desired relation. Remember that by definition $\mu_{r}=\chi +1$.

Added: We have M because we have H. Only in a permanent magnet we have M without an applied field.

5. Apr 8, 2012

### QuantumBunnii

Ahh, I see.
Though, I'm still having some trouble understanding why the H-field doesn't equal 0 in this case.
From Maxwell's equation, we can invoke Stoke's Theorem to obtain the following relation:

$\int$H $\bullet$ dl = I(free)

Moreover, Hemholtz' theorem guarantees a viable solution from the curl alone, since div(H) = 0.
From this, shouldn't we deduce that H=0 everywhere, as there is no free current?
Also, you mentioned that a constant Magnetization would not induce a physical current, as is the case here. However, isn't the definition of Magnetization the uniform flow of electric charge, in order to produce a Magnetic field? Magnetization results from the uniform orientation of each magnetic "dipole" (i.e., atom) to produce a net Magnetization-- but isn't this concomitant with forcing the charges to move in a net direction?
I understand that a changing magnetic flux will produce an additional emf (Faraday's Law), but shouldn't any Magnetized object should exhibit some sort of current?
Thanks. :3

Also, I hope I'm not monopolizing this thread with my questions. Just a bit interested. :)

Edit: Also, the experimental relation between M and H only applies to linear media with particular magnetic susceptibility. How would we invoke these equations to solve this problem if they made no mention of linear media and, therefore, magnetic susceptibility? Or perhaps that was left out of the description. :p

Last edited: Apr 8, 2012
6. Apr 8, 2012

### Hassan2

1. In a magnetized media $\nabla . H=-\nabla . M$ because

$\nabla . B=\mu_{0}\nabla .(M+H)=0$

2. About the equivalent current, I hope one of the science advisers of the forum, known as timy_tim, appear and explain it in detail. But one thing I know is that that current is zero in nonmagnetic material. For example in a coil made of copper, you have no M, so you have no equivalent current density. Also that current is not due to drift of free electron, so it can't do work for us. For example a permanent magnet has such an equivalent current but of course we can't turn on a light with it!

Edit: For a nonlinear media, we need to know the B-H curve or M-H curve or the material which is again experimental. This is in classical electromagnetic. In micromagnetics , the relation between M and H is derived more theoretically, and no $\mu_{r}$ or $\chi_{m}$ is required.

Last edited: Apr 8, 2012
7. Apr 8, 2012

### Hassan2

Perhaps because we rarely deal with M.

Last edited by a moderator: May 5, 2017
8. Apr 8, 2012

### marcusl

Use Hassan's formula to calculate the induced voltage (not current) using your expression of Faraday's Law. Note that you can ignore H since it is constant and has zero derivative in your expression above, leaving the time-varying portion $$\vec{B}(t)=\mu_0 \vec{M}(t)$$ From this you can easily compute the flux.

9. Apr 9, 2012

### simorgh

thanks all, specially Hassan and marcusl.

But I have a problem.
Unit of B shoud be Tesla but in above formula is not.
Unit of M is $\frac{J}{T}$, and unit of $\mu_0$ is $\frac{V.s}{A.m}=\frac{T}{A.m}$, and then unit of B in above relation is $\frac{J}{A.m}=\frac{V.A.s}{A.m}=\frac{V.s}{m}= \frac {T}{m}$.
Have I a mistake?Where is my mistake?

thanks all again.

10. Apr 9, 2012

### Hassan2

You're welcome simorgh,

In SI units, the unit of M is A/m or J/(Tm3). I think you are confusing magnetic moment and magnetization. Magnetization is the volume density of magnetic moment.

11. Apr 9, 2012

### simorgh

OK. But in this situation unit of B in mentioned relation will be $\frac{T}{m^4}$!!!
Should we change relation and use other relation?

12. Apr 9, 2012

### Hassan2

The units are fine,

unit of μ0: $\frac{T.m}{A}$
unit of $M$: $\frac{J}{T.m^{3}}$

then
unit of $B= \frac{T.m}{A} \frac{J}{T.m^{3}}=\frac{J}{A.m^{2}}= \frac{Wb}{m^{2}}=T$

13. Apr 10, 2012

### simorgh

yes, you are right, thank you.

In the NMR, we have a sample that mentioned magnetization is generated from it.
Unit of volume that you use in magnetization is volume of space (and not volume of sample), isnt it?
If it is, then magnetic moment and magnetization will increase if we increase volume of sample (of course if total volume of sample be less than unit of volume), isnt it?

14. Apr 10, 2012

### Hassan2

$M=\frac{\sum \mu_{k}}{\Delta V}$ as $\Delta V\rightarrow 0$ ( not really zero but a small volume)

where $\mu_{k}$ is the atomic magnetic dipol and $\Delta V$ is a small volume around the point for at which the magnetization is defined. With this definition, M does not depends on the total volume but it depends on how parallel the atomic moments of the volume are. Since $\mu_{k}$s have equal magnitude, under a strong field all become parallel and M reaches its maximum which is called saturation magnetization. Beyond saturation, M does not depend on the applied field.

Last edited: Apr 10, 2012
15. Apr 10, 2012

### DrDu

The equations hold independently of the geometry of the sample. In NMR, the sample will at best be weakly paramagnetic and the sample size is macroscopic so that the equations of linear electrostatics are applicable. So the change of magnetic flux can be expressed in terms of M and the problem is solved inside the sample. However if the sample does not fill the entire coil, the field in the surrounding vacuum has to be taken into account, and, as magnetization vanishes there, can't be expressed in terms of magnetization. Rather, one would have to solve Helmholtz equations with the magnetization at the sample surface as boundary conditions.

16. Apr 10, 2012

### Hassan2

For a rectangular prism of iron, when it is magnetized along the long side, the flux density would be larger than if it was magnetized along one of the shorter sides with the same uniform magnetization. Doesn't it hold in NMR?

Also, I'm curious to know what are typical values of $\mu_{r}$ for NMR samples.

Last edited: Apr 10, 2012
17. Apr 10, 2012

### DrDu

I am not sure if I understand your setup. Do you mean the irron being magnetized by bringing the prism into a homogeneous extended B field?

18. Apr 10, 2012

### Hassan2

Yes. We have such a thing as demagnetizing factors which relates M and the field arising from M. This field is added to the applied field and the sum appear in Maxwell's equations. The factors depend on geometry.

Last edited: Apr 10, 2012
19. Apr 10, 2012

### DrDu

Yes, I know, but that is irrelevant for the material equations which relate H, M and B.
It is a problem of solving the Maxwell equations, i.e. finding the magnetization if the field is known e.g. at large distances. In an initial homogeneous field, a homogeneous magnetization will only arise if the object is ellipsoidal.
In the problem given this problem does not arise as M is assumed to be known.

20. Apr 10, 2012

### Hassan2

Yes, the material law doesn't depends on geometry but the second equation

$B=\mu_{0}(M+\frac{M}{\chi})$ is debatable because in general there relation between M and H at a point is not $M=\chi H$ but $M=\chi H_{app}=\chi (H-H_{d})$ where $H_{d}$ is the stray field arising from magnetization istelf and depends on geometry. In other words, susceptibility is the response to applied field while H in Maxwell equation is the sum of the applied field and the stray field.