# What parts of an EMW does a ferrite rod antenna respond to?

richard9678
Hi. Would it be true to say, that a ferrite rod antenna, operating at fairly low frequencies (say 1Mhz) for all intents and purposes, only responds (in terms of voltage output) to the magnetic field part of a "radio wave"? Thanks.

## Answers and Replies

Homework Helper
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Depends what you exactly mean by "only responds". The magnetic field by itself can't move the electrons inside the antenna, it is the electric field that always does the job.

I think it is more correct to say that a ferrite rod antenna amplifies directly the magnetic part of the wave which in turn induces a stronger electric field inside the antenna.

Klystron, vanhees71 and tech99
Homework Helper
Gold Member
Hi. Would it be true to say, that a ferrite rod antenna, operating at fairly low frequencies (say 1Mhz) for all intents and purposes, only responds (in terms of voltage output) to the magnetic field part of a "radio wave"? Thanks.
Yes.
The nice thing about running a loop antenna of typical size (say 1"=10" diameter) at 1 MHz is that the B field is essentially uniform across its area. So you can apply Faraday's law to determine the emf as ## -d\phi/dt # times the number of loops.

vanhees71
Science Advisor
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For an air filled loop, either the electric or magnetic calculation gives the same result. For the electric calculation, the loop is regarded as two vertical conductors. Post #3 describes the underlying physics.

vanhees71 and Delta2
Homework Helper
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Depends what you exactly mean by "only responds". The magnetic field by itself can't move the electrons inside the antenna, it is the electric field that always does the job.

I think it is more correct to say that a ferrite rod antenna amplifies directly the magnetic part of the wave which in turn induces a stronger electric field inside the antenna.
I'm aware of the alleged magnetic field "concentration" effect by the ferrite core, but do you know of any analyses of the quantitative effect? By how much is the B field amplified? By the relative permeability? I have been unable to find a good analysis of this effect.

Science Advisor
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Two papers I wanted have unfortunately been deleted. I remember that the effective permeability is typically much smaller than that of the material, maybe 20 or so, and the mechanism is uncertain. Sometimes the turns-cancelling theory is used, where the inside and outside fields are opposing. In another paper, the radiation resistance was found to depend on the length of the rod, and it was acting as a true magnetic dipole. Both these papers have gone.

Delta2
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By how much is the B field amplified? By the relative permeability?
Yes that's what I had in mind. Of course there must be the quantum mechanical explanation of why materials like ferrite amplify the magnetic field, but unfortunately I am not good in quantum physics.

Homework Helper
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Yes that's what I had in mind. Of course there must be the quantum mechanical explanation of why materials like ferrite amplify the magnetic field, but unfortunately I am not good in quantum physics.
My original idea was that the B field was not augmented at all since then I thought ## \nabla \cdot \bf B = 0 ## would be violated at the boundary between air and the core. I thought all the core did was increase the inductance. But then I never figured out why that was a good idea either - fewer turns needed to get the required inductance, which is non-negotiable (*around ## 0.25 \mu H ## at 1 MHz), ergo lower emf.

Homework Helper
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My original idea was that the B field was not augmented at all since then I thought ## \nabla \cdot \bf B = 0 ## would be violated at the boundary between air and the core. I thought all the core did was increase the inductance. But then I never figured out why that was a good idea either - fewer turns needed to get the required inductance, which is non-negotiable (*around ## 0.25 \mu H ## at 1 MHz), ergo lower emf.
yes well sorry i am puzzled, the ferrite amplifies the B-field and the H field remains the same , or the B-field remains the same and H get reduced?

Science Advisor
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Of course ##\vec{\nabla} \cdot \vec{B}## can NEVER be violated. It's a fundamental law of nature (despite the fact that there might be magnetic monopoles not yet discovered). Further for the ##\vec{H}## field (in Heaviside Lorentz units)
$$\vec{\nabla} \times \vec{H} - \frac{1}{c} \partial_t \vec{D}=\frac{1}{c} \vec{j}_{\text{free}},$$
i.e., by definition ##\vec{H}## is the field due to free charges and currents. The constitutive equations are
$$\vec{B}=\mu \vec{H}, \quad \vec{D}=\epsilon \vec{E}.$$
##\mu## and ##\epsilon## usually depend on the frequency of the electromagnetic field and describe the response to magnetic and electric fields of the material. For (soft) ferrites ##\mu## is large and the material behaves like a paramagnet.

Homework Helper
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yes well sorry i am puzzled, the ferrite amplifies the B-field and the H field remains the same , or the B-field remains the same and H get reduced?
I think the B field remains the same so of course the H field is reduced by ## \mu_r ##. But, as I said, I'm not sure.

Delta2