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

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A ferrite rod antenna primarily responds to the magnetic field component of radio waves, particularly at low frequencies like 1 MHz. While the magnetic field alone cannot induce electron movement, it amplifies the magnetic part of the wave, which in turn enhances the electric field within the antenna. The discussion highlights the complexity of how ferrite materials affect the magnetic field, with references to relative permeability and the potential for quantum mechanical explanations. Participants express confusion over whether the B-field is amplified or if the H-field is reduced, emphasizing the need for clarity on the underlying physics. Overall, the conversation underscores the intricate relationship between magnetic and electric fields in antenna performance.
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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.
 
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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.
 
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richard9678 said:
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.
 
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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.
 
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Delta2 said:
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.
 
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.
 
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rude man said:
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.
 
Delta2 said:
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.
 
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rude man said:
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?
 
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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.
 
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Delta2 said:
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.
 
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