Methods of inducing an EMF

  • #1
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So from Faraday's law, a change in the magnetic flux will induce an emf on a loop of wire.
[itex]{ \phi }_{ B }\quad =\quad BA\\ { \varepsilon }_{ ind }\quad =\quad \frac { d{ \phi }_{ B } }{ dt } [/itex]
(in this case a perpendicular field)
From these equations, it looks like the only way to induce a voltage is to change the magnetic field, change the area, and change the angle between the two. Then why does accelerating the loop through a constant field also induce an emf? The magnetic field and the area is constant in this case.
 

Answers and Replies

  • #2
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Well, if you're in the frame of the loop which is moving then there will be an electric field in addition to a magnetic one.
 
  • #3
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Well, if you're in the frame of the loop which is moving then there will be an electric field in addition to a magnetic one.
Because of the magnetic force on the electrons right? So the emf will be due solely to the electric field generated, and not from Faraday's law right?
 
  • #4
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Well, I was thinking more in terms of relativity. If one has a ##B##-field and no ##E##-field in one frame then a frame moving relative to the first will see both ##E## and ##B##-fields. So, ridding on the wire one would see an emf.
 
  • #5
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I think to answer more directly the "generated" E field is due to the relative motion, so yes.
 
  • #6
Drakkith
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From these equations, it looks like the only way to induce a voltage is to change the magnetic field, change the area, and change the angle between the two. Then why does accelerating the loop through a constant field also induce an emf? The magnetic field and the area is constant in this case.

What do you mean by an EMF here? Is there current flow through the loop (not just charge separation)?
 
  • #7
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What do you mean by an EMF here? Is there current flow through the loop (not just charge separation)?
Yes, there was a slide that my professor went through that showed had a question along the lines of this:
A rectangular loop of wire is moving through a uniform magnetic field. When the loop is moving at constant velocity is there an emf? What about when it's accelerating?
We didn't go over motional emf before this.
 
  • #8
Drakkith
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Honestly I'm not sure. I wouldn't think so, but I haven't gone over anything in my E&M class related to accelerating a loop through a magnetic field either.
 
  • #9
vanhees71
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If the surface you integrate the local Faraday Law,
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}$$
over a time-dependent surface with time-dependent boundary, you have to define the EMF as
$$\mathrm{EMF}=\int_{\partial F} \mathrm{d} \vec{x} \cdot \left[ \vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ],$$
where ##\vec{v}## is the three-velocity of the surface (taken along its boundary ##\partial F## within the integral). Then you have
$$\mathrm{EMF}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.$$
If the loop is just accelerated in a constant ##\vec{B}##, no EMF is induced.
 
  • #10
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If the surface you integrate the local Faraday Law,
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}$$
over a time-dependent surface with time-dependent boundary, you have to define the EMF as
$$\mathrm{EMF}=\int_{\partial F} \mathrm{d} \vec{x} \cdot \left[ \vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ],$$
where ##\vec{v}## is the three-velocity of the surface (taken along its boundary ##\partial F## within the integral). Then you have
$$\mathrm{EMF}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.$$
If the loop is just accelerated in a constant ##\vec{B}##, no EMF is induced.
Haha, I have not learned all that yet. Is that just another form of Faraday's law? I do believe my instructor made a mistake.
 

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