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As I keep reading through texts and online, I keep getting conflicting views when in magnetic fields and electromotive forces are involved.

From what I can see, the emf is the work done per unit charge integrated along a closed loop by a source that is not electrostatic. If the Lorentz force does no work, how can a magnetic field produce an emf?

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  • $\begingroup$ The Lorentz force can do work on a charged particle. I suspect the confusion you have is that you're mixing up magnetostatics and magnetodynamics. The force from a static magnetic field does no work on a charged particle. However, a changing magnetic field can result in work being done. $\endgroup$ – Brionius Jan 31 '15 at 15:58
  • $\begingroup$ Yes, I chose my words carefully. The Lorentz force can do work. A changing magnetic field can result in work being done. I didn't say that the magnetic field can do work. $\endgroup$ – Brionius Feb 9 '15 at 5:12
  • $\begingroup$ @Brionius A static $\vec{B}$ field can produce an emf about a moving circuit, a changing $\vec{B}$ field can induce an electric field that can produce an emf about a circuit (moving or not), and an emf is not work it is the line integral of the force per unit change in the direction of the circuit (not the direction of displacement). In the quasistatic limit the total emf from the total Lorentz force per unit charge equals the change in $\vec{B}$ flux, which is different (for moving circuits) than the flux of the change of the $\vec{B}$ field. $\endgroup$ – Timaeus Feb 10 '15 at 6:39
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The emf is not the work done per unit charge integrated along a closed loop by a source that is not electrostatic, it's just:

$$ \mathscr E =\oint \vec{f}_s \cdot d\vec{\ell},$$

where the integral is around the circuit, and $\vec{f}_s$ is the net force per unit charge on the conduction charges that move about within the circuit element. This might seem the same, but it isn't. The integral for the emf is around a loop at some fixed time. The work done between two points needs to follow the charges in space and time. In statics there isn't much difference, but in dynamics (changing fields or moving wires) it matters.

The Lorentz force can do work, but I think you are specifically asking about how a magnetic field can produce an emf. There are two ways a magnetic field can be responsible for an emf.

The first is cheating. If the magnetic field at a point in space changes, then it is responsible for producing an electric field and that electric field can do work and in fact $\vec{E}$ can be line integrated along the circuit, and the result is equal to the flux of $\vec{\nabla}\times\vec{E}$ through the surface determined by the circuit, so equal to the flux of $-\partial \vec{B}/\partial t$ through the surface determined by the circuit.

The second is if the wire is moving. If there wire is moving, then you can compute $\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}$ where $\vec{v}=\vec{w}+\vec{u}$ is the velocity of the conduction (mobile) charge, and $\vec{w}$ is the velocity of the wire element itself, and $\vec{u}$ is the relative velocity of the mobile charge through the wire. This is the magnetic contribution to the emf because $\vec{v}\times\vec{B}$ is the magnetic force per unit charge on the mobile charges.

We can evaluate this:

$$ \oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\oint \vec{B} \cdot (\vec{v}\times d\vec{\ell})=-\oint \vec{B} \cdot ((\vec{w}+\vec{u})\times d\vec{\ell}).$$

Then notice that $\vec{u}$ is along the wire, so parallel to $d\vec{\ell}$, so

$$ \oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\oint \vec{B} \cdot (\vec{w}\times d\vec{\ell})$$

In a small amount of time, there is a circuit at one place, and then a time $\Delta t$ later the circuit is somewhere else and each part moves $\vec{w}\Delta t$ in space. Imagine a circuit at one time, and at the later time (so a latter circuit, and a former circuit), and there is a ribbon in between. Each part of the ribbon has an area $d\vec{a}=(\vec{w}\times d\vec{\ell})\Delta t$. Make your time interval so small that $\vec {B}$ doesn't change much (in time) from when the circuit is in one place to the other or anywhere inside. Then for a fixed time in that interval $\Delta t$, $\vec{\nabla}\cdot \vec{B}=0$ so the total flux through the surface $S$ bounded by the latter circuit ($C_2$), the former circuit ($C_1$) and the ribbon ($R$) is zero.

$0=\oint_S \vec{B}\cdot d\vec{a}=\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}+\int_{R} \vec{B}\cdot d\vec{a}.$

So $$\Delta t\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\Delta t\oint \vec{B} \cdot (\vec{w}\times d\vec{\ell})=\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}.$$

Where $d\vec{a}$ points outwards in both cases.


So if the $\vec{B}$ field is changing we get an induced electric field and a corresponding electric emf of:

$$\oint_{C_1} \vec{E} \cdot d\vec{\ell}=\int_{C_1} (\vec{\nabla}\times\vec{E})\cdot d\vec{a}=\int_{C_1} (\vec{\nabla}\times\vec{E})\cdot d\vec{a}=\int_{C_1} (-\partial \vec{B}/\partial t)\cdot d\vec{a}.$$

So for a small time interval $\Delta t$:

$$\Delta t\oint_{C_1} \vec{E} \cdot d\vec{\ell}=\Delta t\int_{C_1} (-\partial \vec{B}/\partial t)\cdot d\vec{a}=-\int_{C_1} (\vec{B}(t_0+\Delta t)-\vec{B}(t_0))\cdot d\vec{a}.$$

And if the circuit is moving we get:

$$\Delta t\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}.$$

Where this time both the $d\vec{a}$ vectors point in the direction associated with the direction of the oriented circuit.

Now if you compute the magnetic flux $\Phi=\int_{C} \vec{B}\cdot d\vec{a}$, then it's time derivative has two parts (from the product rule), one from the changing $\vec{B}$ (for fixed circuit) and one from a fixed $\vec{B}$ (and changing circuit).

So putting them together, $$\oint \vec{E} \cdot d\vec{\ell}+\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-d\Phi/dt.$$


Thus the negative of the time rate of change of the magnetic flux is literally equal to the integral of the Lorentz force per unit charge around the circuit, and the electric part of the Lorentz force is due to the parts of the $\vec{B}$ field that are changing at some point, and the magnetic parts of the Lorentz force contribute where the circuit element itself is moving.

So the Lorentz force exactly contributes the emf due to the changing magnetic flux. The magnetic part because of the moving circuit, the electric part because of the induced electric field from the changing magnetic field. Both matter in general.

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  • $\begingroup$ How exactly does the Lorentz force do work? $\endgroup$ – Oscar Flores Feb 10 '15 at 18:46
  • $\begingroup$ @OscarFlores I suggest asking that as a separate question, and that you also ask in the question to compare and contrast work with electromotive force. $\endgroup$ – Timaeus Feb 11 '15 at 3:25

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