Motional Electromotive Force

(a) Translational motion of a conductor

Set-up and diagram

• Consider a C-shaped frame of wires PQRS of area $lx$ situated in a constant magnetic field $\vec{B}$. QR is a conducting wire of length $l$ that slides on the frame, parallel to PS.
• Diagram in words: a rectangular circuit lying in the plane of the paper with $\vec{B}$ into the page (crosses drawn all over); the left arm PS is fixed, the right arm QR is a sliding bar of length $l$; the bar moves to the right with velocity $\vec{v}$ along the $x$-axis, so the enclosed length is $x$ and the area is $lx$.

Derivation from the flux rule

• As the wire QR of length $l$ is moved out with velocity $\vec{v}$ to increase $x$, the area of the loop PQRS increases, so the flux of $\vec{B}$ through the loop increases with time.
• According to the flux rule, the induced emf equals the rate at which the magnetic flux through the conducting circuit changes. The induced emf will cause a current in the loop; it is assumed that there is enough resistance in the wire so that the induced currents are very small and produce a negligible magnetic field of their own.
• As the flux through the frame PQRS is $\phi = Blx$, the magnitude of the induced emf is

$$\lvert e\rvert = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(Blx) = Bl\dfrac{dx}{dt} = Blv \qquad \ldots (1)$$

• where $v$ is the velocity of the wire QR increasing the length $x$ of the wires PQ and SR.

The same result from the Lorentz force

• A charge $q$ carried along by the moving wire QR experiences the Lorentz force

$$\vec{F} = q\left(\vec{v}\times\vec{B}\right)$$

• which is perpendicular to both $\vec{v}$ and $\vec{B}$ and hence is parallel to the wire QR.
• The force $\vec{F}$ is constant along the length $l$ of the wire QR (since $v$ and $B$ are constant) and zero elsewhere, because $v = 0$ for the stationary part RSPQ of the frame.
• When the charge $q$ moves a distance $l$ along the wire, the work done by the Lorentz force is

$$W = F\cdot l = qvB\sin\theta\cdot l$$

• where $\theta$ is the angle between $\vec{B}$ and $\vec{v}$.
• The emf generated is work per unit charge:

$$e = \dfrac{W}{q} = vB\sin\theta\cdot l$$

• For maximum induced emf, $\sin\theta = 1$:

$$e_{\max} = Blv \qquad \ldots (2)$$

• Equations (1) and (2) agree. In general it can be proved that for any circuit whose parts move in a fixed magnetic field, the induced emf is the time derivative of the flux, regardless of the shape of the circuit.

The stationary-loop case

• The flux rule also applies to a wire loop kept stationary while the magnetic field is changed. The Lorentz force on the charges is $\vec{F} = q\left(\vec{E}+\vec{v}\times\vec{B}\right)$; there are no new special forces due to changing magnetic fields, and any force on charges at rest in a stationary wire must come from the $\vec{E}$ term.
• Faraday's observations therefore led to a new law relating electric and magnetic fields: in a region where the magnetic field is changing with time, electric fields are generated. It is this electric field that drives the electrons around the circuit and is responsible for the induced emf in a stationary circuit whenever the flux changes.
• The flux rule holds good so long as the change in flux is due to a change of magnetic field, or to the motion of the circuit, or both.

(b) Motional emf in a rotating bar

Set-up and diagram

• A conducting bar of length $l$ rotates about a pivot at one end, in a plane perpendicular to a uniform magnetic field $\vec{B}$, with angular velocity $\omega$. A rotational emf is induced between the two ends of the bar.
• Diagram in words: a straight rod pivoted at O, sweeping out a circle in the plane of the page, with $\vec{B}$ into the page; a small element $dr$ of the rod is marked at a distance $r$ from the pivot.
• A rotating bar differs from a sliding bar because different points of it move with different speeds: the element at distance $r$ has $v = \omega r$.

Derivation

• Consider a small segment $dr$ of the bar at distance $r$ from the pivot. It is a short length of conductor moving with velocity $v$ in the field $\vec{B}$, so it has an induced emf like a sliding bar:

$$de = Bv\,dr = B\omega r\,dr$$

• All such segments are in series, so the emfs of the individual segments add. The total induced emf in the rotating rod is

$$e = \int de = \int_0^l B\omega r\,dr = B\omega\int_0^l r\,dr = B\omega\dfrac{l^2}{2}$$

$$e = \dfrac{1}{2}B\omega l^2$$

• Compare this with the induced emf in the sliding bar, $e = Blv$: the rotating bar carries the extra factor $\dfrac{1}{2}$.
• With $\omega = 2\pi f$, where $f$ is the number of revolutions per second, this can be written as

$$e = \dfrac{1}{2}B(2\pi f)l^2 = B\pi f l^2$$

Faraday disc dynamo

• The same argument applied to a metal disc of radius $R$ rotating with angular velocity $\omega$ about an axis through its centre and perpendicular to its plane, in a field $B$ parallel to that axis, gives the emf between the rim and the axis:

$$e = \dfrac{1}{2}B\omega R^2$$

• Read as flux cut: a radius of the disc sweeps out an area $\pi R^2 f$ per second, and the flux cut per second is $B\pi R^2 f$, which is numerically the induced emf.

Standard results

• Sliding bar, $\vec{B}$, $\vec{l}$, $\vec{v}$ mutually perpendicular: $\lvert e\rvert = Blv$.
• General orientation: $e = Blv\sin\theta$ with $\theta$ the angle between $\vec{v}$ and $\vec{B}$; $e = 0$ when $\vec{v}$ is parallel to $\vec{B}$.
• Rotating rod about one end: $e = \dfrac{1}{2}B\omega l^2 = B\pi f l^2$.
• Faraday disc: $e = \dfrac{1}{2}B\omega R^2$; area swept per second by a radius $= \pi R^2 f$; flux cut per second $= B\pi R^2 f$.
• Aircraft wings, railway axles and falling wires are all $e = Blv$ problems, using the component of the earth's field perpendicular to both $l$ and $v$.