A rod that turns
A rod moving straight across a field has the emf \(Blv\), because every point of it moves at the same \(v\). Now pivot the rod at one end and turn it at a steady angular speed \(\omega\), in a plane at right angles to the field. The pivot does not move at all. The free end moves fastest, at \(\omega L\). A point at a distance \(r\) from the pivot moves at \(\omega r\).
In Fig 5.1 the velocity arrows grow steadily along the rod, in proportion to \(r\). So cut the rod into short pieces. A piece of length \(dr\) at a distance \(r\) moves at \(\omega r\), and it carries the small emf \(B(\omega r)\,dr\). The pieces are in series along the rod, so their emfs add.
\[ v = \omega r \]
Why. a point at a distance \(r\) from the pivot moves on a circle at the speed \(\omega r\), at right angles to the rod.
\[ d\varepsilon = B\,v\,dr = B\omega r\,dr \]
Why. the piece \(dr\), its velocity and the field are at right angles, so it is a small rod with emf \(Bv\,dr\).
\[ \varepsilon = \int_0^L B\omega r\,dr = \tfrac12 B\omega L^2 \]
Read it. the pieces lie end to end, so their emfs add from the pivot to the free end.
The result is also Faraday's law in another dress. In one turn the rod sweeps a circle of area \(\pi L^2\) in the time \(2\pi/\omega\). So it sweeps area at the rate \(\tfrac12\omega L^2\), and the flux it cuts each second is \(\tfrac12 B\omega L^2\). The factor \(\tfrac12\) is the average speed \(\tfrac12\omega L\) of the rod's points.
For the sign, ask which way a positive charge in the rod is pushed. In Fig 5.1 the field points out of the page and the rod turns anticlockwise. A point's velocity is along the circle, and that velocity crossed with out of the page points outward, along the rod. So positive charge is pushed to the free end, which is positive, and the pivot is negative. Turn the rod the other way and the polarity reverses.
(a) A metal rod \(1\,\text{m}\) long turns at \(10\,\text{rad/s}\) about one end, in a field of \(0.5\,\text{T}\) at right angles to its plane. Find the emf between its ends. (b) The blades of a ceiling fan in Delhi turn in a horizontal plane at \(300\) revolutions per minute. Each metal blade reaches \(0.6\,\text{m}\) from the axis. Find the emf between the axis and a blade tip.
(a) \(\varepsilon = \tfrac12 B\omega L^2\) \(= \tfrac12 \times 0.5 \times 10 \times 1^2\) \(= 2.5\,\text{V}\).
(b) \(\omega = 300 \times 2\pi/60 = 31.4\,\text{rad/s}\). The horizontal blade cuts only the vertical part of the field, so \(\varepsilon = \tfrac12 B_V\omega L^2\) \(= \tfrac12 \times 3.24\times10^{-5} \times 31.4 \times 0.36\) \(= 1.8\times10^{-4}\,\text{V}\), far too small to notice.
Values used: Delhi's \(B_V = 3.24\times10^{-5}\,\text{T}\).
A rod \(0.4\,\text{m}\) long turns about one end at \(20\,\text{rad/s}\), in a field of \(0.5\,\text{T}\) at right angles to the plane in which it turns. What is the emf between its ends, in volts?
The Faraday disc
Replace the rod by a solid metal disc of radius \(R\), turning at \(\omega\) about its axis in a field along that axis. Every radius of the disc is a rotating rod. Each one has the emf \(\tfrac12 B\omega R^2\) between the centre and the rim. The radii are side by side, in parallel, so together they give the same emf. A brush pressed on the axle and another on the rim pick it up.
Every point of the rim is at the same potential, because every radius gives the same emf. So the rim brush can touch the rim anywhere. The disc gives a steady emf, not one that swings back and forth. Faraday built the first such disc in October 1831, two months after his ring. It was the first machine to make a steady current from motion.
A copper disc of radius \(0.2\,\text{m}\) turns at \(50\,\text{rad/s}\) in a field of \(0.4\,\text{T}\) along its axis. Find (a) the emf between the centre O and a point A on the rim. Say (b) whether it depends on which rim point A is. The field points out of the page and the disc turns anticlockwise. Find (c) which is at the higher potential, the centre or the rim.
(a) \(\varepsilon = \tfrac12 B\omega R^2\) \(= \tfrac12 \times 0.4 \times 50 \times 0.04\) \(= 0.4\,\text{V}\).
(b) No. Every radius is the same rotating rod, so all rim points are at the same potential.
(c) A positive charge on a radius moves along the circle, and its velocity crossed with out of the page points outward. Positive charge is pushed to the rim, so the rim is at the higher potential.
A metal disc turns clockwise, seen from the front, in a field that points out of the page towards you. Which part is positive?
A rod turned about its middle
Now pivot a rod of length \(L\) at its midpoint. Each half is a rod of length \(L/2\) turning about one end. So the emf between the centre and either end is \(\tfrac12 B\omega(L/2)^2 = \tfrac18 B\omega L^2\). Both ends are pushed the same way relative to the centre: both positive, or both negative.
The two ends are therefore at the same potential, and the emf between them is zero. The second chip of Fig 5.1 shows it. A rod \(2\,\text{m}\) long about its midpoint has halves of \(1\,\text{m}\), and each end is \(2.5\,\text{V}\) above the centre. A voltmeter joined between the two ends reads nothing.
A rod \(1\,\text{m}\) long turns about its midpoint at \(10\,\text{rad/s}\), in a field of \(0.5\,\text{T}\) at right angles to its plane. What is the emf between the centre and one end, in volts?
An electric field with no charges
So far the charges moved and the field stood still. Now hold a loop of wire still and let the field change. A current flows, so the charges in the wire feel a force. It cannot be the magnetic force \(q\vec v\times\vec B\), because the charges start at rest. Something else must push them: an electric field. A changing magnetic field makes an electric field, the induced electric field.
The field does not need the wire. Take a long solenoid whose field \(B\), out of the page inside a circle of radius \(R\), rises steadily. By symmetry the induced field has the same size at every point of a circle of radius \(r\) about the axis. It runs along the circle. Its work on a unit charge taken once round is \(E \times 2\pi r\), which is the emf round that circle.
\[ \oint \vec E\cdot d\vec l = E\,2\pi r \]
Why. \(E\) is the same all round the circle and points along it.
\[ E\,2\pi r = \pi r^2\,\dfrac{dB}{dt} \quad (r < R) \]
Why. Faraday's law, with the flux \(B\pi r^2\) through a circle that lies inside the field region.
\[ E = \dfrac r2\,\dfrac{dB}{dt} \quad (r < R) \]
Why. divide by \(2\pi r\): inside, \(E\) grows in proportion to \(r\).
\[ E\,2\pi r = \pi R^2\,\dfrac{dB}{dt}, \quad E = \dfrac{R^2}{2r}\,\dfrac{dB}{dt} \quad (r > R) \]
Read it. a circle outside encloses all the flux \(B\pi R^2\) and no more, so \(E\) falls as \(1/r\).
The figure above shows the two parts of the result. Inside, \(E\) grows with \(r\): at \(5.2\,\text{cm}\) it is \(\tfrac12 \times 0.052 \times 0.13\) \(= 3.4\times10^{-3}\,\text{V/m}\). It peaks at the edge, \(5.5\,\text{mV/m}\). Outside it falls as \(1/r\): at \(12.5\,\text{cm}\) it is \(\tfrac12 \times 0.13 \times 0.085^2/0.125\) \(= 3.8\times10^{-3}\,\text{V/m}\). There is an electric field at \(12.5\,\text{cm}\), where there is no magnetic field at all.
The arrows run clockwise. That is Lenz's law again. A wire laid along the circle would carry a current making a field into the page, against the rising flux. Its direction and its closed lines make this field unlike any field from charges. It starts on no charge and ends on none; its lines are closed loops, like those of a magnetic field.
That makes it non-conservative. Take a charge once round a circle: the field does work \(q\,\varepsilon\) on it, not zero. So this field has no potential, and "the potential difference between two points" has no single value near it. This is why a transformer works. The changing flux in the iron core makes an electric field that drives current in the secondary coil. No charge passes between the two coils.
A long solenoid has a radius of \(10\,\text{cm}\). The field inside it rises at \(0.2\,\text{T/s}\). What is the induced electric field \(20\,\text{cm}\) from its axis, in millivolt per metre?
How is the electric field induced by a changing magnetic field different from the field of charges at rest?
In 1940 Donald Kerst, at the University of Illinois, built a machine that sped electrons up with no electrodes at all. It held electrons in a ring-shaped vacuum tube between the poles of an electromagnet. As the magnet's current rose, the changing flux through the ring made an induced electric field round it. That field pushed the electrons on at every turn. Kerst called the machine a betatron, and it gave electrons energies of millions of electronvolts.
- Using \(B\omega L \cdot L\) for a rotating rod.The speed grows from zero at the pivot to \(\omega L\) at the end, so the emf is \(\tfrac12 B\omega L^2\).
- Expecting an emf between the ends of a rod turned about its midpoint.Both ends are \(\tfrac18 B\omega L^2\) from the centre with the same sign, so the emf between them is zero.
- Taking the induced field to be zero outside the changing field.A circle outside encloses all the flux, so \(E = \dfrac{R^2}{2r}\dfrac{dB}{dt}\), which is not zero.
- Speaking of a potential for the induced field.Its work round a closed path is not zero, so it has no potential.