Charges carried through a field
A metal rod moving through a magnetic field carries its free electrons with it. Each electron therefore moves through the field and feels the magnetic force \(q\vec v\times\vec B\) from the chapter Magnetic Force on Moving Charges. Take a rod PQ standing across the page, moving to the right at speed \(v\) in a field \(\vec B\) into the page.
For a positive charge, right crossed with into the page points up the rod, towards P. An electron is negative, so it is pushed down, towards Q. Electrons gather at Q, and Q becomes negative. The top end P is left short of electrons, so it becomes positive. No current flows round any circuit; charge only shifts along the rod.
The separated charge makes an electric field \(\vec E\) inside the rod, pointing from P to Q. That field pushes each electron back up the rod with the force \(eE\). In Fig 4.1 the pile-up grows until \(eE\) equals \(evB\). Then the net force on an electron along the rod is zero, and no more charge moves. In steady state \(E = vB\), \(0.8\,\text{V/m}\) here.
\[ F_{mag} = evB \]
Why. an electron moves with the rod at \(v\), at right angles to \(\vec B\), so it feels \(evB\) along the rod.
\[ F_{el} = eE \]
Why. the charge piled up at the ends makes a field \(E\) inside the rod, which pushes the electron the other way.
\[ eE = evB \quad\Rightarrow\quad E = vB \]
Why. in steady state no more charge moves, so the two forces balance.
\[ \varepsilon = El = Blv \]
Read it. a uniform field \(E\) along a length \(l\) gives a potential difference \(El\) between the ends.
This emf is called a motional emf. The rod behaves like a cell of emf \(Blv\). The magnetic force does the work of the cell's chemistry, pushing charge from one end to the other against the electric field.
A rod \(0.5\,\text{m}\) long moves at \(4\,\text{m/s}\) across a field of \(0.2\,\text{T}\). The rod, its velocity and the field are at right angles to one another. Find (a) the emf between its ends. The rod then moves at \(30°\) to the field, still at right angles to its own length. Find (b) the emf now.
(a) \(\varepsilon = Blv\) \(= 0.2 \times 0.5 \times 4\) \(= 0.4\,\text{V}\).
(b) Only the part of the velocity across the field counts, \(v\sin 30° = 2\,\text{m/s}\). So \(\varepsilon = 0.2 \times 0.5 \times 2 = 0.2\,\text{V}\).
Values used: only the data; \(e = 1.6\times10^{-19}\,\text{C}\) cancels out of \(eE = evB\).
A rod \(0.3\,\text{m}\) long moves at \(5\,\text{m/s}\) across a field of \(0.6\,\text{T}\). The rod, its velocity and the field are at right angles to one another. What is the emf between its ends, in volts?
Which end is positive
The rule is short: the end towards which \(\vec v\times\vec B\) points is positive. For the rod PQ moving right in a field into the page, \(\vec v\times\vec B\) points up, so P is the positive end. Reverse the motion or reverse the field and the polarity reverses. Reverse both and it stays the same.
Join P and Q by a wire outside the field, and a current flows. Like any cell, the rod pushes current out of its positive end. So current flows from P through the outside wire to Q, and back inside the rod from Q to P. Inside the source, current runs from the negative end to the positive end, just as inside a torch cell.
Try the figure above. With the field into the page the upper end is positive, and the current runs up the rod and anticlockwise round the circuit. Reverse the field and every sign reverses. Double the speed and the emf, the current and the drag all double. At \(4\,\text{m/s}\) in \(0.5\,\text{T}\) the rod gives \(1\,\text{V}\) and \(0.5\,\text{A}\), and the field drags it back with \(0.125\,\text{N}\). Lesson 6 follows that drag further.
A rod moving on its own, joined to nothing, is an open circuit. Its electrons shift until \(eE = evB\), and then stop. The emf is there, as a potential difference between the ends, but no current flows along the rod, because charge has no path back.
A rod lies along the page, from end A at the left to end B at the right. It moves up the page in a field out of the page. Which end is positive?
Motional emfs in the Earth's field
The Earth's field is weak but always there. In Delhi its horizontal part is \(B_H = 3.6\times10^{-5}\,\text{T}\), pointing north, and the field dips \(42°\) below the horizontal. So its vertical part is \(B_V = B_H\tan 42° = 3.24\times10^{-5}\,\text{T}\), pointing down. A horizontal rod moving horizontally cuts only the vertical part, and a horizontal rod falling cuts only the horizontal part.
(a) An aircraft with a wingspan of \(40\,\text{m}\) flies level at \(250\,\text{m/s}\) where the vertical part of the Earth's field is \(5\times10^{-5}\,\text{T}\). Find the emf between the wing tips. (b) The Rajdhani runs at \(20\,\text{m/s}\) through Delhi. Its axle is \(1.5\,\text{m}\) long. Find the emf across the axle. (c) A horizontal rod \(1\,\text{m}\) long, pointing east-west, falls at \(10\,\text{m/s}\) where \(B_H = 4\times10^{-5}\,\text{T}\). Find the emf across it.
(a) \(\varepsilon = B_V lv\) \(= 5\times10^{-5} \times 40 \times 250\) \(= 0.5\,\text{V}\). In Delhi's \(B_V\) it would be \(0.32\,\text{V}\).
(b) \(\varepsilon = B_V lv\) \(= 3.24\times10^{-5} \times 1.5 \times 20\) \(= 9.7\times10^{-4}\,\text{V}\), less than a millivolt. The sheet's \(5\times10^{-5}\,\text{T}\) gives \(1.5\times10^{-3}\,\text{V}\).
(c) The rod, its velocity and \(B_H\) are at right angles, so \(\varepsilon\) \(= B_H lv\) \(= 4\times10^{-5} \times 1 \times 10\) \(= 4\times10^{-4}\,\text{V}\). Delhi's \(B_H\) gives \(3.6\times10^{-4}\,\text{V}\).
Values used: Delhi's \(B_H = 3.6\times10^{-5}\,\text{T}\) at a dip of \(42°\), so \(B_V = 3.24\times10^{-5}\,\text{T}\); the worksheet takes \(5\times10^{-5}\,\text{T}\) and \(4\times10^{-5}\,\text{T}\).
Which wing tip is positive? North of the equator the vertical field points down. For a plane flying north, north crossed with down points west, to the pilot's left. For a plane flying east, east crossed with down points north, again to the left. In the northern hemisphere the left wing tip is always the positive one, whichever way the plane flies.
The pilot cannot light a lamp with the wing-tip emf. Wires run from the tips to the lamp, and those wires move with the plane through the same field. They carry the same motional emf, which cancels the wings' emf round the circuit. A voltmeter carried on board reads nothing. Only a circuit whose parts move differently, or a field that changes, gives a net emf.
A train's axle is \(1.5\,\text{m}\) long. The train runs at \(20\,\text{m/s}\) where the vertical part of the Earth's field is \(5\times10^{-5}\,\text{T}\). What is the emf across the axle, in millivolts?
The same answer from the flux
The force on the electrons gave \(Blv\). Faraday's law should agree. Put the rod on two rails closed at the left end, so that the rod and the rails form a circuit. Let \(x\) be the distance of the rod from the closed end. The area inside the circuit is \(lx\), and the flux is \(\Phi = Blx\).
As the rod moves, only \(x\) changes, at the rate \(dx/dt = v\). So \(d\Phi/dt = Blv\), and Faraday's law gives the same emf, \(Blv\). The two pictures are one physics seen two ways. Faraday's law counts the area swept out each second, \(lv\). The force picture finds where the emf sits: in the moving rod. The rails and the closed end, which do not move, carry none.
Now carry a closed rectangular loop to the right, wholly inside a uniform field. Its left and right sides each cut the field and each carries an emf \(Blv\), with the upper ends positive. Round the loop the two emfs drive current in opposite senses, so they cancel. The flux picture says the same thing more quickly: the flux through the loop does not change. There is no current.
Here is a case where only one side cuts the field. A square loop of side \(10\,\text{cm}\) has only its left side in a field of \(2\,\text{T}\), and the circuit's total resistance is \(4\,\Omega\). A steady current of \(1\,\text{mA}\) needs \(Blv = 4\times10^{-3}\,\text{V}\). So \(v\) \(= 4\times10^{-3}/(2 \times 0.1)\) \(= 0.02\,\text{m/s}\), just \(2\,\text{cm/s}\).
A closed rectangular loop moves at a steady velocity, wholly inside a uniform magnetic field. Is there a current in it?
In 1996 a Space Shuttle crew unrolled a conducting tether about \(20\,\text{km}\) long, with a small satellite at its far end. Moving at about \(7.5\,\text{km/s}\) across the Earth's field, the tether carried a motional emf of a few thousand volts. Electrons from the thin gas of the upper atmosphere closed the circuit. An electric arc then burned through the tether, which snapped near its full length. Still, engineers had seen a moving wire act as a source of emf in orbit.
- Using the total speed when the rod moves at an angle to the field.Only the part of \(\vec v\) across both \(\vec B\) and the rod counts.
- Naming the end where the electrons gather as positive.Electrons gather at the negative end. The positive end is the one towards which \(\vec v\times\vec B\) points.
- Sending current from + to − inside the rod.The rod is the source. Inside it, current flows from its negative end to its positive end.
- Using the whole Earth's field for a level wing.A horizontal rod moving horizontally cuts only the vertical part, \(B_V\).