A Delhi Metro train starts without a sound of fuel or fire. Under its coaches sit traction motors, fed with current from the overhead line. Inside each motor, copper windings carrying a large current lie in a strong magnetic field. Nothing touches the windings, yet they are pushed round with enough force to move a full train.
A wire is a crowd of moving charges
In the chapter Magnetic Force on Moving Charges, a charge \(q\) moving with velocity \(\vec v\) in a magnetic field \(\vec B\) felt the force \(\vec F = q\vec v \times \vec B\). A copper wire holds a huge number of free electrons. When a current flows, these electrons drift along the wire. So every one of them feels a magnetic force, and the wire should feel the sum.
Without a current, the free electrons still race about at high speed. But their velocities point every way at random, so their magnetic forces cancel. A current adds a small common velocity, the drift velocity \(\vec v_d\), to every electron. Only this shared part survives the adding, and only it gives a net force.
An electron has charge \(-e\) and drifts opposite to the current. Its force is \((-e)(\vec v_e \times \vec B)\) with \(\vec v_e\) opposite to the current. The two minus signs cancel. So the force is the same as for a positive charge \(+e\) moving along the current. We may therefore picture the carriers as positive charges moving with the current.
\[ N = nA\,dl \]
Why. take a short piece of wire of length \(dl\) and cross-section \(A\), with \(n\) free electrons in each cubic metre.
\[ d\vec F = N\,e\,(v_d\,\hat l \times \vec B) \]
Why. each electron feels \(ev_dB\) in the direction of \(\hat l \times \vec B\), where \(\hat l\) is the unit vector along the current. The signs of charge and drift cancel.
\[ d\vec F = (neAv_d)\,dl\,(\hat l \times \vec B) \]
Why. put \(N = nA\,dl\) and regroup the factors.
\[ d\vec F = I\,d\vec l \times \vec B \]
Read it. \(I = neAv_d\), from the chapter Electric Current and Resistance, and \(d\vec l = dl\,\hat l\) points along the current.
The product \(I\,d\vec l\) is called a current element. It has the size \(I\,dl\) and points along the current at that place. Every force on a wire in this chapter is built from it. Find \(d\vec F\) for each element, then add the forces over the whole wire.
A short piece of wire of length \(dl\) carries a current \(I\) in a magnetic field \(\vec B\). It feels the force \(d\vec F = I\,d\vec l \times \vec B\), where \(d\vec l\) points along the current.
A copper wire of cross-section \(1\,\text{mm}^2\) carries \(3\,\text{A}\). Copper has \(8.5 \times 10^{28}\) free electrons per cubic metre. A \(0.5\,\text{m}\) length lies across a field of \(0.2\,\text{T}\). Find (a) the drift speed and (b) the force on one electron. Find also (c) the number of free electrons in the length and (d) the force on the wire.
(a) \(v_d\) \(= I/neA\) \(= 3/(8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6})\) \(= 2.2 \times 10^{-4}\,\text{m/s}\), about a fifth of a millimetre each second.
(b) \(f\) \(= ev_dB\) \(= 1.6 \times 10^{-19} \times 2.2 \times 10^{-4} \times 0.2\) \(= 7.1 \times 10^{-24}\,\text{N}\).
(c) \(N\) \(= nAL\) \(= 8.5 \times 10^{28} \times 10^{-6} \times 0.5\) \(= 4.25 \times 10^{22}\).
(d) \(F\) \(= Nf\) \(= 4.25 \times 10^{22} \times 7.1 \times 10^{-24}\) \(= 0.30\,\text{N}\). This is exactly \(BIL = 0.2 \times 3 \times 0.5\).
Values used: \(e = 1.6 \times 10^{-19}\,\text{C}\).
The electrons are pushed sideways inside the wire, but the wire itself moves. The pushed electrons crowd a little towards one side of the wire. They pull on the fixed positive ions of the metal lattice, and collisions pass their push on too. So the force on the electrons becomes a force on the whole wire.
A copper wire lies across a magnetic field. Which motion of its free electrons gives the wire a net magnetic force?
A straight wire in a uniform field
Take a straight wire of length \(L\) in a uniform field. Every element points the same way, and \(\vec B\) is the same at every element. So the forces on all the elements are parallel, and they simply add. The sum of the \(d\vec l\) is \(\vec L\), a vector of length \(L\) along the current. The force on the wire is \(\vec F = I\vec L \times \vec B\).
The size of a cross product is the product of the sizes times the sine of the angle between them. So \(F = BIL\sin\theta\), where \(\theta\) is the angle between the wire and the field. Only the part of the wire across the field, \(L\sin\theta\), feels any force. A wire along the field feels nothing.
In the sim, the force is \(0.3\sin\theta\) newton. It is \(0.30\,\text{N}\) at \(90°\), when the wire lies straight across the field. It is \(0.15\,\text{N}\) at \(30°\) and again at \(150°\), since \(\sin 150° = \sin 30°\). At \(0°\) and \(180°\) the wire lies along the field and the force vanishes.
A wire \(0.5\,\text{m}\) long carries \(3\,\text{A}\) in a uniform field of \(0.2\,\text{T}\). Find the force when the wire is (a) perpendicular to the field, (b) at \(30°\) to it, (c) along it. Also (d), a rod \(1\,\text{m}\) long carries \(10\,\text{A}\) at \(37°\) to a field of \(0.1\,\text{T}\). Find the force on it.
(a) \(F = BIL = 0.2 \times 3 \times 0.5 = 0.3\,\text{N}\).
(b) \(F = BIL\sin 30° = 0.3 \times 0.5 = 0.15\,\text{N}\).
(c) \(\sin 0° = 0\), so there is no force at all.
(d) \(F = 0.1 \times 10 \times 1 \times \sin 37° = 1 \times 0.6 = 0.6\,\text{N}\), taking \(\sin 37° = 0.6\).
A straight wire \(0.8\,\text{m}\) long carries \(2\,\text{A}\) in a uniform field of \(0.5\,\text{T}\). The wire makes \(30°\) with the field. Find the size of the force on it, in newtons.
A horizontal wire in a Delhi house runs east to west and carries \(5\,\text{A}\). The Earth's field there is about \(4.8 \times 10^{-5}\,\text{T}\), at \(42°\) below the horizontal, pointing north. Roughly what force does the field put on each metre of the wire?
Which way the wire is pushed
A cross product is perpendicular to both of its factors. So the force on a wire is at right angles to the wire and to the field. To find its sense, point the fingers of the right hand along the current and curl them towards \(\vec B\). The thumb gives \(\vec F\). Reverse the current, or reverse the field, and the force reverses.
Many students use Fleming's left-hand rule instead. Hold the thumb and the first two fingers of the left hand at right angles. The First finger points along the Field, the seCond finger along the Current. Then the thuMb gives the Motion, the direction of the force. Both rules give the same answer, because both describe \(I\vec L \times \vec B\).
(a) A wire carries \(3\,\text{A}\) to the right across a field into the page. Which way is it pushed? (b) A wire carries \(2\,\text{A}\) along \(\vec L = 0.5\,\hat j\,\text{m}\) in a field \(\vec B = 0.4\,\hat i\,\text{T}\). Find the force vector. (c) A wire of length \(l\) carries \(I\) along the \(x\) axis in \(\vec B = B_0(\hat i + \hat j + \hat k)\). Find the size of the force.
(a) Right is \(\hat i\) and into the page is \(-\hat k\), so \(\hat i \times (-\hat k) = \hat j\). The wire is pushed up the page.
(b) \(\vec F\) \(= 2 \times 0.5\,\hat j \times 0.4\,\hat i\) \(= 0.4\,(\hat j \times \hat i)\) \(= -0.4\,\hat k\,\text{N}\), into the page.
(c) \(\hat i \times \hat i = 0\), so only \(\hat j\) and \(\hat k\) of the field count: \(\vec F = IlB_0(\hat k - \hat j)\), of size \(\sqrt2\,IlB_0\).
In Fig 1.1 nothing happens while the switch is open. When it closes, the current flows down the rod, and \((-\hat j) \times (-\hat k) = \hat i\). So the rod is pushed along the rails, away from the cell. With \(2\,\text{A}\) the force is \(BIL = 0.2 \times 2 \times 0.5 = 0.2\,\text{N}\), so the rod gains \(0.5\,\text{m/s}^2\). Double the current and the force doubles. Reverse it and the rod runs the other way.
This is the Metro's secret. A traction motor is a set of such rods, the windings of its rotor, carrying a large current across a strong field. Each winding is pushed sideways, and the rotor turns. The pushes come from the same drifting electrons as in a torch wire, in far greater numbers.
A horizontal wire \(1\,\text{m}\) long, of mass \(0.1\,\text{kg}\), lies across a horizontal field of \(0.5\,\text{T}\) that points north. What current makes the wire float, and which way must it flow? Take \(g = 10\,\text{m/s}^2\).
For the wire to float, the magnetic force must be upward and equal to the weight \(mg = 1\,\text{N}\).
\(BIL = mg\), so \(I = mg/BL\).
\(I = (0.1 \times 10)/(0.5 \times 1) = 2\,\text{A}\).
East \(\times\) north gives up, so the current must flow east. A current flowing west would push the wire down.
A wire on the page carries a current up the page. The magnetic field points out of the page. Which way is the wire pushed?
A horizontal wire \(0.5\,\text{m}\) long, of mass \(20\,\text{g}\), lies at right angles to a horizontal field of \(0.4\,\text{T}\). What current makes it float? Take \(g = 10\,\text{m/s}^2\).
A rod \(0.4\,\text{m}\) long, of mass \(0.1\,\text{kg}\), rests on horizontal rails in a vertical field of \(0.25\,\text{T}\). The friction coefficient is \(0.3\). What is the least current that makes the rod start to slide? Take \(g = 10\,\text{m/s}^2\).
The magnetic force on a single charge does no work, because it is always perpendicular to the charge's velocity. Yet the rod on the rails speeds up. The work is done through the wire, and the energy comes from the cell that drives the current. Moving the rod \(0.1\,\text{m}\) with a force of \(0.4\,\text{N}\) takes \(0.04\,\text{J}\) from the cell.
In September 1821 Michael Faraday hung a wire so that its lower end dipped into a bowl of mercury. A bar magnet stood upright in the middle of the bowl. When he passed a current through the wire, the wire swung round and round the magnet without stopping. He had turned a current into steady motion for the first time. Every motor since, from a mixer to a Metro coach, rests on the force of this lesson.
A loudspeaker has a small coil fixed to its paper or plastic cone. The coil sits in the gap of a permanent magnet, where the field points straight out from the centre. The sound signal is a current that changes hundreds of times each second. The force \(BIL\) on the coil changes with it, pushing the cone in and out. The cone pushes the air, and you hear the song.
- Pushing the wire along the current or along the field.The force \(I\vec L \times \vec B\) is perpendicular to both. Use the right-hand rule or Fleming's left hand.
- Using \(BIL\) when the wire is not across the field.Use \(BIL\sin\theta\), with \(\theta\) the angle between the wire and \(\vec B\). A wire along the field feels no force.
- Reversing the direction for electrons.Use the direction of the current. The electron's negative charge and backward drift cancel.
- Forgetting that the energy comes from the source.The magnetic force on a charge does no work; the cell driving the current supplies the energy that moves the rod.