Many Indian homes of the 1990s watched Doordarshan on a picture-tube television. Children soon found that a magnet held near the screen bent the picture into swirls of colour. With the set switched off, the same magnet did nothing to the glass at all. Behind the glass, a beam of electrons had been painting the picture.
A second kind of field
In the chapter Electric Charge, Coulomb's Law and the Electric Field, one charge acted on another through a field. The first charge set up the electric field \(\vec E\), and the field pushed on the second charge with \(q\vec E\). Magnetism follows the same two steps. Moving charges and currents set up a magnetic field \(\vec B\). That field then pushes on other charges, but only on those that are moving.
This chapter takes the second step: given a field \(\vec B\), what does it do to a moving charge? Where the field comes from is the subject of the chapter Sources of the Magnetic Field. For now a magnet or a coil simply supplies a field of known size and direction.
What experiments show
Careful experiments on beams of charged particles give four facts about the force. It is proportional to the charge \(q\). It is proportional to the speed \(v\) and to the field \(B\). It vanishes when the charge moves along the field, and it is largest when the charge moves across the field. Finally, it always points at right angles to both the velocity and the field.
Only one product of two vectors behaves like this: the cross product, met in the chapter Torque and Equilibrium. Its size is \(vB\sin\theta\), and it points at right angles to both vectors. So the four facts become one line.
A charge \(q\) moving with velocity \(\vec v\) through a magnetic field \(\vec B\) feels the force \(\vec F = q\,\vec v \times \vec B\). This is the magnetic part of the Lorentz force. Its size is \(qvB\sin\theta\), where \(\theta\) is the angle between \(\vec v\) and \(\vec B\).
The tesla
The unit of \(B\) follows from \(B = F/(qv)\) for a charge moving across the field. One tesla is the field that pushes a charge of 1 C, moving at 1 m/s across it, with a force of 1 N. A coulomb per second is an ampere, so \(1\,\text{T} = 1\,\text{N/(A m)}\). The dimensions of \(B\) are \([\text{M}\,\text{T}^{-2}\,\text{A}^{-1}]\).
The tesla is a large unit. The Earth's field at Delhi is about \(4.8\times10^{-5}\,\text{T}\). A small magnet on a fridge door gives a few thousandths of a tesla near its face. The MRI scanners of a large hospital, such as AIIMS in New Delhi, use 1.5 T or 3 T. An older unit, the gauss, is \(10^{-4}\,\text{T}\), so the Earth's field is about half a gauss.
A proton moves at \(2\times10^6\,\text{m/s}\) at right angles to a field of \(0.5\,\text{T}\). Find the magnetic force on it, and compare it with the proton's weight.
The angle is \(90°\), so \(F = qvB = 1.6\times10^{-19}\times2\times10^6\times0.5 = 1.6\times10^{-13}\,\text{N}\).
The weight is \(m_pg = 1.67\times10^{-27}\times10 = 1.67\times10^{-26}\,\text{N}\).
The ratio is \(1.6\times10^{-13}/1.67\times10^{-26} \approx 10^{13}\). Gravity is ignored for charged particles in this chapter, and this ratio is the reason.
Values used: \(e = 1.6\times10^{-19}\,\text{C}\), \(m_p = 1.67\times10^{-27}\,\text{kg}\), \(g = 10\,\text{m/s}^2\).
The same proton, at \(2\times10^6\,\text{m/s}\), now moves at \(30°\) to the \(0.5\,\text{T}\) field. Find the magnetic force on it, in units of \(10^{-14}\,\text{N}\).
Finding the direction
The right-hand rule gives the direction of \(\vec v\times\vec B\). Point the fingers of the right hand along \(\vec v\). Curl them towards \(\vec B\) through the smaller angle. The thumb then points along \(\vec v\times\vec B\). For a positive charge that is the direction of the force. For a negative charge, such as an electron, the force points the opposite way.
A page is flat, but fields often cross it. A field into the page is drawn as a grid of crosses, like the tail feathers of an arrow flying away from you. A field out of the page is drawn as dots, like the tip of an arrow coming towards you. In the figures of this book, crosses mean \(\vec B\) points into the page.
Turn the slider through a full circle. The force turns with the velocity and stays a quarter turn away from it. For a proton in a field into the page, it is always a quarter turn anticlockwise from the velocity. For an electron it is a quarter turn clockwise. Such a force keeps bending a moving charge in one sense, and the next lesson shows that the path is a circle.
Working with components
When the vectors are given in components, the cross product does the work. Use \(\hat i\times\hat j = \hat k\), \(\hat j\times\hat k = \hat i\), \(\hat k\times\hat i = \hat j\). Reversing the order changes the sign, and any unit vector crossed with itself gives zero.
A charge of \(2\,\mu\text{C}\) moves with \(\vec v = (3\hat i + 4\hat j)\times10^4\,\text{m/s}\) in a field \(\vec B = 0.1\hat k\,\text{T}\). Find the magnetic force and its size.
\(\vec F = q\,\vec v\times\vec B\) with \(q = 2\times10^{-6}\,\text{C}\). The field is along \(\hat k\), so only \(\hat i\times\hat k\) and \(\hat j\times\hat k\) are needed.
\(\vec v\times\vec B = 10^4\times0.1\,(3\hat i\times\hat k + 4\hat j\times\hat k) = 10^3(-3\hat j + 4\hat i)\).
\(\vec F = 2\times10^{-6}\times10^3\,(4\hat i - 3\hat j) = (8\hat i - 6\hat j)\times10^{-3}\,\text{N}\). Its size is \(\sqrt{64 + 36}\times10^{-3} = 0.01\,\text{N}\).
Check: \(\vec F\cdot\vec v = (8\times3 - 6\times4)\times10^1 = 0\). The force is perpendicular to the velocity, as it must be.
In Delhi the Earth's field points northwards and downwards, at \(42°\) below the horizontal. An electron beam travels horizontally towards the north. Which way does the magnetic force push it?
The electrons of the last question move at \(3\times10^7\,\text{m/s}\), and the vertical part of the field at Delhi is \(3.24\times10^{-5}\,\text{T}\). Find the size of the force on one electron, in units of \(10^{-16}\,\text{N}\).
A force that does no work
The magnetic force is perpendicular to the velocity at every instant. The power it delivers is \(\vec F\cdot\vec v\), and that is zero for every direction of motion. By the work-energy theorem, the kinetic energy of the charge cannot change. The field can only turn the charge.
\[ dW = \vec F\cdot d\vec r = \vec F\cdot\vec v\,dt \]
Why. in a short time \(dt\) the charge moves \(d\vec r = \vec v\,dt\).
\[ \vec F\cdot\vec v = q\,(\vec v\times\vec B)\cdot\vec v \]
Why. substitute the magnetic force.
\[ (\vec v\times\vec B)\cdot\vec v = 0 \]
Why. a cross product is perpendicular to each of its two vectors, so its dot product with \(\vec v\) vanishes.
\[ dW = 0 \]
Read it. zero work in every interval adds up to zero work on any path, so \(\tfrac12 mv^2\) stays constant.
A magnetic force on a moving charge does no work. It changes the direction of the velocity but never the speed. So the kinetic energy of a charge acted on by a magnetic field alone stays constant.
Two protons sit at rest, one in an electric field and one in a magnetic field. Both are released; then both are fired to the right at the same speed. What happens to the speeds?
In the figure above the speeds are in the figure's own units. The electric force pushes the resting proton up at once, and its speed grows steadily. Fired to the right, the proton bends upward along a curve and still speeds up, reaching \(1.66\,\text{m/s}\) from \(0.90\,\text{m/s}\). In the magnetic field the resting proton stays put. Fired to the right, it circles with its speed fixed at \(0.90\,\text{m/s}\).
"No work" does not mean "no effect". A magnetic force can change the momentum of a charge completely, turning it round to go the other way. Momentum is a vector and changes with direction. Kinetic energy depends only on the speed, so it stays the same.
A charged particle moves through a region with a magnetic field and no electric field. Which of these can the field change?
A student wants to speed up a proton using only a strong magnet held still. Explain why this cannot work, whatever the shape of the magnet.
A still magnet gives only a magnetic field. The magnetic force \(q\,\vec v\times\vec B\) is perpendicular to the proton's velocity at every instant. Its power \(\vec F\cdot\vec v\) is therefore zero, and it does no work on the proton. By the work-energy theorem the kinetic energy, and so the speed, cannot change. The magnet can only bend the path. Speeding the proton up needs an electric field.
An electric motor does work with magnets, which seems to break this rule. The force of the field acts on the moving charges inside a wire, and the wire passes the push on to its metal lattice. The energy comes from the battery that drives the current, not from the magnetic field. The chapter Magnetic Force on Currents and the Magnetic Dipole meets this force on a wire.
Now return to the television. The electrons of the beam were moving, so the magnet's field pushed them sideways and they landed on the wrong spots of the screen. With the set off there was no beam and nothing moving, so the magnet had nothing to push. The picture could move but not brighten, because the magnetic force never changed the electrons' speed or energy.
- Using \(qvB\) when the velocity is at an angle to the field.Only the part of \(\vec v\) across \(\vec B\) counts: \(F = qvB\sin\theta\).
- Applying the right-hand rule to an electron and forgetting to reverse it.The rule gives \(\vec v\times\vec B\). For a negative charge the force points the opposite way.
- Thinking a magnetic field can speed a charge up or slow it down.The force is always perpendicular to \(\vec v\), so it does no work. Only an electric field changes the speed.