Rods and loops in magnetic fields

A rod that becomes a generator

Pull a metal rod along two rails in a magnetic field, and you feel a drag. Let go, and the rod soon stops. The moving rod is a small generator, and the rest of the circuit is its load. Whatever energy the load receives is paid for by the work that you do on the rod.

All the problems in this lesson, about rods, rails, frames and capacitors, come from this one idea. We solve each of them in the same four steps.

The four steps

1. The emf. A conductor of length $l$ that moves at speed $v$ across a field $B$ has $e = Blv$. The field, the length and the velocity are at right angles to each other. If they are not, use the part of each that is.
2. The current. The current is the emf divided by the resistance of the whole circuit: $I = \dfrac{e}{R}$.
3. The force. The current in the rod feels the force $F = BIl$. By Lenz's law, its direction always opposes the motion.
4. The motion. Put this force into Newton's second law. Or use energy: the work that the force does is turned into heat.

Put steps 1 to 3 together, and the opposing force on a rod with a resistor in the circuit is

It grows with the speed. Double the speed and the force doubles. The power that you must supply, $Fv = \dfrac{B^2l^2v^2}{R}$, grows as the square of the speed. It is equal to the heating power $I^2R$ in the circuit.

Example: pulling a rod along rails

A rod $25$ cm long is pulled at a steady $4.0$ m/s along rails in a field of $0.20$ T, at right angles to it. The rails are joined by a resistor, and the whole circuit has a resistance of $0.40\ \Omega$. What force keeps the rod moving, and what power does it take?

1. The emf. $e = Blv = 0.20\ \text{T} \times 0.25\ \text{m} \times 4.0\ \text{m/s} = 0.20\ \text{V}$.
2. The current. $I = \dfrac{e}{R} = \dfrac{0.20\ \text{V}}{0.40\ \Omega} = 0.50\ \text{A}$.
3. The force. $F = BIl = 0.20\ \text{T} \times 0.50\ \text{A} \times 0.25\ \text{m} = 0.025\ \text{N}$. The steady pull is $0.025$ N along the motion.
4. The power. $Fv = 0.025\ \text{N} \times 4.0\ \text{m/s} = 0.10\ \text{W}$. The heat in the circuit is $I^2R = (0.50\ \text{A})^2 \times 0.40\ \Omega = 0.10\ \text{W}$. The two agree.

A rod falling on vertical rails

Now stand the rails upright. A rod slides down them with the field at right angles to the plane of the rails, and the rails are joined at the top. The rod falls under its weight $mg$. As it speeds up, the emf and the current grow, and so does the upward force $\dfrac{B^2l^2v}{R}$.

Newton's second law, with down as positive, is

At the start, $v = 0$ and the rod falls freely. Later, the magnetic force catches up with the weight. Then the acceleration is zero and the speed stops growing. This is the terminal speed. We want $v$, so from $mg = \dfrac{B^2l^2v}{R}$:

The approach to this speed is exponential, $v = v_t\left(1 - e^{-t/\tau}\right)$ with $\tau = \dfrac{mR}{B^2l^2}$. At the terminal speed, all the potential energy that the rod loses each second becomes heat: $mgv_t = I^2R$.

A rod of mass $40$ g falls on rails $36$ cm apart, in a field of $0.50$ T. The whole circuit has a resistance of $0.25\ \Omega$. Its terminal speed is

The time constant is $\tau = \dfrac{0.040\ \text{kg} \times 0.25\ \Omega}{(0.50\ \text{T})^2 \times (0.36\ \text{m})^2} = 0.31$ s. After about one and a half seconds, the rod falls at a steady speed.

The current then is $I = \dfrac{mg}{Bl} = \dfrac{0.392\ \text{N}}{0.50\ \text{T} \times 0.36\ \text{m}} = 2.18\ \text{A}$, found by balancing $BIl$ against $mg$.

A frame crossing a field

A square frame of side $a$ and resistance $R$ is pushed at a steady speed $v$. It goes through a region where there is a field $B$ into the page, and out the other side. The region is wider than the frame. Let us follow the flux.

• Entering. The front edge is in the field, and the rest is outside. The area in the field grows, so the flux grows. Only the front edge cuts the field lines, so $e = Bav$ and $I = \dfrac{Bav}{R}$. By Lenz's law, the current opposes the growth of the flux into the page, so it is anticlockwise. The force on the front edge is $F = BIa = \dfrac{B^2a^2v}{R}$, and it opposes the motion. The forces on the two side edges are equal and opposite, so they cancel.
• Fully inside. The flux is constant, so there is no emf, no current and no force. The frame glides at constant speed.
• Leaving. The flux falls. Now the back edge is in the field, and it cuts the field lines. The emf and the current are as large as before, but the current is clockwise. The force on the back edge still opposes the motion. The frame is pulled back into the field.

So the force is against the motion both when the frame enters and when it leaves. Each stage lasts a time $a/v$. The heat made in one stage is $I^2R \times \dfrac{a}{v} = \dfrac{B^2a^3v}{R}$, and the whole crossing makes twice as much. This equals the work that you do against the force, $F \times a$ for each stage.

The charge. The charge that flows while the frame enters is $q = I \times \dfrac{a}{v} = \dfrac{Ba^2}{R}$. It does not depend on the speed. In general, $q = \dfrac{\Delta\Phi}{R}$, the change in flux divided by the resistance.

Example: a frame entering a field

A square frame has sides of $10$ cm and a resistance of $0.050\ \Omega$. It enters a field of $0.60$ T at $0.80$ m/s. Find the current, the force, and the heat made while it enters.

1. The emf. $e = Bav = 0.60\ \text{T} \times 0.10\ \text{m} \times 0.80\ \text{m/s} = 0.048\ \text{V}$.
2. The current. $I = \dfrac{0.048\ \text{V}}{0.050\ \Omega} = 0.96\ \text{A}$.
3. The force. $F = BIa = 0.60\ \text{T} \times 0.96\ \text{A} \times 0.10\ \text{m} = 0.0576\ \text{N}$.
4. The time. $t = \dfrac{a}{v} = \dfrac{0.10\ \text{m}}{0.80\ \text{m/s}} = 0.125\ \text{s}$.
5. The heat. $I^2Rt = (0.96\ \text{A})^2 \times 0.050\ \Omega \times 0.125\ \text{s} = 5.76\ \text{mJ}$. Check: $F \times a = 0.0576\ \text{N} \times 0.10\ \text{m} = 5.76\ \text{mJ}$.
6. The charge. $q = It = 0.96\ \text{A} \times 0.125\ \text{s} = 0.12\ \text{C}$. Check: $\dfrac{Ba^2}{R} = \dfrac{0.60\ \text{T} \times 0.010\ \text{m}^2}{0.050\ \Omega} = 0.12\ \text{C}$.

Double the speed. The current doubles, but each stage takes half as long. The heat doubles, not quadruples. The charge that flows does not change at all.

A capacitor in place of the resistor

Let the rod fall on vertical rails that end in a capacitor $C$, with no resistor. The emf of the rod is $Blv$. The capacitor charges so that its voltage equals this emf: $q = CBlv$. The current is the rate at which the charge grows:

The magnetic force on the rod is $BIl = CB^2l^2a$. Put it into Newton's second law:

This acceleration is constant. The rod falls with a uniform acceleration that is smaller than $g$. The capacitor behaves like an extra mass $CB^2l^2$ that must be accelerated too. There is no terminal speed, because no energy is turned into heat. The potential energy lost goes into the kinetic energy of the rod and into the energy stored in the capacitor.

For a rod of $80$ g, rails $0.20$ m apart, a field of $0.70$ T and a capacitor of $2.0$ F, the extra mass is $CB^2l^2 = 2.0 \times (0.70)^2 \times (0.20)^2 = 0.0392$ kg. Then $a = \dfrac{0.080 \times 9.8}{0.080 + 0.0392} = 6.58\ \text{m/s}^2$.

A rod joined to a battery

Now put a battery of emf $\mathcal{E}$ in the circuit in place of the resistor, so that the rod is on rails joined to a battery. The current $I = \mathcal{E}/R$ gives a force $BIl$ and the rod starts to move. As it speeds up, it makes its own emf $Blv$, which opposes the battery. This is a back emf. The current falls:

The rod speeds up until the current is zero, at $v = \dfrac{\mathcal{E}}{Bl}$. This is a simple motor. A motor is a generator running backwards, and a generator is a motor driven by an outside force.

Practise with new numbers

Try it in Python

The program lets a rod fall on rails in small steps of time. It prints the speed as it nears the terminal speed, and it adds up where the energy goes.

Common mistakes

• Leaving the opposing force out. A moving rod in a circuit is always slowed by the force $BIl$, even when nothing else touches it.
• Squaring only one of $B$ and $l$. The force is $B^2l^2v/R$, and the terminal speed is $mgR/(B^2l^2)$.
• Using the resistance of the resistor alone. Use the resistance of the whole circuit, the rod and the rails too, when they are given.
• Thinking a frame is slowed all the time it is in the field. The force acts only while the flux changes: while the frame enters and while it leaves.
• Thinking the charge depends on the speed. The charge is $\Delta\Phi/R$. A faster crossing gives a larger current for a shorter time.
• Looking for a terminal speed with a capacitor. There is no heat, and the acceleration is constant.
• Forgetting the back emf of the rod in a battery circuit. The current is $(\mathcal{E} - Blv)/R$, not $\mathcal{E}/R$.

Remember these

• Four steps: $e = Blv$, $I = e/R$, $F = BIl$ (it opposes the motion), then Newton's law or energy.
• A rod with a resistor: $F = \dfrac{B^2l^2v}{R}$ and power $= \dfrac{B^2l^2v^2}{R}$.
• Terminal speed of a falling rod: $v_t = \dfrac{mgR}{B^2l^2}$, and the time constant is $\tau = \dfrac{mR}{B^2l^2}$.
• A frame: emf and force only while the flux changes. Heat for the whole crossing: $\dfrac{2B^2a^3v}{R}$.
• The charge that flows is $q = \Delta\Phi/R$, whatever the speed.
• A capacitor adds a mass $CB^2l^2$ and the acceleration is constant: $a = \dfrac{mg}{m + CB^2l^2}$.
• A rod on a battery has a back emf. It stops speeding up at $v = \mathcal{E}/Bl$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A rod is pulled along rails at a steady speed. The speed is doubled. What happens to the pulling force, and to the power that the person supplies?

2. A rod falls on vertical rails. A stronger magnet doubles the field. What happens to the terminal speed?

3. Why does a square frame that is fully inside a uniform field feel no force from the field, even though the field is there?

4. A square frame of side $20$ cm and resistance $0.10\ \Omega$ enters a field of $0.25$ T at $3.0$ m/s. What is the current while it enters?

5. Why does a rod that falls on rails joined by a capacitor have a constant acceleration, and why is it less than $g$?