Chapter 9: Review

The chapter problem, solved

The spacecraft orbits at a height of 400 km, so the distance from the centre of the Earth is $r = 6.77 \times 10^{6}\,\text{m}$ . The ion drive accelerates xenon ions, charge $1.60 \times 10^{-19}\,\text{C}$ and mass $2.18 \times 10^{-25}\,\text{kg}$ , through 1.20 kV between grids 1.5 mm apart. The 20 m tether moves at the orbital speed through the Earth's field of flux density $3.0 \times 10^{-5}\,\text{T}$ and drives a current of 0.80 A.

1. The orbit. The gravitational field strength is

g = \frac{GM}{r^{2}} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.77 \times 10^{6})^{2}} = 8.69\,\text{N}\,\text{kg}^{-1}

Gravity provides the centripetal force, so

v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^{6}}} = 7.67 \times 10^{3}\,\text{m}\,\text{s}^{-1}

The period is $T = \dfrac{2\pi r}{v} = 5.55 \times 10^{3}\,\text{s}$ , about 92 minutes. The gravitational potential at the orbit is

V = -\frac{GM}{r} = -5.88 \times 10^{7}\,\text{J}\,\text{kg}^{-1}

2. The ion drive. Between the grids the field is uniform:

E = \frac{V}{d} = \frac{1.20 \times 10^{3}}{1.5 \times 10^{-3}} = 8.0 \times 10^{5}\,\text{V}\,\text{m}^{-1}

The force on one ion is $F = QE = 1.28 \times 10^{-13}\,\text{N}$ . The ion gains kinetic energy

E_{k} = QV = 1.60 \times 10^{-19} \times 1.20 \times 10^{3} = 1.92 \times 10^{-16}\,\text{J}

so its speed is

v_{\text{ion}} = \sqrt{\frac{2E_{k}}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-16}}{2.18 \times 10^{-25}}} = 4.20 \times 10^{4}\,\text{m}\,\text{s}^{-1}

3. The magnetic force on an ion. In the Earth's field, an ion moving at right angles feels

F = BQv = 3.0 \times 10^{-5} \times 1.60 \times 10^{-19} \times 4.20 \times 10^{4} = 2.0 \times 10^{-19}\,\text{N}

This is tiny compared with the force from the thruster, so the Earth's field barely bends the beam. The radius of the path would be $r = \dfrac{mv}{BQ} = 1.9 \times 10^{3}\,\text{m}$ , far larger than the spacecraft.

4. The tether. The induced emf is

\begin{aligned} \varepsilon &= BLv \\ &= 3.0 \times 10^{-5} \times 20 \times 7.67 \times 10^{3} \\ &= 4.6\,\text{V} \end{aligned}

The current is 0.80 A, so the electrical power is

P = \varepsilon I = 4.6 \times 0.80 = 3.7\,\text{W}

The current feels a magnetic force

F = BIL = 3.0 \times 10^{-5} \times 0.80 \times 20 = 4.8 \times 10^{-4}\,\text{N}

that opposes the motion. The mechanical power removed is $Fv = 4.8 \times 10^{-4} \times 7.67 \times 10^{3} = 3.7\,\text{W}$ , which is exactly the electrical power. The two powers agree because energy is conserved.

5. The judgement on the tether. The tether supplies about 3.7 W. That is useful for a small instrument or for trickle-charging a battery, but it is not free energy. The 3.7 W comes from the kinetic energy of the spacecraft, so the orbit slowly loses energy and the spacecraft drifts lower. A mission planner must decide whether the power is worth the loss of orbit. The first prediction was right: the emf is a few volts, not a few millivolts or a few kilovolts.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate. Some questions bring back ideas from earlier chapters.

Find the gravitational field strength at a distance of 1.0e7 m from the centre of the Earth.
Find the orbital speed of a satellite in a circular orbit of radius 1.0e7 m around the Earth.
A planet has a mass of 6.0e23 kg and a radius of 3.0e6 m. Find the escape speed from its surface.
A 1500 kg spacecraft moves from a point where the gravitational potential is -6.0e7 J/kg to a point where it is -4.5e7 J/kg. Find the energy needed.
Two point charges, each +1.0 microcoulomb, are 0.10 m apart in air. Find the force between them.
Two parallel plates are 5.0 mm apart with a potential difference of 300 V. Find the electric field strength and the force on an electron between them.
An electron is accelerated from rest through a potential difference of 150 V. Find its speed. The mass of an electron is 9.11e-31 kg.
A straight wire of length 0.40 m carries a current of 3.0 A at right angles to a field of flux density 0.25 T. Find the force on the wire.
A proton moves at 4.0e6 m/s at right angles to a field of flux density 0.20 T. Find the radius of its circular path. The mass of a proton is 1.67e-27 kg.
A coil of 300 turns and area 5.0e-3 m^2 is at right angles to a field of flux density 0.15 T. Find the flux linkage.
A coil of 120 turns has an area of 0.025 m^2. The flux density through it rises steadily from 0.10 T to 0.50 T in 0.30 s. Find the induced emf.
A rod of length 0.15 m moves at 8.0 m/s at right angles to a field of 0.40 T. It completes a circuit of resistance 1.2 ohms. Find the emf and the power dissipated.
A transformer has 500 turns on the primary coil and 25 turns on the secondary coil. The primary is connected to a 230 V alternating supply. Find the secondary voltage.
A charged particle is accelerated from rest by a uniform electric field and then enters a magnetic field at right angles. State which equations you would use to find its speed and the radius of its circular path, and why.
A satellite of mass 800 kg orbits the Earth at a radius of 7.0e6 m. Find the gravitational force on it and its centripetal acceleration.
A photon has a wavelength of 500 nm. Find its energy in joules.

Answers and concise solutions

1. The field strength

Answer: $g = \dfrac{GM}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(1.0 \times 10^{7})^{2}} = 3.98\,\text{N}\,\text{kg}^{-1}$ .

2. The orbital speed

Answer: $v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{3.98 \times 10^{14}}{1.0 \times 10^{7}}} = 6.3 \times 10^{3}\,\text{m}\,\text{s}^{-1}$ .

3. The escape speed

Answer: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{8.0 \times 10^{13}}{3.0 \times 10^{6}}} = 5.2 \times 10^{3}\,\text{m}\,\text{s}^{-1}$ .

4. The energy needed

Answer: $\Delta V = 1.5 \times 10^{7}\,\text{J}\,\text{kg}^{-1}$ , so $E = m\Delta V = 1500 \times 1.5 \times 10^{7} = 2.25 \times 10^{10}\,\text{J}$ .

5. The force

Answer: $F = \dfrac{kQq}{r^{2}} = \dfrac{8.99 \times 10^{9} \times 1.0 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.10^{2}} = 0.90\,\text{N}$ , repulsive.

6. The field strength and the force

Answer: $E = \dfrac{V}{d} = \dfrac{300}{5.0 \times 10^{-3}} = 6.0 \times 10^{4}\,\text{V}\,\text{m}^{-1}$ , and $F = QE = 9.6 \times 10^{-15}\,\text{N}$ .

7. The speed

Answer: The energy gained is $E_{k} = eV = 2.4 \times 10^{-17}\,\text{J}$ , so $v = \sqrt{\dfrac{2E_{k}}{m}} = 7.3 \times 10^{6}\,\text{m}\,\text{s}^{-1}$ .

8. The force

Answer: $F = BIL = 0.25 \times 3.0 \times 0.40 = 0.30\,\text{N}$ .

9. The radius

Answer: $r = \dfrac{mv}{BQ} = \dfrac{1.67 \times 10^{-27} \times 4.0 \times 10^{6}}{0.20 \times 1.60 \times 10^{-19}} = 0.21\,\text{m}$ .

10. The flux linkage

Answer: $N\Phi = NBA = 300 \times 0.15 \times 5.0 \times 10^{-3} = 0.23\,\text{Wb}$ .

11. The induced emf

Answer: The change in flux linkage is $120 \times 0.025 \times 0.40 = 1.2\,\text{Wb}$ , so $\varepsilon = \dfrac{1.2}{0.30} = 4.0\,\text{V}$ .

12. The emf and the power

Answer: $\varepsilon = BLv = 0.40 \times 0.15 \times 8.0 = 0.48\,\text{V}$ . Then $P = \dfrac{\varepsilon^{2}}{R} = \dfrac{0.48^{2}}{1.2} = 0.19\,\text{W}$ .

13. The secondary voltage

Answer: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ , so $V_s = 230 \times \dfrac{25}{500} = 11.5\,\text{V}$ .

14. A charged particle is accelerated from rest by a uniform electric field and then enters a magnetic field at right angles. State which equations you would use to find its speed and the radius of its circular path, and why.

First find the kinetic energy from the electric field: $E_{k} = QV$ .
Then find the speed from $\tfrac{1}{2}mv^{2} = E_{k}$ .
In the magnetic field the force is $F = BQv$ , at right angles to the velocity.
Set $BQv = \dfrac{mv^{2}}{r}$ and rearrange to $r = \dfrac{mv}{BQ}$ .

15. The force and the acceleration

Answer: The force is $F = \dfrac{GMm}{r^{2}} = 6.5 \times 10^{3}\,\text{N}$ . The centripetal acceleration is $a = \dfrac{v^{2}}{r} = \dfrac{GM}{r^{2}} = 8.1\,\text{m}\,\text{s}^{-2}$ , which is the gravitational field strength at that radius.

16. The photon energy

Answer: $E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{500 \times 10^{-9}} = 4.0 \times 10^{-19}\,\text{J}$ , from Section 8.1.

What you can now do

describe gravitational, electric and magnetic fields and their field lines;
use Newton's law of gravitation and Coulomb's law;
define field strength and potential for both gravitational and electric fields, and use the equations for each;
relate field strength to the potential gradient;
find the orbit speed, the period and the escape speed;
compare gravitational and electric fields;
describe the motion of a charged particle in a uniform electric field;
define the tesla, and use $F = BIL$ and $F = BQv$ ;
use $r = \dfrac{mv}{BQ}$ and $T = \dfrac{2\pi m}{BQ}$ for a particle in a magnetic field;
define magnetic flux and flux linkage;
use Faraday's law and Lenz's law, including the conservation of energy;
find the emf induced in a moving conductor and in a rotating coil;
explain how a search coil and a transformer use induction.

Your relevance reflection

Return to the prompt from the chapter opener: think of a time when a magnet moved near a wire, or a wire moved near a magnet, such as a bicycle dynamo, a wireless phone charger, a security tag, a tram, or a wind turbine. Write one more example now that you have finished the chapter. Has the chapter changed how you think about what a moving magnet does to a wire?