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Chapter 22: MagnetismContent check — College Physics, Chapter 22: Magnetism

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 22: Magnetism

48 problems · 13 figures

22-1 §22.4 source ↗

What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in Figure 22.47?

figure a shows magnetic field line direction symbols with solid circles labeled B out; a velocity vector points down; figure b shows B vectors pointing right and v vector pointing up; figure c shows B in and v to the right; figure d shows B vector pointing right and v vector pointing left; figure e shows B vectors up and v vector into the page; figure f shows B vectors pointing left and v vectors out of the page — Figure 22.47 in the book ↗

Answer

(a) Left (West)

(b) Into the page

(c) Up (North)

(d) No force

(e) Right (East)

(f) Down (South)

22-3 §22.4 source ↗

What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in Figure 22.48, assuming it moves perpendicular to

B ?

Figure a shows the force vector pointing up and B out of the page. Figure b shows the F vector pointing up and the B vector pointing to the right. Figure c shows the F vector pointing to the left and the B vector pointing into the page. — Figure 22.48 in the book ↗

Answer

(a) East (right)

(b) Into page

(c) South (down)

22-5 §22.4 source ↗

What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming

B

is perpendicular to

v

?

Figure a shows a force vector pointing toward the left and a velocity vector pointing up. Figure b shows the force vector pointing into the page and the velocity vector pointing down. Figure c shows the force vector pointing up and the velocity vector pointing to the left. — Figure 22.49 in the book ↗

Answer

(a) Into page

(b) West (left)

(c) Out of page

22-7 §22.4 source ↗

What is the maximum magnitude of the force on an aluminum rod with a

0.100 -μC

charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force?

Answer

7 . 50 \times 10^{- 7} N

perpendicular to both the magnetic field lines and the velocity

22-9 §22.4 source ↗

(a) A cosmic ray proton moving toward the Earth at

5.00 \times 10^{7} m/s

experiences a magnetic force of

1 . 70 \times 10^{- 16} N

. What is the strength of the magnetic field if there is a

45º

angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.

Answer

(a) $3 . 01 \times 10^{- 5} T$

(b) This is slightly less then the magnetic field strength of $5 \times 10^{- 5} T$ at the surface of the Earth, so it is consistent.

22-11 §22.4 source ↗

(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than

1 . 00 \times 10^{- 12} N

. What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

Answer

(a) $6 . 67 \times 10^{- 10} C$ (taking the Earth’s field to be $5 . 00 \times 10^{- 5} T$ )

(b) Less than typical static, therefore difficult

22-12 §22.5 source ↗

If you need additional support for these problems, see More Applications of Magnetism.

A cosmic ray electron moves at $7 . 50 \times 10^{6} m/s$ perpendicular to the Earth’s magnetic field at an altitude where field strength is $1 . 00 \times 10^{- 5} T$ . What is the radius of the circular path the electron follows?

Answer

4.27 m

22-14 §22.5 source ↗

If you need additional support for these problems, see More Applications of Magnetism.

(a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at $5 . 00 \times 10^{7} m/s$ in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility?

Answer

(a) 0.261 T

(b) This strength is definitely obtainable with today’s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets.

22-16 §22.5 source ↗

If you need additional support for these problems, see More Applications of Magnetism.

What radius circular path does an electron travel if it moves at the same speed and in the same magnetic field as the proton in Exercise 22.13?

Answer

4 . 36 \times 10^{- 4} m

22-18 §22.5 source ↗

If you need additional support for these problems, see More Applications of Magnetism.

An electron in a TV CRT moves with a speed of $6 . 00 \times 10^{7} m/s$ , in a direction perpendicular to the Earth’s field, which has a strength of $5 . 00 \times 10^{- 5} T$ . (a) What strength electric field must be applied perpendicular to the Earth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)

Answer

(a) 3.00 kV/m

(b) 30.0 V

22-20 §22.5 source ↗

If you need additional support for these problems, see More Applications of Magnetism.

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial ice. (The relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) The ratio of the masses of these two ions is 16 to 18, the mass of oxygen-16 is $2 . 66 \times 10^{- 26} kg,$ and they are singly charged and travel at $5 . 00 \times 10^{6} m/s$ in a 1.20-T magnetic field. What is the separation between their paths when they hit a target after traversing a semicircle?

Answer

0.173 m

22-22 §22.6 source ↗

A large water main is 2.50 m in diameter and the average water velocity is 6.00 m/s. Find the Hall voltage produced if the pipe runs perpendicular to the Earth’s

5 . 00 \times 10^{- 5} -T

field.

Answer

7 . 50 \times 10^{- 4} V

22-24 §22.6 source ↗

(a) What is the speed of a supersonic aircraft with a 17.0-m wingspan, if it experiences a 1.60-V Hall voltage between its wing tips when in level flight over the north magnetic pole, where the Earth’s field strength is

8 . 00 \times 10^{- 5} T?

(b) Explain why very little current flows as a result of this Hall voltage.

Answer

(a) 1.18 × 10 ³ m/s

(b) Once established, the Hall emf pushes charges one direction and the magnetic force acts in the opposite direction resulting in no net force on the charges. Therefore, no current flows in the direction of the Hall emf. This is the same as in a current-carrying conductor—current does not flow in the direction of the Hall emf.

22-26 §22.6 source ↗

Calculate the Hall voltage induced on a patient’s heart while being scanned by an MRI unit. Approximate the conducting path on the heart wall by a wire 7.50 cm long that moves at 10.0 cm/s perpendicular to a 1.50-T magnetic field.

Answer

11.3 mV

22-28 §22.6 source ↗

Using information in Example 20.6, what would the Hall voltage be if a 2.00-T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.0-A current?

Example 20.6 Calculating Resistance: Hot-Filament Resistance in the book ↗

Answer

1. 16 μV

22-30 §22.6 source ↗

A patient with a pacemaker is mistakenly being scanned for an MRI image. A 10.0-cm-long section of pacemaker wire moves at a speed of 10.0 cm/s perpendicular to the MRI unit’s magnetic field and a 20.0-mV Hall voltage is induced. What is the magnetic field strength?

Answer

2.00 T

22-31 §22.7 source ↗

What is the direction of the magnetic force on the current in each of the six cases in Figure 22.50?

Figure a shows the magnetic field B out of the page and the current I downward. Figure b shows B toward the right and I upward. Figure c shows B into the page and I toward the right. Figure d shows B toward the right and I toward the left. Figure e shows B upward and I into the page. Figure f shows B toward the left and I out of the page. — Figure 22.50 in the book ↗

Answer

(a) west (left)

(b) into page

(c) north (up)

(d) no force

(e) east (right)

(f) south (down)

22-33 §22.7 source ↗

What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases in Figure 22.52, assuming

B

is perpendicular to

I

?

Figure a show the current I vector pointing upward and the force F vector pointing left. Figure b shows the current vector pointing down and F directed into the page. Figure c shows the current pointing left and force pointing up. — Figure 22.52 in the book ↗

Answer

(a) into page

(b) west (left)

(c) out of page

22-35 §22.7 source ↗

(a) A DC power line for a light-rail system carries 1000 A at an angle of

30.0º

to the Earth’s

5.00 \times 10^{- 5} -T

field. What is the force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any.

Answer

(a) 2.50 N

(b) This is about half a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns.

22-37 §22.7 source ↗

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?

Answer

1.80 T

22-39 §22.7 source ↗

(a) What is the angle between a wire carrying an 8.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N? (b) What is the force on the wire if it is rotated to make an angle of

90º

with the field?

Answer

(a) $30º$

(b) 4.80 N

22-41 §22.8 source ↗

(a) By how many percent is the torque of a motor decreased if its permanent magnets lose 5.0% of their strength? (b) How many percent would the current need to be increased to return the torque to original values?

Answer

(a) $τ$ decreases by 5.00% if B decreases by 5.00%

(b) 5.26% increase

22-43 §22.8 source ↗

Find the current through a loop needed to create a maximum torque of

9 . 00 N \cdot m.

The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-T magnetic field.

Answer

10.0 A

22-45 §22.8 source ↗

Since the equation for torque on a current-carrying loop is

τ = NIAB sin θ

, the units of

N \cdot m

must equal units of

A \cdot m^{2} T

. Verify this.

Answer

A \cdot m^{2} \cdot T = A \cdot m^{2} \frac{N}{A \cdot m} = N \cdot m

.

22-47 §22.8 source ↗

A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop

0 . 650 \times 10^{- 15} m

in radius with a current of

1 . 05 \times 10^{4} A

(no kidding). Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)

Answer

3 . 48 \times 10^{- 26} N \cdot m

22-49 §22.8 source ↗

Repeat Exercise 22.41, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where the Earth’s field is north, but at an angle

45 . 0º

below the horizontal and with a strength of

6. 00 \times 10^{- 5} T

.

Answer

(a) $0.666 N \cdot m$ west

(b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be alternated to make the loop rotate (otherwise it would oscillate).

22-50 §22.10 source ↗

(a) The hot and neutral wires supplying DC power to a light-rail commuter train carry 800 A and are separated by 75.0 cm. What is the magnitude and direction of the force between 50.0 m of these wires? (b) Discuss the practical consequences of this force, if any.

Answer

(a) 8.53 N, repulsive

(b) This force is repulsive and therefore there is never a risk that the two wires will touch and short circuit.

22-52 §22.10 source ↗

A 2.50-m segment of wire supplying current to the motor of a submerged submarine carries 1000 A and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is the direction and magnitude of the current in the other wire?

Answer

400 A in the opposite direction

22-54 §22.10 source ↗

An AC appliance cord has its hot and neutral wires separated by 3.00 mm and carries a 5.00-A current. (a) What is the average force per meter between the wires in the cord? (b) What is the maximum force per meter between the wires? (c) Are the forces attractive or repulsive? (d) Do appliance cords need any special design features to compensate for these forces?

Answer

(a) $1 . 67 \times 10^{- 3} N/m$

(b) $3 . 33 \times 10^{- 3} N/m$

(c) Repulsive

(d) No, these are very small forces

22-56 §22.10 source ↗

Find the direction and magnitude of the force that each wire experiences in Figure 22.55(a) by, using vector addition.

Figure a shows the cross sections of three wires that are parallel to each other and arranged in an equilateral triangle. The bottom left wire has current of ten point zero amps into the page. The bottom right wire has a current of twenty point zero amps also into the page. The wire at the top of the triangle has current five point zero amps out of the page. The triangle that the wires make with each other is ten point zero centimeters on each side. Figure b shows four parallel wires arranged in a square that is twenty point zero centimeters on a side. The top two wires have current of ten point zero amps out of the page. The bottom two wires have current of five point zero amps into the page. — Figure 22.55 in the book ↗

Answer

(a) Top wire: $2 . 65 \times 10^{- 4} N/m$ s, $10 . 9º$ to left of up

(b) Lower left wire: $3 . 61 \times 10^{- 4} N/m$ , $13 . 9º$ down from right

(c) Lower right wire: $3 . 46 \times 10^{- 4} N/m$ , $30 . 0º$ down from left

22-58 §22.11 source ↗

Indicate whether the magnetic field created in each of the three situations shown in Figure 22.56 is into or out of the page on the left and right of the current.

Figure a shows current I running from bottom to top. Figure b shows an electron moving with velocity v from bottom to top. Figure c shows current I running from top to bottom. — Figure 22.56 in the book ↗

Answer

(a) right-into page, left-out of page

(b) right-out of page, left-into page

(c) right-out of page, left-into page

22-60 §22.11 source ↗

What are the directions of the currents in the loop and coils shown in Figure 22.58?

Figure a shows the magnetic field into the page in the middle of a loop. Figure b shows the magnetic field within a coil running from left to right. Figure c shows B running from right to left within a coil. — Figure 22.58 in the book ↗

Figure a shows the magnetic field into the page in the middle of a loop. Figure b shows the magnetic field within a coil running from left to right. Figure c shows B running from right to left within a coil. — Figure 22.58 in the book ↗

Answer

(a) clockwise

(b) clockwise as seen from the left

(c) clockwise as seen from the right

22-61 §22.11 source ↗

To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop

0 . 650 \times 10^{- 15} m

in radius carrying

1 . 05 \times 10^{4} A

. What is the field at the center of such a loop?

Answer

1 . 01 \times 10^{13} T

22-63 §22.11 source ↗

Nonnuclear submarines use batteries for power when submerged. (a) Find the magnetic field 50.0 cm from a straight wire carrying 1200 A from the batteries to the drive mechanism of a submarine. (b) What is the field if the wires to and from the drive mechanism are side by side? (c) Discuss the effects this could have for a compass on the submarine that is not shielded.

Answer

(a) $4 . 80 \times 10^{- 4} T$

(b) Zero

(c) If the wires are not paired, the field is about 10 times stronger than Earth’s magnetic field and so could severely disrupt the use of a compass.

22-65 §22.11 source ↗

What current is needed in the solenoid described in Exercise 22.64 to produce a magnetic field

10^{4}

times the Earth’s magnetic field of

5 . 00 \times 10^{- 5} T

?

Answer

39.8 A

22-67 §22.11 source ↗

Measurements affect the system being measured, such as the current loop in Figure 22.53. (a) Estimate the field the loop creates by calculating the field at the center of a circular loop 20.0 cm in diameter carrying 5.00 A. (b) What is the smallest field strength this loop can be used to measure, if its field must alter the measured field by less than 0.0100%?

Diagram showing a rectangular loop of wire, one end of which is within a magnetic field that is present within a circular area. The field B is oriented out of the page. The current I runs in the plane of the page, down the left side of the circuit, toward the right at the bottom of the circuit, and upward on the right side of the circuit. The length of the segment of wire that runs left to right at the bottom of the circuit is twenty centimeters long. — Figure 22.53 in the book ↗

Diagram showing a rectangular loop of wire, one end of which is within a magnetic field that is present within a circular area. The field B is oriented out of the page. The current I runs in the plane of the page, down the left side of the circuit, toward the right at the bottom of the circuit, and upward on the right side of the circuit. The length of the segment of wire that runs left to right at the bottom of the circuit is twenty centimeters long. — Figure 22.53 in the book ↗

Answer

(a) $3 . 14 \times 10^{- 5} T$

(b) 0.314 T

22-69 §22.11 source ↗

Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.55(a), using the rules of vector addition to sum the contributions from each wire.

Figure a shows the cross sections of three wires that are parallel to each other and arranged in an equilateral triangle. The bottom left wire has current of ten point zero amps into the page. The bottom right wire has a current of twenty point zero amps also into the page. The wire at the top of the triangle has current five point zero amps out of the page. The triangle that the wires make with each other is ten point zero centimeters on each side. Figure b shows four parallel wires arranged in a square that is twenty point zero centimeters on a side. The top two wires have current of ten point zero amps out of the page. The bottom two wires have current of five point zero amps into the page. — Figure 22.55 in the book ↗

Answer

7 . 55 \times 10^{- 5} T

,

23.4º

22-71 §22.11 source ↗

What current is needed in the top wire in Figure 22.55(a) to produce a field of zero at the point equidistant from the wires, if the currents in the bottom two wires are both 10.0 A into the page?

Figure a shows the cross sections of three wires that are parallel to each other and arranged in an equilateral triangle. The bottom left wire has current of ten point zero amps into the page. The bottom right wire has a current of twenty point zero amps also into the page. The wire at the top of the triangle has current five point zero amps out of the page. The triangle that the wires make with each other is ten point zero centimeters on each side. Figure b shows four parallel wires arranged in a square that is twenty point zero centimeters on a side. The top two wires have current of ten point zero amps out of the page. The bottom two wires have current of five point zero amps into the page. — Figure 22.55 in the book ↗

Answer

10.0 A

22-73 §22.11 source ↗

Integrated Concepts

(a) A pendulum is set up so that its bob (a thin copper disk) swings between the poles of a permanent magnet as shown in Figure 22.60. What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive $0 . 250 μC$ charge and is released from a height of 30.0 cm above its lowest point? The magnetic field strength is 1.50 T. (b) What is the acceleration of the bob at the bottom of its swing if its mass is 30.0 grams and it is hung from a flexible string? Be certain to include a free-body diagram as part of your analysis.

Diagram showing a pendulum swinging between the poles of a magnet. The magnetic field B runs from the north to the south pole. — Figure 22.60 in the book ↗

Answer

(a) $9 . 09 \times 10^{- 7} N$ upward

(b) $3 . 03 \times 10^{- 5} {m/s}^{2}$

22-75 §22.11 source ↗

Integrated Concepts

Find the radius of curvature of the path of a 25.0-MeV proton moving perpendicularly to the 1.20-T field of a cyclotron.

Answer

60.2 cm

22-77 §22.11 source ↗

Integrated Concepts

(a) Using the values given for an MHD drive in Exercise 22.59, and assuming the force is uniformly applied to the fluid, calculate the pressure created in ${N/m}^{2} .$ (b) Is this a significant fraction of an atmosphere?

Answer

(a) $1 . 02 \times 10^{3} {N/m}^{2}$

(b) Not a significant fraction of an atmosphere

22-79 §22.11 source ↗

Integrated Concepts

A current balance used to define the ampere is designed so that the current through it is constant, as is the distance between wires. Even so, if the wires change length with temperature, the force between them will change. What percent change in force per degree will occur if the wires are copper?

Answer

17 . 0 \times 10^{- 4} %/ºC

22-81 §22.11 source ↗

Integrated Concepts

A cyclotron accelerates charged particles as shown in Figure 22.61. Using the results of the previous problem, calculate the frequency of the accelerating voltage needed for a proton in a 1.20-T field.

Diagram of a cyclotron. — Figure 22.61 in the book ↗

Answer

18.3 MHz

22-83 §22.11 source ↗

Integrated Concepts

(a) What is the direction of the force on a wire carrying a current due east in a location where the Earth’s field is due north? Both are parallel to the ground. (b) Calculate the force per meter if the wire carries 20.0 A and the field strength is $3 . 00 \times 10^{- 5} T$ . (c) What diameter copper wire would have its weight supported by this force? (d) Calculate the resistance per meter and the voltage per meter needed.

Answer

(a) Straight up

(b) $6 . 00 \times 10^{- 4} N/m$

(c) $94.1 μm$

(d)2.47 Ω/m, 49.4 V/m

22-85 §22.11 source ↗

Unreasonable Results

(a) Find the charge on a baseball, thrown at 35.0 m/s perpendicular to the Earth’s $5 . 00 \times 10^{- 5} T$ field, that experiences a 1.00-N magnetic force. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?

Answer

(a) 571 C

(b) Impossible to have such a large separated charge on such a small object.

(c) The 1.00-N force is much too great to be realistic in the Earth’s field.

22-87 §22.11 source ↗

Unreasonable Results

An inventor wants to generate 120-V power by moving a 1.00-m-long wire perpendicular to the Earth’s $5 . 00 \times 10^{- 5} T$ field. (a) Find the speed with which the wire must move. (b) What is unreasonable about this result? (c) Which assumption is responsible?

Answer

(a) $2.40 \times 10^{6} m/s$

(b) The speed is too high to be practical $\leq$ 1% speed of light

(c) The assumption that you could reasonably generate such a voltage with a single wire in the Earth’s field is unreasonable

22-89 §22.11 source ↗

Unreasonable Results

A surveyor 100 m from a long straight 200-kV DC power line suspects that its magnetic field may equal that of the Earth and affect compass readings. (a) Calculate the current in the wire needed to create a $5 . 00 \times 10^{- 5} T$ field at this distance. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?

Answer

(a) 25.0 kA

(b) This current is unreasonably high. It implies a total power delivery in the line of 50.0x10^9 W, which is much too high for standard transmission lines.

(c)100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor’s concerns are not a problem for his magnetic field measurements.

22-92 §22.11 source ↗

Critical Thinking A magnetic field of

2.00 \times 10^{−7} T

in the positive x-direction exists. Three electrons pass through the origin, one at a time, each with a velocity of

5.00 \times 10^{7} m/s

. The first electron passes the origin along the x-axis, the second electron along the y-axis, and the third at an angle of

45 °

with respect to both the y- and z-axes in the yz-plane. (a) What is the force acting on each electron? (b) What is the shape of the path of each electron? (c) When, if ever, will each electron return to the origin?

Answer

(a) $F = q v B \sin θ$
Since $\sin θ =$ $0,$ $F =$ $0$ for the first electron.
For the second and third electrons, $\sin θ = 1$ .
So, $F =$ $(1.60 \times 10^{- 19} C) (5.00 \times 10^{7} m/s) (2.00 \times 10^{- 7} T) =$ $1.60 \times 10^{- 18} N$ .

(b) The first electron moves in a straight line. The other two electrons take a circular path.

(c) The first electron will never return to the origin. The other two will return. $\begin{array}{l} r = \frac{(9.11 \times 10^{- 11}) (5.00 \times 10^{7})}{(1.60 \times 10^{- 19}) (2.00 \times 10^{- 7})} m \\ v = \frac{r}{t} \\ t = \frac{(9.11 \times 10^{- 11})}{(1.60 \times 10^{- 19}) (2.00 \times 10^{- 7})} s = 2.85 \times 10^{15} s \end{array}$

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.