---
title: "Magnetism and Magnetic Fields"
book: "High School Physics"
subject: physics
language: en
chapter: 15
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields
---

# Chapter 15 — Magnetism and Magnetic Fields

Sailors steered by magnetized needles for a thousand years before anyone knew why they point north. The answer came in 1820, when a current made a compass twitch: magnetism is made by moving charges. This chapter maps the [magnetic field](#def-g11-magnetic-fields-bfield) — of [magnets](#def-g11-magnetic-fields-magnet), of the Earth, of currents — then closes the loop: the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) pushes back on currents, which is how every electric motor turns.

## 15.1 Magnets and the magnetic field

**Definition 15.1 (Magnet and poles).**

A *magnet* attracts iron and, hung by a thread, turns to face north. Its action concentrates at two *magnetic poles*, *north* (the end that seeks geographic north) and *south*: like poles repel, unlike attract. Sawing a magnet in half yields two complete magnets — no experiment has ever isolated a pole.

**Definition 15.2 (Magnetic field).**

A [magnet](#def-g11-magnetic-fields-magnet) modifies the space around it ([Chapter 14](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#ch-g11-electric-gravitational-fields)): it creates everywhere a *magnetic field* $\vect B$. A compass reads its direction (the needle aligns with $\vect B$, north end forward); a *Hall probe* measures its magnitude, in *teslas* ($\mathrm{T}$).

**Definition 15.3 (Magnetic field lines).**

*Magnetic field lines* are curves tangent to $\vect B$, crowding where the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is strong. Outside a [magnet](#def-g11-magnetic-fields-magnet) they run from north pole to south; unlike [electric field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-efield) lines they have no ends: each closes through the [magnet](#def-g11-magnetic-fields-magnet)’s body.

![Field lines of a bar magnet: out of N, into S, closed through the body. A compass placed anywhere settles along the line through it.](https://one-course.com/images/onecourse/chapters/physics-2/g11-magnetic-fields/fig-d47fec6b5b16.svg)

*[Field lines](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-lines) of a bar [magnet](#def-g11-magnetic-fields-magnet): out of N, into S, closed through the body. A compass placed anywhere settles along the line through it.*

## 15.2 The Earth is a magnet

**Definition 15.4 (The Earth’s magnetic field).**

Currents in the Earth’s molten iron core make the planet a giant [magnet](#def-g11-magnetic-fields-magnet): at the surface $B \approx 50\,\text{µ}\mathrm{T}$, and a free needle aligns with the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field)’s horizontal component — this is the compass. The geomagnetic poles are not the geographic ones; the angle between compass north and true north is the *declination*, corrected on every bearing.

**Example 15.5 (Correcting a bearing).**

Where the [declination](#def-g11-magnetic-fields-earth) is $7{}^{\circ}$ west, a ship steering “compass north” for $5.0\,\mathrm{km}$ drifts $5000 \times \sin7{}^{\circ} \approx
6.1 \times 10^{2}\,\mathrm{m}$ west of its track: to sail true north, steer $7{}^{\circ}$ *east* of the needle.

## 15.3 The magnetic field of a current

**Remark 15.6 (Oersted, 1820).**

During a lecture in 1820, Hans Christian Oersted saw a compass needle swing when he closed a circuit nearby: electricity and magnetism were one subject — *currents create [magnetic fields](#def-g11-magnetic-fields-bfield)*.

**Proposition 15.7 (Field of a straight wire).**

A long straight wire carrying a current $I$ creates a [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) whose lines are circles centered on the wire, in planes perpendicular to it. At distance $d$ from the wire,

$$
B = \frac{\mu_0\, I}{2\pi d},
\qquad \mu_0 = 4\pi\times 10^{-7}\ \mathrm{T}\,\mathrm{m}/\mathrm{A} \approx 1.26 \times 10^{-6}\,\mathrm{T}\,\mathrm{m}/\mathrm{A}.
$$

**Proof.** *Admitted at this level.* ∎

![Circular field lines around a straight wire, seen head-on (: current out of the page; : into it). Grip the wire with the right hand, thumb along the current: the fingers curl the way B turns.](https://one-course.com/images/onecourse/chapters/physics-2/g11-magnetic-fields/fig-7cf5154146fc.svg)

![Circular field lines around a straight wire, seen head-on (: current out of the page; : into it). Grip the wire with the right hand, thumb along the current: the fingers curl the way B turns.](https://one-course.com/images/onecourse/chapters/physics-2/g11-magnetic-fields/fig-9e8ac8bbe8e1.svg)

*Circular [field lines](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-lines) around a straight wire, seen head-on ($\odot$: current out of the page; $\otimes$: into it). Grip the wire with the right hand, thumb along the current: the fingers curl the way $\vect B$ turns.*

**Example 15.8 (A wire is a weak magnet).**

At $d = 1.0\,\mathrm{cm}$ from a wire carrying $I = 15\,\mathrm{A}$: $B = 2 \times 10^{-7} \times 15 / 0.010 = 0.30\,\mathrm{mT}$ — six Earth [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field), from a current that would trip a household breaker. Strong [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) need a better geometry.

**Definition 15.9 (Coil and solenoid).**

Winding the wire concentrates its [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field). A *flat coil* ($N$ turns in a disc) behaves like a [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) [magnet](#def-g11-magnetic-fields-magnet), one face north, the other south; a *solenoid* — a long cylindrical winding, $n$ turns per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) — like a bar [magnet](#def-g11-magnetic-fields-magnet).

**Proposition 15.10 (Field inside a solenoid).**

Inside a long [solenoid](#def-g11-magnetic-fields-solenoid), away from the ends, the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is *uniform*: parallel to the axis and the same at every interior point, of magnitude $B = \mu_0\, n\, I$, independent of the bore’s width. Outside, the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is weak.

**Proof.** *Admitted at this level.* ∎

**Remark 15.11.**

Both formulas are derived honestly in the Year 1 volume, from the law relating a [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field)’s circulation to the current it encircles. Note what matters: $n$, turns *per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit)* — wind tighter, not longer.

**Method 15.12 (The right hand, twice).**

- *Wire* : thumb = current; curled fingers = turning of the lines.
- *[Coil](#def-g11-magnetic-fields-solenoid), [solenoid](#def-g11-magnetic-fields-solenoid)* : curled fingers = current in the turns; thumb = $\vect B$ inside, pointing at the north face.

![A solenoid cut lengthwise (/: the turns crossing the page): inside, a uniform B = _0 n I along the axis; outside, a weak, spread-out return field (dashed).](https://one-course.com/images/onecourse/chapters/physics-2/g11-magnetic-fields/fig-006e5fb48b1e.svg)

*A [solenoid](#def-g11-magnetic-fields-solenoid) cut lengthwise ($\odot$/$\otimes$: the turns crossing the page): inside, a uniform $B = \mu_0 n I$ along the axis; outside, a weak, spread-out return [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) (dashed).*

**Example 15.13 (Solenoid numbers).**

At $10$ turns per centimetre ($n = 1000\,\mathrm{m}^{-1}$) and $I = 5.0\,\mathrm{A}$: $B = 1.26 \times 10^{-6} \times 1000 \times 5.0 \approx
6.3\,\mathrm{mT}$ — a hundred Earth [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field), adjustable by a knob.

## 15.4 Electromagnets

**Definition 15.14 (Electromagnet).**

An *electromagnet* is a [solenoid](#def-g11-magnetic-fields-solenoid) wound on a soft iron core: the iron magnetizes in the [coil](#def-g11-magnetic-fields-solenoid)’s [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) and multiplies it, typically by $100$ to $1000$. Unlike a permanent [magnet](#def-g11-magnetic-fields-magnet), it switches off with the current.

**Remark 15.15 (Three machines).**

The scrapyard crane lifts a car with an [electromagnet](#def-g11-magnetic-fields-electromagnet) and — the whole point — *drops* it by opening a switch. A relay’s small [electromagnet](#def-g11-magnetic-fields-electromagnet) pulls a contact closed: a weak current switches a strong one ([Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power)). An MRI scanner holds a uniform $1.5\,\mathrm{T}$ around the patient with a superconducting [solenoid](#def-g11-magnetic-fields-solenoid).

Orders of magnitude, from the galaxy to a dead star:

| interstellar space | $\sim 10^{-10}\,\mathrm{T}$ |
| --- | --- |
| the Earth’s surface [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) | $5 \times 10^{-5}\,\mathrm{T}$ |
| a fridge [magnet](#def-g11-magnetic-fields-magnet) | $\sim 5 \times 10^{-3}\,\mathrm{T}$ |
| a neodymium [magnet](#def-g11-magnetic-fields-magnet), at its pole | $\sim 0.5\,\mathrm{T}$ |
| an MRI [solenoid](#def-g11-magnetic-fields-solenoid) | $1.5\,\mathrm{T}$ |
| surface of a neutron star | $\sim 10^{8}\,\mathrm{T}$ |

## 15.5 The Laplace force

**Proposition 15.16 (Laplace force).**

A straight wire of length $L$, carrying a current $I$ in a [uniform field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-capacitor) $\vect B$ *perpendicular* to the wire, feels the *Laplace force* of magnitude $F = I\,L\,B$, perpendicular to both wire and $\vect B$: flat right hand, thumb along the current, straight fingers along $\vect B$ — the palm pushes along $\vect F$. A wire *parallel* to the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) feels no [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force).

**Proof.** *Admitted at this level.* ∎

**Remark 15.17 (The jumping rail).**

The Year 1 volume derives this from the magnetic [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on each moving charge. In the laboratory: a copper rod rests across two horizontal rails between the poles of a [magnet](#def-g11-magnetic-fields-magnet) ([field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) into the page in the figure); close the switch, and the rod shoots along the rails. Reverse the current, or flip the [magnet](#def-g11-magnetic-fields-magnet): it shoots the other way.

![The rail experiment: current down the rod, field into the page () — the Laplace force F = ILB drives the rod along the rails.](https://one-course.com/images/onecourse/chapters/physics-2/g11-magnetic-fields/fig-4f0415657935.svg)

*The rail experiment: current down the rod, [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) into the page ($\otimes$) — the [Laplace force](#prop-g11-magnetic-fields-laplace) $F = ILB$ drives the rod along the rails.*

**Example 15.18 (Laplace numbers).**

A rod of length $L = 5.0\,\mathrm{cm}$ carrying $I = 8.0\,\mathrm{A}$ across $B = 0.50\,\mathrm{T}$: $F = 8.0 \times 0.050 \times 0.50 = 0.20\,\mathrm{N}$ — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of $20\,\mathrm{g}$, on a wire weighing a few grams.

**Remark 15.19 (The DC motor).**

Bend the wire into a loop in the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field): the two sides perpendicular to $\vect B$ carry opposite currents, so their [Laplace forces](#prop-g11-magnetic-fields-laplace) are opposite — a pair that spins the loop. Half a turn later the pair would spin it back, so a rotating switch (the *commutator*) reverses the current every half-turn. Every fan and drill is this loop, multiplied.

## 15.6 Exercises

**Exercise 15.1 ★.**

A bar [magnet](#def-g11-magnetic-fields-magnet) is sawn in half between its poles. What poles does each piece carry? Brought back together, do the freshly cut faces attract or repel? Can any cutting scheme isolate the north pole?

**Solution of Exercise 15.1.**

Each piece is a complete [magnet](#def-g11-magnetic-fields-magnet) with a north and a south pole: a new pole appears on each cut face. The fresh faces are an N and an S, so they attract — the pieces try to reassemble. No scheme isolates a pole: every cut creates a pair.

**Exercise 15.2 ★.**

True or false, with one reason each: (a) two [magnetic field lines](#def-g11-magnetic-fields-lines) cross where the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is strong; (b) outside a [magnet](#def-g11-magnetic-fields-magnet), [field lines](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-lines) run from north pole to south pole; (c) every [magnetic field line](#def-g11-magnetic-fields-lines) is a closed loop; (d) a needle settles perpendicular to the line through it.

**Solution of Exercise 15.2.**

(a) False: the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) has one direction at each point. (b) True. (c) True: each line closes through the [magnet](#def-g11-magnetic-fields-magnet)’s body. (d) False: the needle settles *along* the line, tangent to it.

**Exercise 15.3 ★.**

Compute the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) $2.0\,\mathrm{cm}$ from a straight wire carrying $10\,\mathrm{A}$. How many times the Earth’s $50\,\text{µ}\mathrm{T}$ is that?

**Solution of Exercise 15.3.**

$B = 2 \times 10^{-7} \times 10 / 0.020 = 1.0 \times 10^{-4}\,\mathrm{T}$ — twice the Earth’s [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field).

**Exercise 15.4 ★.**

A [solenoid](#def-g11-magnetic-fields-solenoid) of $800$ turns wound over $40\,\mathrm{cm}$ carries $I = 1.5\,\mathrm{A}$. Compute $n$, then $B$ inside; what does $B$ become if the current doubles?

**Solution of Exercise 15.4.**

$n = 800/0.40 = 2000\,\mathrm{m}^{-1}$; $B = 1.26 \times 10^{-6} \times 2000 \times 1.5 \approx 3.8\,\mathrm{mT}$. $B \propto I$: doubling the current gives $7.5\,\mathrm{mT}$.

**Exercise 15.5 ★.**

A wire segment of length $10\,\mathrm{cm}$ carries $5.0\,\mathrm{A}$ perpendicular to a $0.20\,\mathrm{T}$ [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field). Compute the [Laplace force](#prop-g11-magnetic-fields-laplace); the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of what mass equals it?

**Solution of Exercise 15.5.**

$F = ILB = 5.0 \times 0.10 \times 0.20 = 0.10\,\mathrm{N}$ — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of $m = F/g = 0.10/9.81 \approx 10\,\mathrm{g}$.

**Exercise 15.6 ★★.**

Where a hiker stands, the [declination](#def-g11-magnetic-fields-earth) is $8{}^{\circ}$ west. She walks $2.0\,\mathrm{km}$ “north by compass”: how far, and to which side of true north, does she drift? What heading would have walked her true north?

**Solution of Exercise 15.6.**

*1.* Her track points $8{}^{\circ}$ west of true north: $2000 \times \sin8{}^{\circ} \approx 2.8 \times 10^{2}\,\mathrm{m}$ to the west.

*2.* Steer $8{}^{\circ}$ *east* of the needle’s north.

**Exercise 15.7 ★★.**

At what distance from a straight wire carrying $20\,\mathrm{A}$ does its [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) equal the Earth’s $50\,\text{µ}\mathrm{T}$? What does this say about trusting a compass near live cables?

**Solution of Exercise 15.7.**

$d = \mu_0 I / (2\pi B) = 2 \times 10^{-7} \times 20 / 5.0 \times 10^{-5}
= 8.0\,\mathrm{cm}$. Within a decimetre of such a cable the compass reads the wire, not the planet — keep it away from live conductors.

**Exercise 15.8 ★★.**

Design a [solenoid](#def-g11-magnetic-fields-solenoid) producing $B = 10\,\mathrm{mT}$ with $I = 4.0\,\mathrm{A}$: compute the required $n$, then the turn count for a $25\,\mathrm{cm}$ [coil](#def-g11-magnetic-fields-solenoid).

**Solution of Exercise 15.8.**

$n = B/(\mu_0 I) = 0.010 / (1.26 \times 10^{-6} \times 4.0)
\approx 2.0 \times 10^{3}\,\mathrm{m}^{-1}$; over $25\,\mathrm{cm}$, $N = 1989 \times 0.25 \approx 500$ turns.

**Exercise 15.9 ★★.**

In the rail experiment, the rod (mass $20\,\mathrm{g}$, length $L = 12\,\mathrm{cm}$ between the rails) carries $10\,\mathrm{A}$ in a $0.50\,\mathrm{T}$ perpendicular [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field). Compute the [Laplace force](#prop-g11-magnetic-fields-laplace), then the rod’s initial acceleration; compare with $g$.

**Solution of Exercise 15.9.**

$F = 10 \times 0.12 \times 0.50 = 0.60\,\mathrm{N}$; $a = F/m = 0.60/0.020 = 30\,\mathrm{m}/\mathrm{s}^{2}$ — about $3g$: the rod genuinely jumps.

**Exercise 15.10 ★★.**

A horizontal wire of mass $5.0\,\mathrm{g}$ and length $20\,\mathrm{cm}$ sits in a horizontal [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) $B = 0.10\,\mathrm{T}$ perpendicular to it. What current makes the [Laplace force](#prop-g11-magnetic-fields-laplace) balance the wire’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight), so that it levitates? Which way must the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) point?

**Solution of Exercise 15.10.**

Balance: $ILB = mg$, so $I = 5.0 \times 10^{-3} \times 9.81 / (0.20 \times 0.10) \approx
2.5\,\mathrm{A}$. The [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) must point up; with $\vect B$ horizontal and perpendicular to the wire, the flat right hand (palm up) fixes the required current direction.

**Exercise 15.11 ★★.**

Give the direction of the [Laplace force](#prop-g11-magnetic-fields-laplace) (or say it vanishes): (a) current to the right of the page, $\vect B$ into the page; (b) current toward the top of the page, $\vect B$ out of the page; (c) current parallel to $\vect B$. Use the flat right hand.

**Solution of Exercise 15.11.**

Flat right hand: (a) toward the top of the page; (b) to the right; (c) no [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) — the wire is parallel to the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field).

**Exercise 15.12 ★★★.**

A horizontal wire runs geographic north–south, $3.0\,\mathrm{cm}$ above a compass; the Earth’s horizontal component there is $20\,\text{µ}\mathrm{T}$. The wire carries $I = 5.0\,\mathrm{A}$: compute its [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) at the compass, give its direction, then the needle’s deflection ([fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) add as vectors).

**Solution of Exercise 15.12.**

*1.* $B = 2 \times 10^{-7} \times 5.0 / 0.030 \approx
33\,\text{µ}\mathrm{T}$; the lines circle the wire, so directly below it the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is horizontal, perpendicular to the wire: east–west.

*2.* The needle follows the vector sum: $\tan\theta = 33/20 =
1.67$, so $\theta \approx 59{}^{\circ}$ away from north.

**Exercise 15.13 ★★★.**

A scrapyard [electromagnet](#def-g11-magnetic-fields-electromagnet) is a [coil](#def-g11-magnetic-fields-solenoid) of $400$ turns over $20\,\mathrm{cm}$, [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) $2.0\,\Omega$, on a $12\,\mathrm{V}$ supply; its iron core multiplies the bare [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) by $100$. Compute the current, the [bare-coil](#def-g11-magnetic-fields-solenoid) [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field), the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) with the core, and the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) the [coil](#def-g11-magnetic-fields-solenoid) dissipates ([Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power)). Why does the scrap fall the instant the switch opens?

**Solution of Exercise 15.13.**

*1.* $I = U/R = 12/2.0 = 6.0\,\mathrm{A}$; $n = 400/0.20 = 2000\,\mathrm{m}^{-1}$; $B_0 = 1.26 \times 10^{-6} \times 2000 \times 6.0 \approx 15\,\mathrm{mT}$; with the core, $100 B_0 \approx 1.5\,\mathrm{T}$.

*2.* $P = UI = 12 \times 6.0 = 72\,\mathrm{W}$.

*3.* No current, no [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) — and soft iron does not stay magnetized: the load releases instantly, by design.

**Exercise 15.14 ★★★.**

A motor’s rectangular loop has $N = 50$ turns carrying $I = 2.0\,\mathrm{A}$; its two sides of length $5.0\,\mathrm{cm}$ are perpendicular to a [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) $B = 0.30\,\mathrm{T}$. Compute the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on each of these sides; why are the two [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) opposite, and what do they do to the loop? Why do the other two sides sometimes feel no [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force)? Why must the current be reversed every half-turn?

**Solution of Exercise 15.14.**

*1.* $F = NILB = 50 \times 2.0 \times 0.050 \times 0.30 =
1.5\,\mathrm{N}$ on each side. The current runs opposite ways in the two sides, so the [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) are opposite: a couple that rotates the loop.

*2.* When parallel to $\vect B$, a wire feels no [Laplace force](#prop-g11-magnetic-fields-laplace).

*3.* After half a turn the couple would reverse and undo the rotation; the [commutator](#rem-g11-magnetic-fields-motor) flips the current each half-turn so the loop keeps spinning one way.

**Exercise 15.15 ★★★.**

How close to a cable carrying $100\,\mathrm{A}$ would you have to be for its [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) to reach $1\,\mathrm{T}$? Compare with the cable’s millimetre radius, and conclude why tesla-strength [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) are built from [solenoids](#def-g11-magnetic-fields-solenoid), not single wires. Then: a neutron star’s surface [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) of $10^{8}\,\mathrm{T}$ is how many [powers](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) of ten above the Earth’s?

**Solution of Exercise 15.15.**

*1.* $d = 2 \times 10^{-7} \times 100 / 1 = 2.0 \times 10^{-5}\,\mathrm{m} =
20\,\text{µ}\mathrm{m}$ — deep *inside* the millimetre-thick copper. No accessible point near a single wire reaches $1\,\mathrm{T}$: strong [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) stack thousands of turns (a [solenoid](#def-g11-magnetic-fields-solenoid)), plus iron or superconductors.

*2.* $10^{8} / 5 \times 10^{-5} = 2 \times 10^{12}$: about twelve [powers](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) of ten.

## 15.7 Problem: The compass, the crane and the MRI

**Problem 15.1.**

Weekend problem — three magnets at work: a compass crossing an ocean, an electromagnet lifting a car, and an MRI solenoid — one field, spanning ten powers of ten, doing three jobs

The same vector $\vect B$ steers a sailboat with $50\,\text{µ}\mathrm{T}$, lifts scrap iron with $1\,\mathrm{T}$, and images a knee with $1.5\,\mathrm{T}$ — three machines for this chapter’s three formulas.

**Part I — The compass.** At the ship’s position, the Earth’s [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) has horizontal component $20\,\text{µ}\mathrm{T}$ and vertical component $44\,\text{µ}\mathrm{T}$; the [declination](#def-g11-magnetic-fields-earth) is $6{}^{\circ}$ west.

1. Why does a compass needle point (magnetic) north at all?
2. Compute the total [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) ’s magnitude and its angle with the horizontal.
3. Steering “compass north” for $30\,\mathrm{km}$ , how far west of the true-north track does the ship end up?
4. Steel cargo deflects the needle a further $4{}^{\circ}$ west: same question.
5. Near the geomagnetic pole a compass is useless: which component is to blame, and why?

**Part II — The crane.** The scrapyard [electromagnet](#def-g11-magnetic-fields-electromagnet) is a [solenoid](#def-g11-magnetic-fields-solenoid) of $600$ turns wound over $30\,\mathrm{cm}$, of [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) $1.5\,\Omega$, fed by a $12\,\mathrm{V}$ supply; its iron core multiplies the bare [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) by $50$.

6. Compute $n$ , the turn density.
7. Compute the current in the [coil](#def-g11-magnetic-fields-solenoid) ( [Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power) ).
8. Compute the [bare-coil](#def-g11-magnetic-fields-solenoid) [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) $\mu_0 n I$ .
9. Compute the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) with the core; compare with neodymium’s $0.5\,\mathrm{T}$ .
10. Compute the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) dissipated in the [coil](#def-g11-magnetic-fields-solenoid) , then the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) for a $10\,\mathrm{min}$ lifting shift.
11. A permanent [magnet](#def-g11-magnetic-fields-magnet) this strong exists. Why an [electromagnet](#def-g11-magnetic-fields-electromagnet) anyway?

**Part III — The winch motor.** The crane’s winch is a DC motor: a rectangular loop of $N = 100$ turns, sides of length $4.0\,\mathrm{cm}$ perpendicular to $B = 0.25\,\mathrm{T}$, carrying $I = 3.0\,\mathrm{A}$.

12. Compute the [Laplace force](#prop-g11-magnetic-fields-laplace) on each side bundle perpendicular to $\vect B$ .
13. The two [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) are equal and opposite, yet their effect does not cancel. What do they produce, and why must they be *opposite* ?
14. What [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) acts on the other two sides, when parallel to $\vect B$ ?
15. After half a turn, the same [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) would undo the rotation. What does the [commutator](#rem-g11-magnetic-fields-motor) do, and when?
16. Give three separate design changes that each double the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) pair.

**Part IV — The MRI.** The scanner’s [solenoid](#def-g11-magnetic-fields-solenoid) holds a uniform $1.5\,\mathrm{T}$ with a current $I = 500\,\mathrm{A}$.

17. How many times the Earth’s [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is $1.5\,\mathrm{T}$ ?
18. Compute the turn density $n$ needed *without* any iron core.
19. If the winding had [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) $0.20\,\Omega$ , what [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) would it dissipate? What property of the actual winding avoids this, at the price of extreme cold?
20. Punchline: place the ship’s, the crane’s and the MRI’s [fields](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) on the powers-of-ten ladder from interstellar space ( $10^{-10}\,\mathrm{T}$ ) to a neutron star ( $10^{8}\,\mathrm{T}$ ); what changed between the machines — the physics, or the [amperes](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-current) ?

**Solution of Problem 15.1.**

**1.** The needle is itself a small [magnet](#def-g11-magnetic-fields-magnet); the Earth’s [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) aligns it, north end along the horizontal component of $\vect B$, i.e. toward magnetic north.

**2.** $B = \sqrt{20^2 + 44^2} = \sqrt{2336} \approx
48\,\text{µ}\mathrm{T}$, tilted $\tan^{-1}(44/20) \approx 66{}^{\circ}$ below the horizontal.

**3.** The track runs $6{}^{\circ}$ west of true north: $30 \times \sin6{}^{\circ} \approx 3.1\,\mathrm{km}$ west.

**4.** Errors add: $10{}^{\circ}$, so $30 \times \sin10{}^{\circ} \approx
5.2\,\mathrm{km}$ — five kilometres for ten degrees.

**5.** The horizontal component: near the pole the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) is almost vertical, the horizontal part shrinks toward zero, and the needle has nothing left to align with.

**6.** $n = 600/0.30 = 2000\,\mathrm{m}^{-1}$.

**7.** $I = U/R = 12/1.5 = 8.0\,\mathrm{A}$.

**8.** $B_0 = \mu_0 n I = 1.26 \times 10^{-6} \times 2000 \times 8.0
\approx 20\,\mathrm{mT}$.

**9.** $50 \times 20\,\mathrm{mT} = 1.0\,\mathrm{T}$ — twice the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) at a neodymium pole, over a far larger face.

**10.** $P = UI = 12 \times 8.0 = 96\,\mathrm{W}$; in $10\,\mathrm{min}$: $E = 96 \times 600 \approx 58\,\mathrm{kJ}$.

**11.** Because it lets go: open the switch, the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) dies, the car drops exactly where wanted. A permanent [magnet](#def-g11-magnetic-fields-magnet) grips forever.

**12.** $F = NILB = 100 \times 3.0 \times 0.040 \times 0.25 =
3.0\,\mathrm{N}$.

**13.** A couple: applied on opposite sides of the axis, the two opposite [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) both turn the loop the same way. Equal *parallel* [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) would only push the loop sideways, not spin it.

**14.** None: a wire parallel to the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) feels no [Laplace force](#prop-g11-magnetic-fields-laplace).

**15.** It reverses the current in the loop at each half-turn, just as the couple would change sign — so the torque always drives the same rotation.

**16.** Double $N$, or double $I$, or double $B$: the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) pair $F = NILB$ is proportional to each.

**17.** $1.5 / 5.0 \times 10^{-5} = 3.0 \times 10^{4}$: thirty thousand Earths.

**18.** $n = B/(\mu_0 I) = 1.5 / (1.26 \times 10^{-6} \times 500)
\approx 2.4 \times 10^{3}\,\mathrm{m}^{-1}$ — and every one of those turns carries $500\,\mathrm{A}$.

**19.** $P = RI^2 = 0.20 \times 500^2 = 50\,\mathrm{kW}$ — a neighbourhood’s worth of heating. The real winding is superconducting: zero [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance), zero dissipation, provided it is kept a few degrees above absolute zero.

**20.** Compass $5 \times 10^{-5}$, crane $1$, MRI $1.5$: the ship sits five rungs below the two machines, which share a decade — between interstellar $10^{-10}$ and neutron-star $10^{8}$. Nothing changed but the [amperes](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-current) (and the turns, iron and superconductor that multiply them): one [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field), three jobs.