Magnets, the magnetic dipole and the magnetic field
Why does a compass needle in the forest always point north, but swing away the moment a magnet is brought next to it?
Prior knowledge
- Perpendicular vectors are added with the Pythagorean theorem and the direction is found from the tangentchapter Force diagrams and the combined effect of forces
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
An orienteer lays a compass on the map, and the needle obediently turns to point north. When she puts a torch with a small magnet in it beside the compass, the needle swings to a completely different direction. When the torch is taken away, the needle returns to north.
What turns the compass needle, and why does the magnet win that contest when it is brought close?
Optional Revise: the sum of vectors
Predict
Make your predictions before investigating 2 questions
Prediction 1: distance
Prediction 2: at the side of the magnet
Investigate
In the simulation a bar magnet lies on a table seen from above, north upwards. Move the compass around the magnet and watch where the needle turns.
Simulation
How to use
The position of the compass is given as the distance $r$ from the middle of the magnet and the direction $\varphi$, measured anticlockwise from east: $\varphi = 0^\circ$ is on the axis of the magnet on the east side and $\varphi = 90^\circ$ is north of the magnet. You can also drag the compass with the mouse or a finger. Magnet turns the magnet so that its north pole is in the east or in the west. Earth's field adds or removes the horizontal component of the Earth's field, which points north. The checkboxes show the field lines of the magnet itself and a compass grid whose small needles show the direction of the total field at every point. Shake the needle gives the needle a nudge, and Speed changes only the pace of the motion of the needle. In the readings $B_\text{m}$ is the field of the magnet, $B_\text{h}$ the horizontal component of the Earth's field, $B$ the total field and $\theta$ the direction of the needle from north: positive towards the east, negative towards the west.
Assumptions of the model
The values are model values calculated by the simulation. The magnet is modelled as two poles $4\,\text{cm}$ from the middle of the magnet. The strength of the field is chosen so that on the axis of the magnet $10\,\text{cm}$ from the middle $B_\text{m} = 30.0\,\mu\text{T}$. The horizontal component of the Earth's field is the value for southern Finland, $14\,\mu\text{T}$, and its vertical component does not turn a horizontal needle. The field of the compass itself does not change the field of the magnet. The needle turns with damping to point along the total field; the reading $\theta$ is the direction in which the needle settles.
How the calculation works
The field of each pole weakens with the square of the distance, and the field $\overline{B}_\text{m}$ of the magnet is the vector sum of the fields of the poles. The total field is $\overline{B} = \overline{B}_\text{m} + \overline{B}_\text{h}$. The direction of the needle from north is the angle $\theta$ for which $\tan\theta$ is the ratio of the east component to the north component of the total field.
Experiment 1: the field weakens as the distance grows
The initial state places the compass on the axis of the magnet on the east side ($\varphi = 0^\circ$), the north pole of the magnet towards the east and the Earth's field on. Change the distance $r$ and record the direction $\theta$ of the needle and the field $B_\text{m}$ of the magnet. Record at least the distances $10\,\text{cm}$, $12\,\text{cm}$, $14\,\text{cm}$ and $20\,\text{cm}$.
Initial state: direction $\varphi$ = $0^\circ$, magnet: north pole towards the east, Earth's field: on, distance $r$ = $10\,\text{cm}$. Kept constant: direction $\varphi$, magnet, Earth's field. Varied: distance $r$.
| distance $r$ | $\theta$ $(^\circ)$ | field of the magnet $B_m$ $(\mu\text{T})$ |
|---|
From your table, calculate how many times greater the field $B_\text{m}$ of the magnet is at a distance of $10\,\text{cm}$ than at a distance of $20\,\text{cm}$.
Hint 1: a guiding question
Find the rows $r = 10\,\text{cm}$ and $r = 20\,\text{cm}$ in the table.
Hint 2: what is the connection
The question "how many times greater" is answered by dividing: the greater field is divided by the smaller.
Hint 3: an intermediate step
$30.0\,\mu\text{T} : 2.9\,\mu\text{T}$
Show the solution
Given
from the table the model values $B_\text{m} = 30.0\,\mu\text{T}$ when $r = 10\,\text{cm}$ and $B_\text{m} = 2.9\,\mu\text{T}$ when $r = 20\,\text{cm}$
Find
ratio of the fields $B_\text{m}(10\,\text{cm}) : B_\text{m}(20\,\text{cm})$, a plain number
Principle
The question "how many times greater" is answered by dividing the greater value by the smaller. Both fields were measured on the same line on the axis of the magnet, so they are in the same direction and their magnitudes can be compared directly.
Formula
$k = \dfrac{B_1}{B_2}$, where $B_1$ is the field at $10\,\text{cm}$ and $B_2$ the field at $20\,\text{cm}$
Conversions
Not needed: both fields are in microtesla, and the unit cancels.
Substitution
Answer
about $10$, to two significant figures like the reading $2.9\,\mu\text{T}$; with more accurate model values the ratio is $10.4$.
Reasonableness
If the field weakened with the square of the distance, the ratio would be $2^2 = 4$, and in direct proportion $2$. The $10$ obtained is close to $2^3 = 8$ but a little greater, because at $10\,\text{cm}$ the nearer pole of the magnet still acts clearly more strongly than the farther one. The needle tells the same: the angle decreased from $65^\circ$ to about $12^\circ$.
Experiment 2: the direction of the field around the magnet
The initial state removes the Earth's field, so the needle shows the direction of the field of the magnet alone. The compass is $14\,\text{cm}$ from the middle of the magnet. Move the compass round the magnet by changing the direction $\varphi$ and record the direction of the needle. Record at least the directions $0^\circ$, $45^\circ$, $90^\circ$, $135^\circ$ and $180^\circ$. It is worth keeping the field lines visible.
Initial state: distance $r$ = $14\,\text{cm}$, magnet: north pole towards the east, Earth's field: off, direction $\varphi$ = $0^\circ$. Kept constant: distance $r$, magnet, Earth's field. Varied: direction $\varphi$.
| direction $\varphi$ | $\theta$ $(^\circ)$ | field of the magnet $B_m$ $(\mu\text{T})$ |
|---|
Conclusion
Using your measurements, explain where a compass needle turns near a magnet and why the effect of the magnet on the needle weakens when the compass is taken farther away. Use the words field line, magnetic field and Earth's field.
Model answer
A compass needle is a small magnet that turns to point along the total field: its north end points in the direction of the field, that is along the field line. In experiment 2 the Earth's field had been removed. On the axis east of the magnet the needle pointed east, away from the north pole, and on the west side also east, towards the south pole. At the side of the magnet the needle pointed west, that is parallel to the magnet towards the south pole. So the needle followed the field lines, which run outside the magnet from the north pole to the south pole. In experiment 1 the field of the magnet in the model values was $30.0\,\mu\text{T}$ at $10\,\text{cm}$, $9.1\,\mu\text{T}$ at $14\,\text{cm}$ and only $2.9\,\mu\text{T}$ at $20\,\text{cm}$. The needle turned $65^\circ$, $33^\circ$ and $12^\circ$ from north. Close to the magnet its field is much greater than the horizontal component $14\,\mu\text{T}$ of the Earth's field and determines the direction; farther away the Earth's field wins and the needle turns back towards north.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
The poles of a magnet
The effect of a magnet is strongest at its ends, which are called the poles of the magnet. When a bar magnet is hung from a thread so that it can turn freely, it settles in the north–south direction. The end that turns north is the north pole $\text{N}$ of the magnet and the end that turns south is the south pole $\text{S}$. Like poles of two magnets repel each other and unlike poles attract each other. Magnets act on each other even without touching, and the effect weakens quickly as the magnets move apart.
The magnetic dipole
When a bar magnet is cut in two, each piece becomes a complete magnet with both a north pole and a south pole. The same happens however many times the pieces are cut smaller. The poles of a magnet therefore always occur in pairs: a magnet is a magnetic dipole. This is where magnetic interaction differs from electric interaction. Positive and negative electric charge can be separated from each other, but a single magnetic pole has never been observed.
Ferromagnetism and magnetic domains
A magnet attracts iron, nickel, cobalt and many of their alloys, such as steel. These materials are called ferromagnetic. Each of their atoms is a small magnetic dipole. In a ferromagnetic material the dipoles of neighbouring atoms line up in the same direction, and the material typically divides into magnetic domains a few micrometres across, in each of which all the atomic dipoles point the same way. In unmagnetised iron the directions of the domains vary, so their effects cancel and the piece is not a magnet.
When a piece of iron is brought near a magnet, the field of the magnet turns the directions of the domains and makes the domains that point along the field grow. The iron becomes magnetised: it becomes a magnet whose end nearest the magnet has the pole opposite to the magnet's pole. This is why a magnet attracts even an unmagnetised iron nail. Soft iron loses almost all of its magnetisation when the magnet is taken away, but in hard steel and in the alloy of a neodymium magnet the domains stay aligned. These are made into permanent magnets.
Heating to a high enough temperature, for iron about $770\,{}^\circ\text{C}$, destroys the alignment, and the ferromagnetism disappears.
Most other materials, such as aluminium, copper, wood, plastic and water, do not stick to a magnet: their magnetic effects are so weak that ordinary magnets do not reveal them.
The magnetic field and field lines
A magnet changes the space around it so that another magnet or a piece of iron experiences a force in it. The space that forms around a magnet in this way is called a magnetic field. The direction of the field at a point is the direction in which the north end of a compass needle placed at that point points. A compass is therefore a small test magnet with which the direction of the field can be found one point at a time.
The field is visualised with field lines. The direction of a field line at each point is the direction of the field. Outside a magnet the field lines run from the north pole to the south pole, and inside the magnet from the south pole to the north pole, so each line is a closed loop. Where the lines are dense, the field is strong. Field lines never cross each other, because the field has only one direction at each point.
The Earth's magnetic field
The Earth has a magnetic field that resembles the field of a bar magnet. It arises from the motions of molten iron in the Earth's outer core and the electric currents associated with them. The north end of a compass needle turns towards geographic north. Because unlike poles attract each other, the Earth's magnetic south pole $\text{S}$ is near the geographic North Pole, and the magnetic north pole $\text{N}$ is near the geographic South Pole.
The magnetic poles are not exactly at the geographic poles, and they wander slowly. A compass therefore deviates from true north: in Finland at present about $10^\circ$ east in the south and about $15^\circ$ east in Lapland. In Finland the field lines also dip steeply towards the ground, about $73^\circ$–$79^\circ$ below the horizontal. This angle is called the inclination.
Combining magnetic fields
The strength and direction of a magnetic field are described by the vector quantity magnetic flux density $\overline{B}$. Its direction is the direction of the field, and its magnitude $B$ tells how strong the field is. The unit of magnetic flux density is the tesla $\text{T}$. The definition of the tesla is based on the magnetic force on a moving charge, and it is introduced in the chapter A charged particle in electric and magnetic fields. The tesla is a large unit, so everyday fields are often given in millitesla, $1\,\text{mT} = 10^{-3}\,\text{T}$, or in microtesla, $1\,\mu\text{T} = 10^{-6}\,\text{T}$.
| Object | $B$ |
|---|---|
| Earth, southern Finland | about $52\,\mu\text{T}$ |
| fridge magnet | about $5\,\text{mT}$ |
| neodymium magnet | about $0.5\,\text{T}$ |
| magnetic resonance imaging (MRI) | $1.5$–$3\,\text{T}$ |
| laboratory magnet | about $45\,\text{T}$ |
The value of the Earth's field ($52$–$53\,\mu\text{T}$ in southern Finland) is from the World Magnetic Model 2025 (NOAA and the British Geological Survey), the fields of hospital MRI scanners are usually $1.5\,\text{T}$ or $3\,\text{T}$, and continuous fields of about $45\,\text{T}$ are produced in research laboratories, for example with the hybrid magnet of the National High Magnetic Field Laboratory in the United States. The values for magnets are orders of magnitude at the surface of the magnet.
When two magnetic fields act at the same point, the total field is their vector sum:
$$\overline{B} = \overline{B}_1 + \overline{B}_2$$
where $\overline{B}_1$ and $\overline{B}_2$ are the magnetic flux densities of the fields (T) and $\overline{B}$ is the total field (T). A compass needle turns to point along the total field. When the fields are perpendicular to each other, the magnitude of the total field is given by the Pythagorean theorem and the direction by the tangent:
$$\begin{aligned} B &= \sqrt{B_1^2 + B_2^2} \\ \tan\theta &= \frac{B_2}{B_1} \end{aligned}$$
where $\theta$ is the angle between the total field and the field $\overline{B}_1$.
This is used in a compass measurement. At the compass, the horizontal component $\overline{B}_\text{h}$ of the Earth's field points north. When a magnet is placed so that its field $\overline{B}_\text{m}$ at the compass points east, the needle turns from north towards east by an angle $\theta$ for which $\tan\theta = B_\text{m} : B_\text{h}$. The field of a magnet can therefore be measured with a compass when the horizontal component of the Earth's field is known.
In Finland the Earth's field points obliquely into the ground. A compass needle turns only in the horizontal plane, so only the horizontal component of the field turns it. When the magnitude of the field is $B$ and the inclination is $I$, the components are
$$\begin{aligned} B_\text{h} &= B\cos I \\ B_\text{v} &= B\sin I \end{aligned}$$
where $B_\text{h}$ is the horizontal component and $B_\text{v}$ the vertical component. In southern Finland $B \approx 52\,\mu\text{T}$ and $I \approx 74^\circ$, so the horizontal component is only about $14\,\mu\text{T}$ and the vertical component about $50\,\mu\text{T}$. These are the course's rounded values: a more accurate model gives a horizontal component of about $15\,\mu\text{T}$ in Helsinki.
Common misconception
Many people think that the Earth's magnetic north pole is near the geographic North Pole, because the compass points there. However, the north end of a compass needle is the north pole of a magnet, and it is attracted by an unlike pole. So the magnetic pole near the geographic North Pole is a south pole.
Another mistake is to think that the poles of a magnet can be separated by cutting the magnet in the middle. Two new poles appear at the cut, and each piece becomes a complete dipole. A third mistake is to think that a magnet attracts all metals. An aluminium can or a copper coin does not stick to a magnet, because they are not ferromagnetic.
In calculations the most common mistake is to add perpendicular fields like plain numbers. Fields of $14\,\mu\text{T}$ north and $18\,\mu\text{T}$ east give a total field of $\sqrt{(14\,\mu\text{T})^2 + (18\,\mu\text{T})^2} \approx 23\,\mu\text{T}$, not $32\,\mu\text{T}$.
Summary
- A magnet always has two poles, a north pole $\text{N}$ and a south pole $\text{S}$: it is a magnetic dipole. Like poles repel and unlike poles attract each other, and a single pole has never been observed.
- In ferromagnetic materials, such as iron, the atomic dipoles form magnetic domains. The field of a magnet aligns them, so the material becomes magnetised and the magnet attracts it.
- The direction of a magnetic field is the direction in which the north end of a compass needle points. Outside a magnet the field lines run from the north pole to the south pole, and their density tells the strength of the field.
- The Earth's magnetic south pole is near the geographic North Pole. In Finland the Earth's field is about $52\,\mu\text{T}$ and points steeply into the ground.
- Magnetic flux density $\overline{B}$ is a vector whose unit is the tesla $\text{T}$. Fields add as vectors: for perpendicular fields $B = \sqrt{B_1^2 + B_2^2}$ and $\tan\theta = B_2 : B_1$.
Optional Return to your predictions
Compare your predictions with the results of the simulation. What did you learn about the direction of the field lines and about how quickly the field of a magnet weakens? Return also to the orienteer at the beginning: why does the magnet in the torch turn the needle only when it is close?
Would you change your prediction?
Apply
Worked example
Problem
A compass lies on a horizontal table in southern Finland, where the horizontal component of the Earth's magnetic field is $14\,\mu\text{T}$. A bar magnet is placed east of the compass, in the east–west direction, with its south pole $\text{S}$ towards the compass. The compass needle turns $52^\circ$ from north towards east. What is the magnetic flux density of the magnet at the compass?
Step 1/7
Given and find. The horizontal component of the Earth's field is $B_\text{h} = 14\,\mu\text{T}$ towards the north and the angle of deflection of the needle is $\theta = 52^\circ$ towards the east. We are asked for the magnetic flux density $B_\text{m}$ of the magnet at the compass.
Step 2/7
Principle. The compass needle turns to point along the total field. At the compass, the field of the magnet points towards the south pole of the magnet, that is east, so it is perpendicular to the horizontal component of the Earth's field, which points north. The total field is the diagonal of a rectangle, and the tangent of the angle $\theta$ is the ratio of the opposite side to the adjacent side.
Step 3/7
Formula solved.
$$\begin{aligned} \tan\theta &= \frac{B_\text{m}}{B_\text{h}} \\ B_\text{m} &= B_\text{h}\tan\theta \end{aligned}$$
Step 4/7
Unit conversions. Not needed: $\tan\theta$ is a plain number, so $B_\text{m}$ comes out in the same unit as $B_\text{h}$, that is in microtesla. If needed, $1\,\mu\text{T} = 10^{-6}\,\text{T}$.
Step 5/7
Substitution. $$\begin{aligned} B_\text{m} &= B_\text{h}\tan\theta \\ &= 14\,\mu\text{T} \cdot \tan 52^\circ \\ &= 14\,\mu\text{T} \cdot 1.280 \\ &= 17.9\,\mu\text{T} \end{aligned}$$
Step 6/7
Result. The magnetic flux density of the magnet at the compass is about $18\,\mu\text{T} = 1.8 \cdot 10^{-5}\,\text{T}$. The result is given to two significant figures, because the horizontal component $14\,\mu\text{T}$ and the angle $52^\circ$ are known to two figures.
Step 7/7
Reasonableness. At an angle of $45^\circ$ the field of the magnet would equal $B_\text{h}$. The angle is greater than $45^\circ$, so $B_\text{m}$ is greater than $14\,\mu\text{T}$, as the result is. The total field is
and $B_\text{h} : \cos 52^\circ$ gives the same. The field is of the order of magnitude of the Earth's field, which fits a magnet about ten centimetres away.
Find the error
Apply to a new situation
On a table there is a compass and two magnets. The field of magnet A at the compass is $20\,\mu\text{T}$ towards the east and the field of magnet B is $12\,\mu\text{T}$ towards the north. The horizontal component of the Earth's field is $14\,\mu\text{T}$ towards the north.
Through how many degrees does the compass needle turn from north towards east?
Hint 1: a guiding question
Which of the fields are in the same direction? They can be added as plain numbers.
Hint 2: what is the connection
The fields pointing north add up to $12\,\mu\text{T} + 14\,\mu\text{T} = 26\,\mu\text{T}$. The field pointing east is $20\,\mu\text{T}$.
Hint 3: an intermediate step
when $\theta$ is measured from north.
Show the solution
Given
field of magnet A $B_\text{A} = 20\,\mu\text{T}$ towards the east, field of magnet B $B_\text{B} = 12\,\mu\text{T}$ towards the north, horizontal component of the Earth's field $B_\text{h} = 14\,\mu\text{T}$ towards the north
Find
angle of deflection $\theta$ of the needle from north towards east in degrees
Principle
The needle turns to point along the total field, and the total field is the vector sum of the fields. The field of magnet B and the Earth's field point in the same direction, so they are added as plain numbers. The field pointing east is perpendicular to them, so the angle is found from the tangent.
Formula
so
Conversions
Not needed: all fields are in microtesla, and the unit cancels in the ratio.
Substitution
Answer
$38^\circ$
From north towards east, to two significant figures like the fields.
Reasonableness
The north component $26\,\mu\text{T}$ is greater than the east component $20\,\mu\text{T}$, so the needle must turn less than $45^\circ$, as it does. Without magnet B the angle would be $55^\circ$: magnet B pulls the needle back towards north.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- How is the direction of a magnetic field at a point found with a compass?
Answer
The compass is placed at the point, and the direction of the field is the direction in which the north end of the needle points. The needle settles along the field line.
- The field of a fridge magnet is about $5\,\text{mT}$ at the surface of the magnet. How many microtesla is it, and how many times greater is it than the Earth's field in southern Finland ($52\,\mu\text{T}$)?
Answer
$5\,\text{mT}$$= 5 \cdot 1000\,\mu\text{T}$$= 5000\,\mu\text{T}$The ratio is $5000\,\mu\text{T} : 52\,\mu\text{T} \approx 96$, so at the surface of the magnet the field is about a hundred times the Earth's field.
- Why can two field lines never cross each other?
Answer
The direction of a field line tells the direction of the field. At a crossing point the field would have two different directions, but the field has only one direction at each point.
1. Concept
2. Field line picture
3. Explain
Using magnetic domains, explain why a magnet attracts an unmagnetised iron nail but not an aluminium nail.
Model answer
Iron is a ferromagnetic material. Its atoms are small magnetic dipoles that line up in the same direction in magnetic domains. In an unmagnetised nail the directions of the domains vary, so their effects cancel. When a magnet is brought close, its field turns the domains and makes those that point along the field grow. The nail becomes magnetised: if the north pole of the magnet is near, the nearer end of the nail becomes a south pole. Unlike poles attract each other, so the magnet attracts the nail. Aluminium is not ferromagnetic. It has no magnetic domains, and it does not become magnetised enough for the attraction to be noticed.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Magnetic field
- A region in which a force acts on a magnetic material or a moving charge.
- Magnetic flux density
- The quantity $B$ that describes the strength of a magnetic field; its unit is the tesla.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 1-1Core
A magnetic resonance imaging scanner in a hospital has a magnetic flux density of $3.0\,\text{T}$. The magnetic flux density of the Earth's magnetic field is about $50\,\mu\text{T}$.
How many times greater is the field of the scanner than the Earth's field?
Answer
$6.0 \cdot 10^{4}$
Show the solution
Given
magnetic flux density of the MRI scanner $B_1 = 3.0\,\text{T}$, magnetic flux density of the Earth's field $B_2 = 50\,\mu\text{T}$
Find
ratio of the fields $k = B_1 : B_2$, a plain number
Principle
The question "how many times greater" is answered by dividing the greater value by the smaller. A ratio can be calculated only when the fields are in the same unit, so the microtesla is converted to tesla first.
Formula
$k = \dfrac{B_1}{B_2}$
Conversions
$B_2$$= 50\,\mu\text{T}$$= 50 \cdot 10^{-6}\,\text{T}$Substitution
$k$$= \dfrac{3.0\,\text{T}}{50 \cdot 10^{-6}\,\text{T}}$$= 0.060 \cdot 10^{6}$$= 60\,000$Answer
$6.0 \cdot 10^{4}$
Times, two significant figures like the field of the scanner $3.0\,\text{T}$ and the Earth's field $50\,\mu\text{T}$.
Reasonableness
One tesla is a million microtesla, so $3\,\text{T}$ is $3\,000\,000\,\mu\text{T}$. Dividing this by $50$ gives $60\,000$. The ratio is enormous, which explains why iron objects must not be taken into an MRI room.
- Exercise 1-2Core
In Kuopio the magnetic flux density of the Earth's magnetic field is about $53\,\mu\text{T}$, and the field points $75^\circ$ below the horizontal.
What is the horizontal component of the field in microtesla?
Answer
$14\,\mu\text{T}$
Show the solution
Given
magnetic flux density of the Earth's field $B = 53\,\mu\text{T}$, inclination $I = 75^\circ$ below the horizontal
Find
horizontal component $B_\text{h}$ of the field in microtesla
Principle
The field points obliquely into the ground. The horizontal component is the leg adjacent to the angle $I$ in a right-angled triangle whose hypotenuse is the whole field.
Formula
$B_\text{h} = B\cos I$
Conversions
Not needed: the result comes out in the same unit as $B$, that is in microtesla.
Substitution
$B_\text{h}$$= 53\,\mu\text{T} \cdot \cos 75^\circ$$= 53\,\mu\text{T} \cdot 0.2588$$= 13.72\,\mu\text{T}$Answer
$14\,\mu\text{T}$
Two significant figures like the field $53\,\mu\text{T}$ and the angle $75^\circ$.
Reasonableness
The field is almost vertical, so the horizontal component must be a small part of the whole field: $\cos 75^\circ$ is about a quarter, and $53 : 4 \approx 13$. The result is of the same order of magnitude as the $14\,\mu\text{T}$ of southern Finland, as expected from the location of Kuopio.
- Exercise 1-3
At a compass the horizontal component of the Earth's magnetic field is $14\,\mu\text{T}$ towards the north and the field of a magnet on the table is $48\,\mu\text{T}$ towards the east.
What is the horizontal total field in microtesla?
Answer
$50\,\mu\text{T}$
Show the solution
Given
horizontal component of the Earth's field $B_\text{h} = 14\,\mu\text{T}$ towards the north, field of the magnet $B_\text{m} = 48\,\mu\text{T}$ towards the east
Find
magnitude $B$ of the horizontal total field in microtesla
Principle
The total field is the vector sum of the fields. Fields pointing north and east are perpendicular to each other, so they are the legs of a right-angled triangle and the total field is the hypotenuse.
Formula
$B = \sqrt{B_\text{h}^2 + B_\text{m}^2}$
Conversions
Not needed: both fields are in microtesla.
Substitution
$B$$= \sqrt{(14\,\mu\text{T})^2 + (48\,\mu\text{T})^2}$$= \sqrt{196 + 2304}\,\mu\text{T}$$= \sqrt{2500}\,\mu\text{T}$$= 50\,\mu\text{T}$Answer
$50\,\mu\text{T}$
Two significant figures like the fields.
Reasonableness
The hypotenuse is longer than either leg but shorter than their sum $62\,\mu\text{T}$. The result $50\,\mu\text{T}$ is only a little greater than $48\,\mu\text{T}$, because the field of the magnet is much greater than the Earth's field.
- Exercise 1-4
A bar magnet is cut in two in the middle.
What poles does each piece have? Do the pieces attract or repel each other when the cut faces are brought back together?
Answer
Each piece has a north pole and a south pole. The cut faces have unlike poles, so the pieces attract each other.
Show the solution
Find
the poles of the pieces and whether the cut faces attract or repel each other
Principle
A magnet is a magnetic dipole: its poles always occur in pairs. Every part of a magnet consists of atomic dipoles lined up in the same direction, so every piece is also a dipole. Unlike poles attract and like poles repel each other.
Model answer
Each piece becomes a complete magnet with a north pole and a south pole. If the north pole of the original magnet was on the right, the north pole of each piece is on the right. So a north pole appears on the cut face of the left piece and a south pole on the cut face of the right piece. When the cut faces are brought back together, they have unlike poles, so the pieces attract each other and form one magnet again.
- Exercise 1-5
A compass is placed on a table on the axis of a bar magnet in front of the north pole $\text{N}$ of the magnet. The Earth's field is ignored.
Which way does the north end of the compass needle point, and why?
Answer
Away from the magnet: the field of the magnet points away from the north pole, and the north pole of the needle repels the north pole of the magnet.
Show the solution
Find
the direction of the north end of the compass needle in front of the north pole of the magnet, and the reason
Principle
The direction of the field is the direction in which the north end of a compass needle points. Outside a magnet the field lines start from the north pole. The north end of the needle is itself a north pole, and like poles repel each other.
Model answer
The north end of the needle points directly away from the magnet, along the axis of the magnet. On the axis in front of the north pole the field lines leave the pole straight outwards, so the field points away from the magnet. The same follows from the poles: the north pole of the magnet repels the north end of the needle and attracts its south end, so the south end turns towards the magnet and the north end away from it.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 1-6Core
The field of a magnet at a compass is $25\,\mu\text{T}$ towards the east, and the horizontal component of the Earth's magnetic field is $14\,\mu\text{T}$ towards the north.
Through how many degrees does the compass needle turn from north towards east?
Answer
$61^\circ$
Show the solution
Given
field of the magnet $B_\text{m} = 25\,\mu\text{T}$ towards the east, horizontal component of the Earth's field $B_\text{h} = 14\,\mu\text{T}$ towards the north
Find
angle of deflection $\theta$ of the needle from north towards east in degrees
Principle
The needle turns to point along the horizontal total field. The fields are perpendicular to each other, so the total field is the diagonal of a rectangle. For the angle measured from north, the opposite leg is the field pointing east and the adjacent leg is the field pointing north.
Formula
$\tan\theta$$= \dfrac{B_\text{m}}{B_\text{h}}$so
$\theta$$= \tan^{-1}\dfrac{B_\text{m}}{B_\text{h}}$Conversions
Not needed: both fields are in microtesla, and the unit cancels.
Substitution
$\theta$$= \tan^{-1}\dfrac{25\,\mu\text{T}}{14\,\mu\text{T}}$$= \tan^{-1} 1.786$$= 60.75^\circ$Answer
$61^\circ$
From north towards east, two significant figures like the fields.
Reasonableness
The field of the magnet is greater than the Earth's field, so the needle must turn more than $45^\circ$ but less than $90^\circ$. A check in the other direction: $14\,\mu\text{T} \cdot \tan 60.75^\circ = 25\,\mu\text{T}$.
- Exercise 1-7Core
The magnetometer application of a smartphone shows the components of the Earth's magnetic field $B_x = 12\,\mu\text{T}$, $B_y = -5.0\,\mu\text{T}$ and $B_z = -48\,\mu\text{T}$. The components are perpendicular to each other.
What is the magnetic flux density of the Earth's magnetic field in microtesla?
Answer
$50\,\mu\text{T}$
Show the solution
Given
components from the magnetometer $B_x = 12\,\mu\text{T}$, $B_y = -5.0\,\mu\text{T}$ and $B_z = -48\,\mu\text{T}$, perpendicular to each other
Find
magnetic flux density $B$ of the Earth's magnetic field in microtesla
Principle
The field is the vector sum of its components. The sum of two perpendicular components is given by the Pythagorean theorem, and when a third perpendicular component is added, the same theorem gives the sum of the squares of the three components. The sign gives only the direction, and squaring removes it.
Formula
$B^2 = B_x^2 + B_y^2 + B_z^2$, so $B = \sqrt{B_x^2 + B_y^2 + B_z^2}$
Conversions
Not needed: all components are in microtesla.
Substitution
$B^2$$= (12\,\mu\text{T})^2 + (-5.0\,\mu\text{T})^2 + (-48\,\mu\text{T})^2$$= 2473\,(\mu\text{T})^2$so
$B$$= \sqrt{2473}\,\mu\text{T}$$= 49.73\,\mu\text{T}$Answer
$50\,\mu\text{T}$
Two significant figures like the components $12\,\mu\text{T}$ and $48\,\mu\text{T}$.
Reasonableness
The result is greater than the greatest component $48\,\mu\text{T}$ and close to the value of about $52\,\mu\text{T}$ for Finland. The large $z$-component fits the fact that in Finland the field points steeply into the ground when the phone is horizontal.
- Exercise 1-8
In Rovaniemi the vertical component of the Earth's magnetic field is about $52\,\mu\text{T}$, and the field points $77^\circ$ below the horizontal.
What is the magnetic flux density of the field in microtesla?
Answer
$53\,\mu\text{T}$
Show the solution
Given
vertical component of the Earth's field $B_\text{v} = 52\,\mu\text{T}$, inclination $I = 77^\circ$ below the horizontal
Find
magnetic flux density $B$ of the field in microtesla
Principle
The whole field is the hypotenuse of a right-angled triangle and the vertical component is the leg opposite to the angle $I$, so $B_\text{v} = B\sin I$. The whole field is solved from the formula.
Formula
$B_\text{v} = B\sin I$, so $B = \dfrac{B_\text{v}}{\sin I}$
Conversions
Not needed: the result comes out in microtesla like $B_\text{v}$.
Substitution
$B$$= \dfrac{52\,\mu\text{T}}{\sin 77^\circ}$$= \dfrac{52\,\mu\text{T}}{0.9744}$$= 53.4\,\mu\text{T}$Answer
$53\,\mu\text{T}$
Two significant figures like the vertical component $52\,\mu\text{T}$.
Reasonableness
The hypotenuse is longer than the leg, so $B$ is greater than $52\,\mu\text{T}$. The field is almost vertical, so the difference is small. The horizontal component left is $53.4\,\mu\text{T} \cdot \cos 77^\circ \approx 12\,\mu\text{T}$, less than in southern Finland, as it should be further north.
- Exercise 1-9
A compass is on the axis of a magnet, and the field of the magnet is perpendicular to the horizontal component of the Earth's field. When the compass is $10\,\text{cm}$ from the magnet, the needle turns $60^\circ$ from north. When it is $20\,\text{cm}$ away, the needle turns $10^\circ$.
How many times greater is the field of the magnet at $10\,\text{cm}$ than at $20\,\text{cm}$?
Answer
$9.8$
Show the solution
Given
at a distance of $10\,\text{cm}$ the angle of the needle is $\theta_1 = 60^\circ$, at a distance of $20\,\text{cm}$ the angle is $\theta_2 = 10^\circ$; the field of the magnet is perpendicular to the horizontal component $B_\text{h}$ of the Earth's field
Find
ratio of the fields $k = B_1 : B_2$, a plain number
Principle
When the field of the magnet is perpendicular to the horizontal component of the Earth's field, the angle of the needle satisfies $\tan\theta = B_\text{m} : B_\text{h}$, so $B_\text{m} = B_\text{h}\tan\theta$. The Earth's field is the same at both positions, so it cancels from the ratio and need not be known.
Formula
$B_1 = B_\text{h}\tan\theta_1$ and $B_2 = B_\text{h}\tan\theta_2$, so
$k$$= \dfrac{B_1}{B_2}$$= \dfrac{\tan\theta_1}{\tan\theta_2}$Conversions
Not needed: the ratio is a plain number, and the distances do not appear in the calculation.
Substitution
$k$$= \dfrac{\tan 60^\circ}{\tan 10^\circ}$$= \dfrac{1.732}{0.1763}$$= 9.82$Answer
about $9.8$ times, to two significant figures like the angles $60^\circ$ and $10^\circ$.
Reasonableness
When the distance doubles, the field weakens by much more than a half or a quarter, as the field of a magnet does in the simulation of the chapter. The ratio of the angles $60 : 10 = 6$ would be a wrong answer, because the angle is not directly proportional to the field.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 1-10
A bar magnet lies on a table in the north–south direction with its north pole $\text{N}$ towards the north. A compass is placed east of the magnet at the level of the middle of the magnet, where the field of the magnet is $25\,\mu\text{T}$. The horizontal component of the Earth's magnetic field is $14\,\mu\text{T}$ towards the north. Calculate the north component of the horizontal total field in microtesla, with north as the positive direction and south as the negative.
Which way does the north end of the compass needle point?
Answer
$-11\,\mu\text{T}$
Show the solution
Given
north pole of the magnet towards the north, compass east of the magnet at the level of its middle, field of the magnet there $B_\text{m} = 25\,\mu\text{T}$, horizontal component of the Earth's field $B_\text{h} = 14\,\mu\text{T}$ towards the north
Find
north component $B_y$ of the horizontal total field in microtesla (north positive) and the direction of the needle
Principle
Outside the magnet the field lines run from the north pole to the south pole, so at the side of the magnet at the level of its middle the field is parallel to the magnet and points from the north-pole end to the south-pole end, that is towards the south. It is opposite to the Earth's field, so the fields are added together with their signs. The needle turns to point along the total field, so the sign of the component tells whether the needle points north or south.
Formula
with north positive, $B_y = B_\text{h} - B_\text{m}$
Conversions
Not needed: both fields are in microtesla.
Substitution
$B_y$$= 14\,\mu\text{T} - 25\,\mu\text{T}$$= -11\,\mu\text{T}$Answer
$-11\,\mu\text{T}$
That is $11\,\mu\text{T}$ towards the south, two significant figures like the fields. The sign is negative, so the north end of the needle points south: the needle has turned round.
Reasonableness
The field of the magnet is greater than the Earth's field, so it wins and the total field points in the direction of the field of the magnet. If the field of the magnet were exactly $14\,\mu\text{T}$, the horizontal total field would be zero and the needle would settle in no direction.
- Exercise 1-11
A steel paper clip is drawn several times in the same direction along a pole of a magnet, after which it picks up small pins. When the clip is heated in a flame hot enough and allowed to cool, it no longer picks up pins. Explain both observations using magnetic domains.
Answer
The field of a magnet aligns the magnetic domains of the steel, and in hard steel they stay aligned, so the clip is a magnet. Heating to a high enough temperature mixes up the alignment of the domains, so the clip is no longer a magnet.
Show the solution
Find
why a steel clip drawn along a magnet becomes a magnet and why heating removes the magnetisation
Principle
Steel is ferromagnetic. Its atomic dipoles form magnetic domains whose directions vary in an unmagnetised piece. An external field aligns the domains, and in hard steel the alignment is kept. At a high enough temperature the thermal motion of the atoms destroys the alignment of the atomic dipoles.
Model answer
At first the domains of the clip point in different directions, and their effects cancel. When a pole of the magnet is drawn along the clip, always in the same direction, the field of the magnet turns the domains and makes those that point along the field grow. In hard steel the domains stay aligned when the magnet is taken away, so the clip becomes a permanent magnet with its own north pole and south pole. It magnetises the pins and attracts them. When the clip is heated hot enough (the Curie temperature of iron is about $770\,{}^\circ\text{C}$), the thermal motion of the atoms mixes up the alignment of the atomic dipoles and the domains disappear. When the clip cools, the domains form again but in random directions, so the clip is no longer a magnet.
- Exercise 1-12
According to the model of the Earth's magnetic field (World Magnetic Model 2025) the field in Helsinki is about $53\,\mu\text{T}$ and points about $74^\circ$ below the horizontal. In Longyearbyen in Svalbard the field is about $55\,\mu\text{T}$ and points about $82.5^\circ$ below the horizontal.
How many times greater is the horizontal component of the field in Helsinki than in Longyearbyen? Why does a compass work worse in Longyearbyen, although the whole field is greater there?
Answer
$2.0$
Show the solution
Given
Helsinki $B_1 = 53\,\mu\text{T}$ and $I_1 = 74^\circ$, Longyearbyen $B_2 = 55\,\mu\text{T}$ and $I_2 = 82.5^\circ$ below the horizontal
Find
ratio of the horizontal components $k = B_\text{h1} : B_\text{h2}$, a plain number, and the reason why a compass works worse
Principle
A compass needle turns only in the horizontal plane, so it is steered only by the horizontal component of the field, $B_\text{h} = B\cos I$. The more steeply the field points into the ground, the smaller the horizontal part of it.
Formula
$k$$= \dfrac{B_\text{h1}}{B_\text{h2}}$$= \dfrac{B_1\cos I_1}{B_2\cos I_2}$Conversions
Not needed: both fields are in microtesla, and the unit cancels.
Substitution
$k$$= \dfrac{53\,\mu\text{T} \cdot \cos 74^\circ}{55\,\mu\text{T} \cdot \cos 82.5^\circ}$$= \dfrac{14.6\,\mu\text{T}}{7.18\,\mu\text{T}}$$= 2.03$Answer
about $2.0$ times, two significant figures like the fields $53\,\mu\text{T}$ and $55\,\mu\text{T}$. In Longyearbyen the field is almost vertical, so its horizontal component is only about half of the value in Helsinki.
Model answer (why)
Only the horizontal component turns the needle. In Longyearbyen it is only about half of that in Helsinki, even though the whole field is greater, because the field points almost straight into the ground. The field that turns the needle is weak, so the needle settles slowly and even small disturbances, such as an iron object nearby, turn it easily. The answer is correct if it says that a compass is steered by the horizontal component and not by the whole field.
Reasonableness
The whole field is a little greater in Longyearbyen, but $\cos 82.5^\circ \approx 0.13$ is much smaller than $\cos 74^\circ \approx 0.28$, so the ratio is close to two. At the magnetic pole the field would be vertical, the horizontal component zero and the compass would not work at all.
Photographs
- Horseshoe magnet metal shavings. Maciej J. Mrowinski. source, CC BY-SA 4.0
- Aurora borealis over Saana fell. WikiLucas00. source, CC BY-SA 4.0
