Electricity and Magnetism 203-SN2-RE • CEGEP, Montreal

Revision sheet: sources of the magnetic field (203-SN2-RE)

Having studied how magnetic fields act on charges and currents, we now learn to produce them. The chapter follows the structure of electrostatics, an elementary law, Biot-Savart, and a global theorem, Ampère's, but the geometry is not the same: the field lines do not leave the source, they surround it.

This sheet gathers the mistakes that cost the most on tests: the field drawn like an electric field, the formulas for an arc and a whole loop confused, the $n$ of the solenoid taken for $N$ , the enclosed current miscounted in a composite conductor, and the finite wire treated as infinite.

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The thread of the chapter

A magnetic field never points toward its source: it CIRCLES around it. Until that reflex is acquired, no superposition is right, and no Amperian loop is well chosen.

This chapter is part of 203-SN2-RE, Electricity and Magnetism

The essentials

The field circles around the current

• Infinite straight wire: $B=\dfrac{\mu_{0}I}{2\pi d}$ with $\mu_{0}=4\pi\times 10^{-7}$ T·m/A, so $\dfrac{\mu_{0}}{2\pi}=2\times 10^{-7}$ in SI units.
• The field lines are CIRCLES centred on the wire: the field is tangent to these circles, hence PERPENDICULAR to the line joining the wire to the point, never directed along that line.
• Right hand: thumb in the direction of the current, fingers curl in the direction of the field.
• Biot-Savart: $dB=\dfrac{\mu_{0}}{4\pi}\dfrac{I\,dl\sin\theta}{r^{2}}$ . A wire element whose EXTENSION passes through the point under study does not contribute, since $\sin\theta=0$ .
• Superposition: the fields of several sources add VECTORIALLY, never as magnitudes.

The field is tangent to the circles centred on the wire: at every point it is perpendicular to the line joining the wire. An electric field, by contrast, would lie ALONG that line.

The difference from electrostatics is not a nuance: an electric field leaves or reaches its source, a magnetic field encircles it. That is also why magnetic field lines always close on themselves.

Two parallel wires: same direction, they attract

• Each wire sits in the field of the other: at wire 2, $B_{1}=\dfrac{\mu_{0}I_{1}}{2\pi d}$ , perpendicular to the plane of the two wires.
• The force per unit length is $\dfrac{F}{L}=\dfrac{\mu_{0}I_{1}I_{2}}{2\pi d}$ , the same on each wire and in opposite directions: this is Newton's third law.
• Currents in the same direction: attraction. Currents in opposite directions: repulsion. This is the reverse of what electric charges suggest.
• Point of zero field: between the wires if the currents are in the same direction; outside, on the side of the weaker one, if they are opposite.

The field

\vec{B}_{1}

of wire 1 is tangent to the circle through wire 2. Same direction: the forces face each other, attraction. Current 2 reversed: they point away from each other, repulsion.

This force defined the ampere until 2019: $2\times 10^{-7}$ N per metre between two wires one metre apart, each carrying one ampere.

The catalogue, with its conditions

• Centre of a loop of radius $R$ : $B=\dfrac{\mu_{0}I}{2R}$ , and $N$ times more for a flat coil of $N$ turns.
• Circular arc of angle $\theta$ in radians: $B=\dfrac{\mu_{0}I\theta}{4\pi R}$ , that is the FRACTION $\dfrac{\theta}{2\pi}$ of the field of the whole loop.
• On the axis of a loop, at a distance $x$ : $B=\dfrac{\mu_{0}IR^{2}}{2(R^{2}+x^{2})^{3/2}}$ .
• Long solenoid: $B=\mu_{0}nI$ inside, with $n=\dfrac{N}{L}$ in turns PER METRE, and a nearly zero field outside. The radius plays no role.
• Toroid of $N$ turns: $B=\dfrac{\mu_{0}NI}{2\pi r}$ inside the toroid, STRICTLY zero in the central hole and outside. Finite-length wire: the infinite-wire result multiplied by $\sin\theta_{max}$ , hence always smaller.

Loop cut by a plane through its axis, current out at the top and in at the bottom: the field crosses it left to right and leaves by the north face. Fingers along the current, thumb to the north.

Every formula carries its geometry in its statement. Writing “loop” or “solenoid” in the margin next to the result avoids applying $\mu_{0}nI$ to a flat coil, which makes no sense.

The solenoid: a straight magnet you can switch on

• In a long solenoid, the field lines are parallel and evenly spaced: the field is uniform, $B=\mu_{0}nI$ , and nearly zero outside.
• The solenoid behaves like a bar magnet: its lines leave by one end, the north pole, and re-enter by the other, the south pole.
• Right hand: fingers curled in the direction of the current in the turns, the extended thumb indicates the interior field, hence the north pole.

Cross-section of a solenoid, current coming out in the top turns: the interior field points to the right, the north pole is on the right. Outside, the lines spread out and the field collapses.

Reversing the current swaps north and south; an iron core multiplies the field by $\mu_{r}$ up to saturation: that is the electromagnet.

Ampère: when it calculates, and when it stays silent

• $\oint\vec{B}\cdot d\vec{l}=\mu_{0}I_{enc}$ , where $I_{enc}$ is the total current ENCLOSED by the chosen closed loop.
• The law is always true, but it only lets you CALCULATE if the symmetry lets you pull $B$ out of the integral: wire, coaxial cable, solenoid, toroid, current sheet.
• Enclosed current is counted with its DIRECTION: two opposite currents enclosed by the same loop subtract.
• In a solid conductor, only the fraction of current crossing the loop counts: $I_{enc}=I\dfrac{r^{2}}{a^{2}}$ for a uniform density.
• In matter, $B=\mu_{r}B_{0}$ , with $\mu_{r}$ from a few hundred to a few thousand for a ferromagnet, but collapsing at saturation, near $2$ T for iron.

Toroid seen from above:

C_{2}

, inside the toroid, encloses the

N

turns, hence

B=\dfrac{\mu_{0}NI}{2\pi r}

C_{1}

, in the hole, encloses none;

C_{3}

, outside, encloses each turn twice in opposite senses:

B=0

Ampère is to $\vec{B}$ what Gauss is to $\vec{E}$ , with the same limit: without symmetry, the equality remains true but silent. A question that gives no symmetry therefore expects an answer about the CIRCULATION, not about the field.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Drawing the magnetic field like an electric field

the whole superposition question, 6 marks

What not to write

“The field created by the wire at point P points from the wire TOWARD P, as for a charge.”

What to write

“The magnetic field is TANGENT to the circle centred on the wire and passing through P, hence perpendicular to the line wire-P. Its direction is found with the right hand, thumb in the direction of the current.”

Why: A wrong direction makes every vector addition that follows wrong, without any calculation being at fault. This is the mistake that separates papers that understood the structure of the chapter from those that copy formulas.

2. Applying the whole-loop formula to an arc

3 marks, and a factor of 2 on everything that follows

What not to write

“The semicircle of radius $5.0$ cm carrying $2.0$ A creates at its centre $B=\dfrac{\mu_{0}I}{2R}=2.5\times 10^{-5}$ T.”

What to write

“A semicircle creates only HALF the field of the complete loop: $B=\dfrac{1}{2}\times\dfrac{\mu_{0}I}{2R}=1.3\times 10^{-5}$ T. The two radial segments, for their part, do not contribute at all.”

Only the arc counts: the two segments point toward

P

, so their

\sin\theta

is zero. The field at the centre is only the fraction

\dfrac{\theta}{2\pi}

of that of the loop.

Why: Biot-Savart is integrated along the wire: the contribution is proportional to the arc length, hence to the angle. And a segment whose extension passes through the point has $\sin\theta=0$ , hence zero contribution.

3. Taking the total number of turns for the number of turns per metre

the whole question, here a factor of 4

What not to write

“The solenoid has $500$ turns and carries $3.0$ A: $B=\mu_{0}NI=4\pi\times 10^{-7}\times 500\times 3.0=1.9\times 10^{-3}$ T.”

What to write

“ $n=\dfrac{N}{L}=\dfrac{500}{0.25}=2000$ turns per metre, so $B=\mu_{0}nI=4\pi\times 10^{-7}\times 2000\times 3.0=7.5\times 10^{-3}$ T.”

Same number of turns per metre, same current: the interior field is the same, even though the second solenoid has twice as many turns. Only the density

n=\dfrac{N}{L}

counts.

Why: The error factor is exactly $\dfrac{1}{L}$ : it goes unnoticed when the length is close to $1$ m and becomes huge on a short solenoid. The check is dimensional, $\mu_{0}NI$ does not have the units of a field.

4. Believing that the field of a solenoid depends on its radius

2 marks, often on a discussion question

What not to write

“The radius of the solenoid is doubled, so the interior field is halved.”

What to write

“ $B=\mu_{0}nI$ contains only the turn density and the current: doubling the radius changes NOTHING about the interior field, as long as the solenoid stays long compared with its diameter.”

Why: Ampère's law dictates it: the rectangular loop encloses a number of turns proportional to its length, regardless of the cross-section. The $\dfrac{1}{r}$ reasoning of the wire does not apply here.

5. Counting all of the cable's current in an outer Amperian loop

the whole question, 6 marks

What not to write

“Outside the coaxial cable, the loop encloses the current $I$ of the central conductor, so $B=\dfrac{\mu_{0}I}{2\pi r}$ .”

What to write

“The loop ALSO encloses the return current $-I$ of the sheath: $I_{enc}=I-I=0$ , so $B=0$ outside the cable.”

The loop

C_{3}

, outside the cable, encloses the core current AND the return current of the sheath:

I_{enc}=I-I=0

, so

B=0

. Only

C_{1}

and

C_{2}

see a field.

Why: That is exactly why a coaxial cable is used: it does not radiate and cannot be disturbed. The enclosed current is counted with its signs, as an algebraic sum, never as absolute values.

6. Looking for a field in the central hole of a toroid

3 marks, and a lost discussion question

What not to write

“The toroid has $800$ turns, so the field at the centre of the hole is strong.”

What to write

“A circular loop drawn in the hole encloses NO current: $B=0$ strictly, in the hole as well as outside the toroid. The field exists only in the toroid itself, where $B=\dfrac{\mu_{0}NI}{2\pi r}$ .”

Why: The toroid is the only device in the course whose field is PERFECTLY confined, with no leakage. That is what makes it the magnetic circuit of transformers.

7. Treating a finite-length wire as an infinite wire

2 marks, more if the question asks you to justify the approximation

What not to write

“The wire is $20$ cm long and the point is $5.0$ cm away: $B=\dfrac{\mu_{0}I}{2\pi d}$ .”

What to write

“On the perpendicular bisector of a finite wire, the infinite-wire result is multiplied by $\sin\theta_{max}=\dfrac{L}{\sqrt{L^{2}+d^{2}}}=0.89$ : the field is therefore $89\,\%$ of the infinite-wire value, and always less.”

Why: The missing portions of the infinite wire would contribute a field in the same direction: removing them can only decrease the result. A value higher than that of the infinite wire is therefore always wrong.

8. Treating relative permeability as a constant of the material

3 marks on the discussion question, the one that asks for a sentence

What not to write

“The iron core has $\mu_{r}=2000$ , so doubling the current doubles the field: $2000$ times the vacuum field, always.”

What to write

“ $\mu_{r}$ is not a constant: iron SATURATES near $2$ T. Beyond that, increasing the current hardly increases the field any more, and the multiplying factor collapses.”

Why: The magnetic domains all end up aligned: there is nothing left to align. This is the physical limit of electromagnets, and the area of the hysteresis loop measures what this back-and-forth costs in energy on each cycle.

Which method to choose

Biot-Savart or Ampère, depending on the symmetry

The geometry of the conductor, before any formula

If infinite wire, coaxial cable, long solenoid, toroid, current sheet → Ampère's law, with the loop that follows the symmetry

Example: circle around a wire, rectangle straddling a solenoid
If loop, circular arc, finite-length wire → Biot-Savart, or directly the integrated formula from the catalogue

Example: arc of angle $\theta$ : the fraction $\dfrac{\theta}{2\pi}$ of the field of the loop
If several conductors → superposition: direction of each field with the right hand, THEN vector sum

Example: two parallel wires, Helmholtz coils
If a point INSIDE a solid conductor → Ampère with the partial enclosed current, $I_{enc}=I\dfrac{r^{2}}{a^{2}}$

Example: the field grows linearly with $r$ in the core of a coaxial cable
If no usable symmetry → Ampère remains true but does not calculate: the question is then about the circulation or the flux

Example: field near a magnet of arbitrary shape
If a core of magnetic material is present → calculate $B_{0}$ in vacuum, then multiply by $\mu_{r}$ , checking for saturation

Example: a solenoid with an iron core hardly exceeds $2$ T

The choice is made on the SYMMETRY, never on apparent difficulty. A statement that mentions a solenoid or a coaxial cable announces Ampère; a statement that mentions a loop or a segment announces Biot-Savart and the catalogue.

Which formula for which geometry

The exact word used by the statement

If “long straight wire”, “infinite conductor” → $B=\dfrac{\mu_{0}I}{2\pi d}$ , circular field lines

Example: $2\times 10^{-7}\dfrac{I}{d}$ in SI units, to be remembered in this form
If “at the centre of the loop”, “flat coil of $N$ turns” → $B=\dfrac{N\mu_{0}I}{2R}$

Example: $R$ is the radius, not the diameter
If “on the axis, at a distance $x$ ” → $B=\dfrac{\mu_{0}IR^{2}}{2(R^{2}+x^{2})^{3/2}}$ , which gives back the centre formula for $x=0$

Example: Helmholtz coils, separated by their radius
If “solenoid”, “long coil” → $B=\mu_{0}nI$ with $n$ in turns per metre, independent of the radius

Example: $500$ turns over $25$ cm give $n=2000$
If “toroid”, “wound ring” → $B=\dfrac{\mu_{0}NI}{2\pi r}$ inside the toroid, zero everywhere else

Example: $N$ is here the TOTAL number of turns
If “arc”, “semicircle”, “quarter circle” → the corresponding fraction of the field of the loop, and the radial segments do not count

Example: a semicircle gives half

Only two formulas contain an $N$ : the flat coil and the toroid, and it is the TOTAL number of turns. Everywhere else, an $n$ means turns per metre. This distinction is worth several marks per exam.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Applying Ampère's law, written for the marker

When to use it: The statement announces a wire, a coaxial cable, a solenoid or a toroid, and asks for the field at a given distance.

1 State the symmetry and deduce the shape of the field lines: “cylindrical symmetry requires circular field lines centred on the axis, with $B$ constant on such a circle.”
2 Choose the loop and SAY so: circle of radius $r$ coaxial with the conductor, or rectangle with one side inside the solenoid and the opposite side outside.
3 Justify pulling $B$ out of the integral: $\vec{B}$ is parallel to $d\vec{l}$ and constant on the loop, hence $\oint\vec{B}\cdot d\vec{l}=B\times 2\pi r$ .
4 Count the ENCLOSED current with its signs, region by region if the conductor is composite.
5 Isolate $B$ , then check continuity at the boundaries between regions and the behaviour at infinity.

Concluding sentence

“By cylindrical symmetry, $\vec{B}$ is tangent to the circle of radius $r$ and of constant magnitude on that circle. Ampère's law gives $B\times 2\pi r=\mu_{0}I_{enc}$ , hence $B=\dfrac{\mu_{0}I_{enc}}{2\pi r}$ , with $I_{enc}$ the current actually enclosed by this loop.”

The trap: Writing $I_{enc}=I$ by reflex. In a solid conductor, only a fraction of the current is enclosed; in a coaxial cable seen from outside, the sum is zero.

Marking: In general 1 mark for the stated symmetry, 1 for the loop, 2 for the circulation correctly written, 2 for the enclosed current, 1 for the result.

Finding where the field of two wires vanishes

When to use it: Two parallel wires carry given currents, and the statement asks where the resulting field is zero.

1 Determine with the right hand the DIRECTION of the field of each wire in the three regions: to the left, between the two, to the right.
2 Eliminate the regions where the two fields have the same direction: they cannot cancel there, and saying so earns the reasoning mark.
3 In the remaining region, write the equality of the MAGNITUDES: $\dfrac{\mu_{0}I_{1}}{2\pi x}=\dfrac{\mu_{0}I_{2}}{2\pi(d-x)}$ , where $\mu_{0}$ and $2\pi$ cancel.
4 Solve, then check that the solution falls in the announced region: if not, it is the region that was wrongly chosen.

Concluding sentence

“The two currents being in the same direction, their fields are opposite ONLY BETWEEN the two wires. The equality $\dfrac{6.0}{x}=\dfrac{2.0}{0.20-x}$ gives $x=0.15$ m from the first wire, a value indeed between $0$ and $0.20$ m.”

The trap: Solving the equation before determining the region. Algebra always provides a solution, including in a region where the field physically cannot vanish.

Marking: 2 marks for the discussion of directions and the choice of region, 2 for the equation, 1 for the solution, 1 for checking that it belongs to the region.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The direction of the field, redrawn with the right hand

On each figure, check that the field drawn is PERPENDICULAR to the line joining the source to the point. A field aligned with that line is an error of kind, not of calculation.

An arrow pointing from the wire toward the point: an electric field has been drawn.
The factor $2\times 10^{-7}$

$\dfrac{\mu_{0}}{2\pi}=2\times 10^{-7}$ exactly: the field of a wire is computed in your head by multiplying $2\times 10^{-7}$ by $\dfrac{I}{d}$ . An order of magnitude that strays from it betrays a power-of-ten error.

$5.0$ A at $2.5$ mm give $4.0\times 10^{-4}$ T, eight times Earth's field: plausible.
The units of the solenoid

Check that the number used in $\mu_{0}nI$ really is in turns per metre. A value below a thousand for a laboratory solenoid is suspect.

An $n=500$ for $500$ turns over $25$ cm: the division by the length was forgotten.
Continuity at the boundaries

In a composite conductor, the expressions for neighbouring regions must give the SAME value at their common boundary.

$\dfrac{\mu_{0}Ir}{2\pi a^{2}}$ and $\dfrac{\mu_{0}I}{2\pi r}$ do coincide at $r=a$ : the calculation is consistent.
The order of magnitude compared with Earth's field

Earth's field is about $5\times 10^{-5}$ T. A laboratory result is counted in tenths or hundredths of a tesla, never in teslas for a simple wire.

A field of $3$ T calculated for a wire at $5$ cm: four orders of magnitude too many.

The typical problem, taken apart

The coaxial cable, region by region

A coaxial cable consists of a solid cylindrical core of radius $a=1.0$ mm carrying a current $I=5.0$ A uniformly distributed, and a conducting sheath between $b=3.0$ mm and $c=4.0$ mm, carrying the same current in the OPPOSITE direction.

Determine the magnetic field in the four regions: $r<a$ , $a<r<b$ , $b<r<c$ and $r>c$ .

The field grows with

r

in the core, decreases as

\dfrac{1}{r}

in the insulator, falls gradually to zero in the sheath, and stays strictly zero beyond: the cable does not radiate.

Step 1

Symmetry and loop: cylindrical symmetry requires circular field lines centred on the axis, with $B$ constant on each. We therefore take as loop a circle of radius $r$ , and $\oint\vec{B}\cdot d\vec{l}=B\times 2\pi r$ in all regions.

Why

The same sentence serves four times: only the value of $I_{enc}$ changes from one region to the next. Writing it once and for all at the top of the paper saves time and earns the method mark.

Step 2

Region $r<a$ : the loop encloses only a fraction of the current, $I_{enc}=I\dfrac{r^{2}}{a^{2}}$ , hence $B=\dfrac{\mu_{0}Ir}{2\pi a^{2}}$ . At $r=0.50$ mm: $I_{enc}=1.25$ A and $B=5.0\times 10^{-4}$ T.

Why

With a uniform current density, the enclosed current is proportional to the AREA, hence to the square of the radius. The field therefore grows linearly, the opposite of the outside behaviour.

Step 3

Region $a<r<b$ : all the core current is enclosed, $I_{enc}=I=5.0$ A, hence $B=\dfrac{\mu_{0}I}{2\pi r}$ . At $r=2.5$ mm: $B=2\times 10^{-7}\times\dfrac{5.0}{2.5\times 10^{-3}}=4.0\times 10^{-4}$ T.

Why

This is the infinite-wire result: seen from outside, a cylindrical conductor behaves exactly like a wire placed on its axis, as a charged sphere behaves like a point charge.

Step 4

Region $b<r<c$ : the loop encloses the WHOLE core and part of the sheath, in the opposite direction: $I_{enc}=I-I\dfrac{r^{2}-b^{2}}{c^{2}-b^{2}}$ . At $r=3.5$ mm: $I_{enc}=5.0-2.32=2.68$ A and $B=1.5\times 10^{-4}$ T.

Why

The two currents subtract because they are in opposite directions: this is where the point of the question is won or lost. The field therefore decreases faster than $\dfrac{1}{r}$ and reaches zero exactly at $r=c$ .

Step 5

Region $r>c$ : $I_{enc}=I-I=0$ , hence $B=0$ , whatever the distance.

Why

This is the property that justifies using a coaxial cable: it radiates nothing outward and is not disturbed by its surroundings. No calculation is needed, only the right count.

Step 6

Continuity check: at $r=a$ , the first expression gives $\dfrac{\mu_{0}I}{2\pi a}$ and so does the second. At $r=b$ and $r=c$ , the neighbouring expressions also coincide, the last giving indeed $0$ .

Why

The four pieces must join: it is the only check available here, and it catches a fraction error in $I_{enc}$ without redoing any calculation.

Step 7

Plausibility check: $4.0\times 10^{-4}$ T is about eight times Earth's field. A consistent order of magnitude for a few amperes at a few millimetres.

Why

The factor $2\times 10^{-7}$ makes this check instantaneous: $\dfrac{I}{d}$ is here $2000$ , so $B$ is $4\times 10^{-4}$ T, in your head, with no calculator.

The conclusion, written out

“The field is $\dfrac{\mu_{0}Ir}{2\pi a^{2}}$ in the core, where it grows linearly to $1.0\times 10^{-3}$ T at $r=a$ ; $\dfrac{\mu_{0}I}{2\pi r}$ in the insulator, where it decreases to $3.3\times 10^{-4}$ T at $r=b$ ; it falls gradually to zero in the sheath since the return current subtracts; and it is strictly ZERO beyond the cable, where $I_{enc}=0$ .”

The classic mistake on this problem: Writing $B=\dfrac{\mu_{0}I}{2\pi r}$ outside the cable, forgetting the return current. The result is a decreasing but nonzero field, which contradicts the very reason for the coaxial cable's existence.

Learn by heart

• $\dfrac{\mu_{0}}{2\pi}=2\times 10^{-7}$ : the field of a wire is $2\times 10^{-7}\dfrac{I}{d}$ , in your head.
• Magnetic field lines are CIRCLES around the current: the field is tangent, never radial.
• Centre of a loop: $\dfrac{\mu_{0}I}{2R}$ . Arc of angle $\theta$ : the fraction $\dfrac{\theta}{2\pi}$ of that field.
• A segment whose extension passes through the point does not contribute.
• Solenoid: $B=\mu_{0}nI$ with $n$ in turns PER METRE, independent of the radius.
• Toroid: $B=\dfrac{\mu_{0}NI}{2\pi r}$ inside, zero in the hole and outside.
• Ampère: $I_{enc}$ is the ENCLOSED current, counted with its signs; it is zero around a complete coaxial cable.
• Finite wire: the infinite wire multiplied by $\sin\theta_{max}$ , hence always less.
• Earth's field $\approx 5\times 10^{-5}$ T, saturation of iron near $2$ T.

Frequently asked questions

Which direction does the magnetic field of a wire point?

It is tangent to the circles centred on the wire, hence perpendicular to the line joining the wire to the point in question. You find it with the right hand: thumb in the direction of the current, fingers curl in the direction of the field. Unlike the electric field, it never points toward its source or away from it.

What is the difference between Ampère's law and the Biot-Savart law?

Biot-Savart gives the contribution of each small piece of circuit and is used for loops, arcs and finite-length wires. Ampère's law relates the circulation of the field around a closed loop to the enclosed current: it is always true but only allows a calculation in highly symmetric geometries.

Does the field of a solenoid depend on its radius?

No. The field inside equals the permeability of free space times the number of turns per metre times the current. Neither the radius nor the total number of turns appears, as long as the solenoid stays long compared with its diameter. A wider solenoid gives the same field, over a larger volume.

Why is the field zero outside a coaxial cable?

Because a loop drawn outside encloses both the outgoing current of the core and the return current of the sheath, which are equal and opposite. The total enclosed current is therefore zero, and so is the field. This is exactly what makes the cable insensitive to disturbances and unable to cause any.

What is the field at the centre of a semicircle carrying current?

Half the field of a complete loop of the same radius, so the permeability of free space times the current, divided by four times the radius. The straight portions that bring the current in, if they point toward the centre, do not contribute at all, because the angle between the current element and the direction of the point is zero there.

Why does an iron core eventually saturate?

Because iron amplifies the field by aligning its magnetic domains. Once they are all aligned, near two teslas, there is nothing left to align and increasing the current hardly increases the field any more. The relative permeability is therefore not a constant of the material, but a value that depends on the applied field.

Practise it

Corrected exercises: Sources of the magnetic field, 203-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Electromagnetic induction Next sheet RC, RL and RLC circuits

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys2-magnetic-field-sources. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: sources of the magnetic field (203-SN2-RE)

The essentials

The field circles around the current

Two parallel wires: same direction, they attract

The catalogue, with its conditions

The solenoid: a straight magnet you can switch on

Ampère: when it calculates, and when it stays silent

The mistakes that cost marks

1. Drawing the magnetic field like an electric field

2. Applying the whole-loop formula to an arc

3. Taking the total number of turns for the number of turns per metre

4. Believing that the field of a solenoid depends on its radius

5. Counting all of the cable's current in an outer Amperian loop

6. Looking for a field in the central hole of a toroid

7. Treating a finite-length wire as an infinite wire

8. Treating relative permeability as a constant of the material

Which method to choose

Biot-Savart or Ampère, depending on the symmetry

Which formula for which geometry

How the answer is expected to be written

Applying Ampère's law, written for the marker

Finding where the field of two wires vanishes

Check before you hand in

The typical problem, taken apart

The coaxial cable, region by region

Learn by heart

Frequently asked questions

Which direction does the magnetic field of a wire point?

What is the difference between Ampère's law and the Biot-Savart law?

Does the field of a solenoid depend on its radius?

Why is the field zero outside a coaxial cable?

What is the field at the centre of a semicircle carrying current?

Why does an iron core eventually saturate?

Corrected exercises: Sources of the magnetic field, 203-SN2-RE

See also

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