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Electromagnetic field · The magnetic field

Fields of wires, loops, solenoids and toroids

In 1820 Hans Christian Oersted discovered, almost by accident, that an electric current deflected the needle of a nearby compass. It was the first proof that electricity and magnetism are the same thing seen from two different angles: every moving charge creates a magnetic field.

Before you start, make sure you know: Electric and magnetic phenomena (3rd ESO)
Before you start, make sure you know: Magnitude and direction of a vector. Applications (1st Bachillerato)

The straight wire and the right-hand rule

If we sprinkle iron filings on a sheet of card pierced by a wire carrying a current, they neither fall towards the wire nor fly away from it, but trace perfect closed circles around it: how does the strength of that invisible field depend on the current and on the distance from the wire?

Straight vertical current-carrying wire with current I upwards passing through a horizontal sheet of card. On the card, the magnetic field lines are circles concentric with the wire, travelling anticlockwise when viewed from above. The vector B is tangent to them at a distance r from the wire. An inset shows the view from above and the right-hand rule. r B I conducting wire very long card perpendicular to the wire field lines: concentric circles B = μ₀·I / (2π·r) view from above I comes out towards you B: anticlockwise right hand: thumb → I, fingers → B
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An upward electric current flows through a vertical wire. a) State in your own words the right-hand rule that relates the direction of the current to the direction of…

An upward electric current flows through a vertical wire.

  1. a) State in your own words the right-hand rule that relates the direction of the current to the direction of the magnetic field lines it creates.
  2. b) Apply it: viewed from above, do the field lines rotate clockwise or anticlockwise? Draw the situation in your exercise book.

You can do it on paper and send a photo with your answer.

Solve
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A current $I = 10\ \text{A}$ flows through a very long straight wire. The magnetic field it creates at a distance $r$ is $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$, with…

A current $I = 10\ \text{A}$ flows through a very long straight wire. The magnetic field it creates at a distance $r$ is $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$, with $\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$. Calculate the field at $r = 5\ \text{cm}$ from the wire. Remember to convert the distance to metres.

Solve
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The field created by a very long straight wire is $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$. Without doing numerical calculations, reason using the formula: a) What happens…

The field created by a very long straight wire is $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$. Without doing numerical calculations, reason using the formula:

  1. a) What happens to $B$ if the current is doubled while keeping the distance constant?
  2. b) What if the distance is doubled while keeping the current constant?
  3. c) Where is the field stronger, near or far from the wire? Relate this to how its field lines spread apart with distance.
Solve

The field depends on $I$ and on $r$

Magnetic field of a straight wire: at distance r the field is B(r) and at distance 2r it is half as strong, with field lines further apart as you move away from the wire. wire B is perpendicular to the plane; its values are shown r 2r B(r) B(2r) = B(r)/2
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A laboratory probe measures the magnetic field at $r = 4\ \text{cm}$ from a very long straight wire and obtains $B = 2\cdot10^{-4}\ \text{T}$. a) Rearrange for the…

A laboratory probe measures the magnetic field at $r = 4\ \text{cm}$ from a very long straight wire and obtains $B = 2\cdot10^{-4}\ \text{T}$.

  1. a) Rearrange for the current in the expression $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$.
  2. b) Calculate its value ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
Solve
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The pantograph cable of an electric train can carry currents of $I = 500\ \text{A}$. a) Rearrange for the distance $r$ in the formula for the field of a wire,…

The pantograph cable of an electric train can carry currents of $I = 500\ \text{A}$.

  1. a) Rearrange for the distance $r$ in the formula for the field of a wire, $B = \frac{\mu_0\cdot I}{2\pi\cdot r}$.
  2. b) Calculate at what distance from the cable the field it creates equals the Earth's magnetic field, $B_T \approx 5\cdot10^{-5}\ \text{T}$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
  3. c) Calculate the field at $10\ \text{m}$ from the cable and comment on whether it could affect a compass placed at that distance.
Solve
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In the laboratory, the magnetic field created by a very long wire is measured at several distances, with the following results: at $r = 20\ \text{cm}$ we obtain…

In the laboratory, the magnetic field created by a very long wire is measured at several distances, with the following results: at $r = 20\ \text{cm}$ we obtain $B = 2\cdot10^{-5}\ \text{T}$; at $r = 10\ \text{cm}$, $B = 4\cdot10^{-5}\ \text{T}$; at $r = 5\ \text{cm}$, $B = 8\cdot10^{-5}\ \text{T}$.

  1. a) Plot $B$ against $\frac{1}{r}$ in your notebook and explain what type of relationship appears.
  2. b) The gradient of the above line is $\frac{\mu_0\cdot I}{2\pi}$; estimate it from the data and deduce the value of the current $I$ flowing through the wire ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
Solve

The field at the centre of a loop

When we bend a straight wire into a circle, the circular field lines from each section build up and pass together through the central gap in the same direction: how strong is the field at the centre of that loop?

Circular loop seen in perspective, carrying a current I anticlockwise, with the magnetic field B perpendicular to the plane and coming out at its centre. I R B (coming out, towards you) centre

If you curl the fingers of your right hand in the direction of the current, the thumb gives the direction of the field at the centre: this tells you, without any calculation, which face of the loop acts as a north pole.

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A circular loop of radius $R = 10\ \text{cm}$ carries a current $I = 2\ \text{A}$. The magnetic field at its centre is $B = \frac{\mu_0\cdot I}{2\cdot R}$, with…

A circular loop of radius $R = 10\ \text{cm}$ carries a current $I = 2\ \text{A}$. The magnetic field at its centre is $B = \frac{\mu_0\cdot I}{2\cdot R}$, with $\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$.

  1. a) Calculate its value.
  2. b) Express the result in standard form, rounded to two significant figures.
Solve
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An anticlockwise current, viewed from above, flows through a circular loop placed horizontally. The loop, viewed from above, is shown in the figure. a) Apply the…

An anticlockwise current, viewed from above, flows through a circular loop placed horizontally. The loop, viewed from above, is shown in the figure.

I B = ? top view
  1. a) Apply the right-hand rule and state the direction of the magnetic field at the centre of the loop: upwards or downwards?
  2. b) Reason which face of the loop (the top face or the bottom face) acts as a north pole.
Solve
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A flat coil with $N = 50$ turns has a radius $R = 8\ \text{cm}$ and carries a current $I = 0,3\ \text{A}$. The field at its centre is…

A flat coil with $N = 50$ turns has a radius $R = 8\ \text{cm}$ and carries a current $I = 0,3\ \text{A}$. The field at its centre is $B = \frac{\mu_0\cdot N\cdot I}{2\cdot R}$.

  1. a) Calculate its value ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
  2. b) How many times greater is it than the field of a single loop of the same radius carrying the same current? Reason your answer using the formula.
Solve

The solenoid: a uniform field inside

When many loops are lined up to form a tube, iron filings reveal a pattern of straight, parallel and very dense lines inside it, identical to that of a cylindrical magnet: what does the strength of that uniform internal field depend on?

Longitudinal section of a solenoid: the current comes out of the page in the top row of loops and goes into the page in the bottom row. Inside, the field lines are straight, parallel and dense, and the field is uniform. Outside, the lines close up far apart and the field is almost zero. outside: field almost zero (lines far apart) ⊙ I comes out of the page ⊗ I goes into the page inside: uniform B, B = μ₀·n·I
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A solenoid has $n = 1000\ \text{espiras/m}$ and carries a current $I = 0,5\ \text{A}$. The magnetic field inside it is $B = \mu_0\cdot n\cdot I$, with…

A solenoid has $n = 1000\ \text{espiras/m}$ and carries a current $I = 0,5\ \text{A}$. The magnetic field inside it is $B = \mu_0\cdot n\cdot I$, with $\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$.

  1. a) Calculate its value.
  2. b) Express it in milliteslas ($1\ \text{mT} = 10^{-3}\ \text{T}$).
Solve
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Describe the magnetic field inside a long solenoid carrying a current: a) What are the field lines like in the central area of the interior: curved, parallel or…

Describe the magnetic field inside a long solenoid carrying a current:

  1. a) What are the field lines like in the central area of the interior: curved, parallel or converging?
  2. b) As we move through the interior, does the field increase, decrease or remain practically constant?
  3. c) What happens to the field on the outside compared to the inside?
Solve
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In the laboratory, $N = 400$ turns are wound around a tube of length $L = 20\ \text{cm}$ and a current $I = 1,5\ \text{A}$ is passed through it. a) Calculate the turn…

In the laboratory, $N = 400$ turns are wound around a tube of length $L = 20\ \text{cm}$ and a current $I = 1,5\ \text{A}$ is passed through it.

  1. a) Calculate the turn density $n = \frac{N}{L}$, in turns per metre.
  2. b) Calculate the field inside the solenoid, $B = \mu_0\cdot n\cdot I$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
Solve
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We want to construct a solenoid that generates a field of $B = 2,5\cdot10^{-3}\ \text{T}$ inside it with a maximum current of $I = 2\ \text{A}$. a) Rearrange for the…

We want to construct a solenoid that generates a field of $B = 2,5\cdot10^{-3}\ \text{T}$ inside it with a maximum current of $I = 2\ \text{A}$.

  1. a) Rearrange for the turn density $n$ in the expression $B = \mu_0\cdot n\cdot I$.
  2. b) Calculate how many turns per metre are required ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
  3. c) If the tube on which it is wound measures $40\ \text{cm}$, how many turns will need to be wound in total?
Solve

The toroid

Top view of a toroid: a ring-shaped core with loops of current I wound around it. The field lines B are closed circles inside the core, at a distance r from the central axis; outside the toroid the field is zero. central axis r B I I loops (current I) toroid core field line B outside: B = 0 Since B depends on 1/r, it is slightly stronger near the inner edge

Unlike the straight solenoid, the toroid has no "ends" where the field weakens: the approximation is excellent throughout its interior, although the field is not perfectly uniform (it is slightly stronger near the inner edge than near the outer one, because $r$ is smaller there).

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A toroid is a coil wound in the shape of a ring. The field inside it, at a distance $r$ from the axis, is $B = \frac{\mu_0\cdot N\cdot I}{2\pi\cdot r}$, where $N$ is the…

A toroid is a coil wound in the shape of a ring. The field inside it, at a distance $r$ from the axis, is $B = \frac{\mu_0\cdot N\cdot I}{2\pi\cdot r}$, where $N$ is the total number of turns. A toroid with $N = 500$ turns carries a current $I = 2\ \text{A}$. Calculate the field at points $r = 10\ \text{cm}$ from the axis ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).

Solve
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A toroid with $N = 1000$ turns has a mean radius of $15\ \text{cm}$. a) Rearrange for the current in the expression $B = \frac{\mu_0\cdot N\cdot I}{2\pi\cdot r}$. b)…

A toroid with $N = 1000$ turns has a mean radius of $15\ \text{cm}$.

  1. a) Rearrange for the current in the expression $B = \frac{\mu_0\cdot N\cdot I}{2\pi\cdot r}$.
  2. b) Calculate the current required for the field inside to be $B = 4\cdot10^{-3}\ \text{T}$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).
Solve
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A toroid with $N = 600$ turns has an inner radius of $8\ \text{cm}$ and an outer radius of $12\ \text{cm}$, and carries a current $I = 5\ \text{A}$. The field inside it,…

A toroid with $N = 600$ turns has an inner radius of $8\ \text{cm}$ and an outer radius of $12\ \text{cm}$, and carries a current $I = 5\ \text{A}$. The field inside it, at a distance $r$ from the axis, is $B = \frac{\mu_0\cdot N\cdot I}{2\pi\cdot r}$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$). Observe the cross-section of the toroid in the figure.

axis r = 8 cm r = 12 cm mean radius = 10 cm B = ? N = 600 turns · I = 5 A
  1. a) Calculate the field next to the inner edge ($r = 8\ \text{cm}$) and next to the outer edge ($r = 12\ \text{cm}$).
  2. b) Calculate the field at the mean radius ($r = 10\ \text{cm}$) and check that it lies between the two previous values.
  3. c) In light of the results, is the field inside a toroid uniform? Why, then, is the mean radius approximation used?
Solve

Superposition: several fields at once

Two parallel wires with currents in the same direction, coming out of the plane, and the opposite magnetic fields B₁ and B₂ that they create at the midpoint P, where they cancel. I₁ I₂ comes out of the plane comes out of the plane P B₁ B₂ d d P: Btotal = B₁ + B₂ (vectors) = 0
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Two straight parallel conductors, separated by $20\ \text{cm}$, carry currents in the same direction $I_1 = I_2 = 10\ \text{A}$ (…

Two straight parallel conductors, separated by $20\ \text{cm}$, carry currents in the same direction $I_1 = I_2 = 10\ \text{A}$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$). The arrangement, in the plane perpendicular to the wires, is shown in the figure.

· I₁ = 10 A · I₂ = 10 A midpoint B total = ? d = 20 cm · current out of the plane
  1. a) Calculate the magnitude of the field created by each wire at the midpoint of the line segment joining them.
  2. b) Using the right-hand rule, determine the direction of each of these fields at the midpoint and justify why the total field there is zero.
  3. c) If the currents were different, would the field at the midpoint still be zero? Reason your answer.
Solve
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Two straight parallel conductors, separated by $d = 40\ \text{cm}$, carry currents in the same direction, $I_1 = 30\ \text{A}$ and $I_2 = 10\ \text{A}$ (…

Two straight parallel conductors, separated by $d = 40\ \text{cm}$, carry currents in the same direction, $I_1 = 30\ \text{A}$ and $I_2 = 10\ \text{A}$ ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$). Observe the diagram:

· I₁ = 30 A · I₂ = 10 A P x = ? d = 40 cm · current out of the plane
  1. a) Using the right-hand rule, explain in which region (between the wires or outside them) their magnetic fields have opposite directions.
  2. b) Set up the condition $B_1 = B_2$ for a point located at a distance $x$ from the first wire, within the segment joining them, and solve the resulting equation.
  3. c) Check the consistency of the result: is the point closer to the wire carrying the smaller current? Why must this be the case?
Solve
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A very long straight wire, lying in the plane of a circular loop of radius $R = 10\ \text{cm}$, passes at $d = 5\ \text{cm}$ from its centre. A current of…

A very long straight wire, lying in the plane of a circular loop of radius $R = 10\ \text{cm}$, passes at $d = 5\ \text{cm}$ from its centre. A current of $I_1 = 10\ \text{A}$ flows through the wire and $I_2 = 4\ \text{A}$ through the loop, with directions chosen such that their magnetic fields at the centre of the loop have opposite directions ($\mu_0 = 4\pi\cdot10^{-7}\ \text{T}\cdot\text{m/A}$).

Diagram of a coplanar straight wire and circular loop carrying electric currentsd = 5 cmR = 10 cmOBtotal = ?I₁ = 10 AStraight wireI₂ = 4 ACircular loop
  1. a) Calculate the magnitude of the wire's magnetic field at the centre of the loop.
  2. b) Calculate the magnitude of the loop's magnetic field at its centre.
  3. c) Justify why both fields are perpendicular to the plane of the loop and calculate the total magnetic field at the centre.
Solve

Back to the beginning: Oersted's compass needle lined up perpendicular to the wire because the field lines form circles concentric with the current; the needle simply aligned itself with the direction tangent to that circle.

Summary

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