Magnetic flux density

Definition

Magnetic flux density

Magnetic flux density is a measure of the strength of a magnetic field, equal to the force per unit current per unit length acting on a conductor placed at right angles to the field.

Definition

Tesla

The tesla is the unit of magnetic flux density, equal to one newton per ampere metre.

A field-line diagram shows where a field is stronger, but it gives no number. Magnetic flux density, given the symbol $B$ , is the measured quantity that puts a value on that strength.
The name comes from the field-line picture. A field with a high flux density has many field lines packed through each square metre, which is exactly what closely spaced lines represent.
Its unit is the tesla, symbol $T\text{T}$ , which can also be written as the newton per ampere metre: $m)1\ \text{T} = 1\ \text{N}/(\text{A}\,\text{m})$
That second form is worth reading as a sentence. A field of $T1\ \text{T}$ pushes on a $m1\ \text{m}$ length of conductor carrying $A1\ \text{A}$ with a force of $N1\ \text{N}$ .
One tesla is a very strong field. The Earth's field is roughly $T0.00005\ \text{T}$ , a school horseshoe magnet gives about $T0.1\ \text{T}$ , and a hospital scanner magnet reaches a few tesla.
Magnetic flux density is not a force and not an energy. It measures the field itself, and a force only appears once a current is put into that field.

The force equation

For a conductor lying at right angles to a magnetic field, the force on it is given by: $F = B I l$
In this equation:
1. $F$ is the force on the conductor, in newtons, $N\text{N}$
2. $B$ is the magnetic flux density, in tesla, $T\text{T}$
3. $I$ is the current in the conductor, in amperes, $A\text{A}$
4. $l$ is the length of conductor inside the field, in metres, $m\text{m}$
The symbol $l$ is a lower-case letter L, not the digit $1$ . Write it clearly, because a badly formed $l$ costs marks in a substitution line.
All three quantities are multiplied together, so each one is directly proportional to the force while the other two are held constant.
Doubling any one of $B$ , $I$ or $l$ doubles the force. Doubling two of them at once multiplies the force by four.
The two rearranged forms you may need are: $\dfrac{F}{Il} \qquad I = \dfrac{F}{Bl} \qquad l = \dfrac{F}{BI}$
The equation applies only when the conductor is at right angles to the field. A conductor set at some other angle feels a smaller force, and one lying along the field feels none.
Only the length of conductor actually inside the field goes into $l$ . A $cm60\ \text{cm}$ wire passing through a magnet with $cm4\ \text{cm}$ pole faces contributes only $m0.04\ \text{m}$ .
Every length must be converted to metres before substituting, so $cm4\ \text{cm}$ becomes $m0.04\ \text{m}$ and $mm85\ \text{mm}$ becomes $m0.085\ \text{m}$ .
The equation gives the size of the force only. Its direction is a separate question, answered by a hand rule rather than by the arithmetic.

Example

Calculating the force on a wire

A wire carries a current of $A3.0\ \text{A}$ . A $m0.40\ \text{m}$ length of it lies at right angles to a field of magnetic flux density $T0.25\ \text{T}$ .
The force is the unknown, so the equation is used as it stands.
$F = B I l$
Every quantity is already in SI units, so the values can be substituted directly.
$\times 3.0 \times 0.40$
$0.30\ \text{N}$
The force on the wire is $N0.30\ \text{N}$ , which is roughly the weight of a $g30\ \text{g}$ mass.

Example

Rearranging to find flux density

A $m0.20\ \text{m}$ length of wire carrying $A4.0\ \text{A}$ at right angles to a field experiences a force of $N0.48\ \text{N}$ .
$B$ is the unknown, so start from the equation and rearrange before substituting.
$\qquad \text{so} \qquad B = \dfrac{F}{Il}$
$\dfrac{0.48}{4.0 \times 0.20}$
$\dfrac{0.48}{0.80} = 0.60\ \text{T}$
The magnetic flux density is $T0.60\ \text{T}$ .

Example

Using only the length inside the field

A $cm75\ \text{cm}$ wire passes through the gap of a magnet whose pole faces are $cm5.0\ \text{cm}$ long. The flux density in the gap is $T0.080\ \text{T}$ and the current is $A2.5\ \text{A}$ .
Only the part of the wire inside the field is pushed, so $l$ is the length of the pole faces and not the length of the wire.
$5.0\ \text{cm} = 0.050\ \text{m}$
$\times 2.5 \times 0.050$
$0.010\ \text{N}$
Using the full $m0.75\ \text{m}$ would have given $N0.15\ \text{N}$ , fifteen times too large, which is why the length inside the field has to be identified first.

Practical

Measuring the force on a current-carrying conductor

Aim: to measure the force on a conductor in a magnetic field, to show that the force is directly proportional to the current, and to use the results to find the magnetic flux density between the poles.
Apparatus: a digital top-pan balance reading to $g0.01\ \text{g}$ , two slab magnets on a steel yoke, a stiff copper wire bent into a rigid rectangle, a clamp stand, a low-voltage direct current supply, a rheostat, an ammeter, a switch, a $cm30\ \text{cm}$ ruler and connecting leads.
Variables: the current is the independent variable, the change in the balance reading is the dependent variable, and the magnets, the length of wire in the field, the position of the wire in the gap and its right-angled orientation are all kept the same.
Method, setting up:
- Stand the yoke and magnets on the balance pan so that the two pole faces face each other across a horizontal gap, and measure the length of the pole faces with the ruler.
- Clamp the wire rectangle so that its lower horizontal section hangs in the middle of the gap, at right angles to the field and touching nothing.
- Check by eye and with the ruler that the wire is horizontal and centred, because a tilted wire produces a sideways force the balance cannot register.
- Connect the wire in series with the switch, ammeter, rheostat and supply, keeping the leads slack so they cannot press on the magnets.
- Zero the balance with the magnets in place and the switch open, so the reading measures only the extra force produced by the current.
Method, taking readings:
- Close the switch, set the current to $A1.0\ \text{A}$ with the rheostat, read the balance, then open the switch again.
- Check that the balance returns to zero each time. If it does not, the wire is touching the magnets or the leads are pulling on the pan.
- Repeat for currents of $A2.0\ \text{A}$ , $A3.0\ \text{A}$ , $A4.0\ \text{A}$ and $A5.0\ \text{A}$ , taking three readings at each current and calculating a mean.
- Reverse the supply connections and repeat one reading. The balance reading changes sign, confirming that the direction of the force follows the direction of the current.
Processing: convert each mean reading in grams to a force using $F = m g$ with $10\ \text{N/kg}$ , so a reading of $g0.35\ \text{g}$ gives $\times 10 = 0.0035\ \text{N}$ .
Analysis: plot $F$ against $I$ . A straight line through the origin shows that $F$ is directly proportional to $I$ , and comparing $F = B I l$ with $y = m x$ shows that the gradient equals $B l$ , so $B$ is found by dividing the gradient by the measured length.
Expected pattern: the balance reading rises steadily with current, the graph is a straight line through the origin, and a typical pair of school slab magnets gives a flux density of the order of $T0.1\ \text{T}$ .
Watch out: forgetting to zero the balance with the magnets already on the pan, a wire that touches the poles, stiff leads that lift or press the assembly, measuring the whole wire instead of the length in the gap, and the current drifting downwards as the wire warms up.
Safety: the wire heats quickly at several amperes, so close the switch only for as long as it takes to read the balance, keep the rheostat in circuit, and handle the slab magnets carefully because they snap together hard enough to trap fingers.

Exam technique

Setting out a force calculation

Write $F = B I l$ on its own line first. The selection mark is available even if the arithmetic afterwards goes wrong.
List the values with their units before substituting, and do any centimetre to metre conversion on that list rather than in your head.
Rearrange the equation before you put numbers in, not after. Substituting first and then trying to unpick the arithmetic is where most working goes wrong.
Give the answer to the same number of significant figures as the data and always attach a unit: $N\text{N}$ for force, $T\text{T}$ for flux density, $A\text{A}$ for current and $m\text{m}$ for length.
Use the phrase directly proportional only when a graph is a straight line through the origin, and quote the gradient as $B l$ when asked to find $B$ from one.

Common Mistake

Do not substitute a length in centimetres or millimetres. Convert to metres first.
Do not use the total length of the wire when only part of it lies in the field. Use the length inside the field.
Do not give the force in joules or watts. Force is measured in newtons, $N\text{N}$ .
Do not apply this equation when the conductor is not at right angles to the field, since it holds only in that case.
Do not treat $B$ as a force. It measures the strength of the field, and the force also depends on the current and the length.

Self review

State what magnetic flux density measures and give its unit in both accepted forms.
Write down the equation for the force on a conductor at right angles to a magnetic field and give the unit of each quantity.
Calculate the force on a $m0.15\ \text{m}$ length of wire carrying $A6.0\ \text{A}$ at right angles to a field of $T0.20\ \text{T}$ .
Explain what happens to the force when the current is trebled and the length is halved.
State which length must be used for $l$ when a long wire passes through a short magnet.
Describe how a graph of force against current can be used to find the magnetic flux density.

Magnetic flux density

Definition

Magnetic flux density

Magnetic flux density is a measure of the strength of a magnetic field, equal to the force per unit current per unit length acting on a conductor placed at right angles to the field.

Definition

Tesla

The tesla is the unit of magnetic flux density, equal to one newton per ampere metre.

A field-line diagram shows where a field is stronger, but it gives no number. Magnetic flux density, given the symbol $B$ , is the measured quantity that puts a value on that strength.
The name comes from the field-line picture. A field with a high flux density has many field lines packed through each square metre, which is exactly what closely spaced lines represent.
Its unit is the tesla, symbol $T\text{T}$ , which can also be written as the newton per ampere metre: $m)1\ \text{T} = 1\ \text{N}/(\text{A}\,\text{m})$
That second form is worth reading as a sentence. A field of $T1\ \text{T}$ pushes on a $m1\ \text{m}$ length of conductor carrying $A1\ \text{A}$ with a force of $N1\ \text{N}$ .
One tesla is a very strong field. The Earth's field is roughly $T0.00005\ \text{T}$ , a school horseshoe magnet gives about $T0.1\ \text{T}$ , and a hospital scanner magnet reaches a few tesla.
Magnetic flux density is not a force and not an energy. It measures the field itself, and a force only appears once a current is put into that field.

The force equation

For a conductor lying at right angles to a magnetic field, the force on it is given by: $F = B I l$
In this equation:
1. $F$ is the force on the conductor, in newtons, $N\text{N}$
2. $B$ is the magnetic flux density, in tesla, $T\text{T}$
3. $I$ is the current in the conductor, in amperes, $A\text{A}$
4. $l$ is the length of conductor inside the field, in metres, $m\text{m}$
The symbol $l$ is a lower-case letter L, not the digit $1$ . Write it clearly, because a badly formed $l$ costs marks in a substitution line.
All three quantities are multiplied together, so each one is directly proportional to the force while the other two are held constant.
Doubling any one of $B$ , $I$ or $l$ doubles the force. Doubling two of them at once multiplies the force by four.
The two rearranged forms you may need are: $\dfrac{F}{Il} \qquad I = \dfrac{F}{Bl} \qquad l = \dfrac{F}{BI}$
The equation applies only when the conductor is at right angles to the field. A conductor set at some other angle feels a smaller force, and one lying along the field feels none.
Only the length of conductor actually inside the field goes into $l$ . A $cm60\ \text{cm}$ wire passing through a magnet with $cm4\ \text{cm}$ pole faces contributes only $m0.04\ \text{m}$ .
Every length must be converted to metres before substituting, so $cm4\ \text{cm}$ becomes $m0.04\ \text{m}$ and $mm85\ \text{mm}$ becomes $m0.085\ \text{m}$ .
The equation gives the size of the force only. Its direction is a separate question, answered by a hand rule rather than by the arithmetic.

Example

Calculating the force on a wire

A wire carries a current of $A3.0\ \text{A}$ . A $m0.40\ \text{m}$ length of it lies at right angles to a field of magnetic flux density $T0.25\ \text{T}$ .
The force is the unknown, so the equation is used as it stands.
$F = B I l$
Every quantity is already in SI units, so the values can be substituted directly.
$\times 3.0 \times 0.40$
$0.30\ \text{N}$
The force on the wire is $N0.30\ \text{N}$ , which is roughly the weight of a $g30\ \text{g}$ mass.

Example

Rearranging to find flux density

A $m0.20\ \text{m}$ length of wire carrying $A4.0\ \text{A}$ at right angles to a field experiences a force of $N0.48\ \text{N}$ .
$B$ is the unknown, so start from the equation and rearrange before substituting.
$\qquad \text{so} \qquad B = \dfrac{F}{Il}$
$\dfrac{0.48}{4.0 \times 0.20}$
$\dfrac{0.48}{0.80} = 0.60\ \text{T}$
The magnetic flux density is $T0.60\ \text{T}$ .

Example

Using only the length inside the field

A $cm75\ \text{cm}$ wire passes through the gap of a magnet whose pole faces are $cm5.0\ \text{cm}$ long. The flux density in the gap is $T0.080\ \text{T}$ and the current is $A2.5\ \text{A}$ .
Only the part of the wire inside the field is pushed, so $l$ is the length of the pole faces and not the length of the wire.
$5.0\ \text{cm} = 0.050\ \text{m}$
$\times 2.5 \times 0.050$
$0.010\ \text{N}$
Using the full $m0.75\ \text{m}$ would have given $N0.15\ \text{N}$ , fifteen times too large, which is why the length inside the field has to be identified first.

Practical

Measuring the force on a current-carrying conductor

Aim: to measure the force on a conductor in a magnetic field, to show that the force is directly proportional to the current, and to use the results to find the magnetic flux density between the poles.
Apparatus: a digital top-pan balance reading to $g0.01\ \text{g}$ , two slab magnets on a steel yoke, a stiff copper wire bent into a rigid rectangle, a clamp stand, a low-voltage direct current supply, a rheostat, an ammeter, a switch, a $cm30\ \text{cm}$ ruler and connecting leads.
Variables: the current is the independent variable, the change in the balance reading is the dependent variable, and the magnets, the length of wire in the field, the position of the wire in the gap and its right-angled orientation are all kept the same.
Method, setting up:
- Stand the yoke and magnets on the balance pan so that the two pole faces face each other across a horizontal gap, and measure the length of the pole faces with the ruler.
- Clamp the wire rectangle so that its lower horizontal section hangs in the middle of the gap, at right angles to the field and touching nothing.
- Check by eye and with the ruler that the wire is horizontal and centred, because a tilted wire produces a sideways force the balance cannot register.
- Connect the wire in series with the switch, ammeter, rheostat and supply, keeping the leads slack so they cannot press on the magnets.
- Zero the balance with the magnets in place and the switch open, so the reading measures only the extra force produced by the current.
Method, taking readings:
- Close the switch, set the current to $A1.0\ \text{A}$ with the rheostat, read the balance, then open the switch again.
- Check that the balance returns to zero each time. If it does not, the wire is touching the magnets or the leads are pulling on the pan.
- Repeat for currents of $A2.0\ \text{A}$ , $A3.0\ \text{A}$ , $A4.0\ \text{A}$ and $A5.0\ \text{A}$ , taking three readings at each current and calculating a mean.
- Reverse the supply connections and repeat one reading. The balance reading changes sign, confirming that the direction of the force follows the direction of the current.
Processing: convert each mean reading in grams to a force using $F = m g$ with $10\ \text{N/kg}$ , so a reading of $g0.35\ \text{g}$ gives $\times 10 = 0.0035\ \text{N}$ .
Analysis: plot $F$ against $I$ . A straight line through the origin shows that $F$ is directly proportional to $I$ , and comparing $F = B I l$ with $y = m x$ shows that the gradient equals $B l$ , so $B$ is found by dividing the gradient by the measured length.
Expected pattern: the balance reading rises steadily with current, the graph is a straight line through the origin, and a typical pair of school slab magnets gives a flux density of the order of $T0.1\ \text{T}$ .
Watch out: forgetting to zero the balance with the magnets already on the pan, a wire that touches the poles, stiff leads that lift or press the assembly, measuring the whole wire instead of the length in the gap, and the current drifting downwards as the wire warms up.
Safety: the wire heats quickly at several amperes, so close the switch only for as long as it takes to read the balance, keep the rheostat in circuit, and handle the slab magnets carefully because they snap together hard enough to trap fingers.

Exam technique

Setting out a force calculation

Write $F = B I l$ on its own line first. The selection mark is available even if the arithmetic afterwards goes wrong.
List the values with their units before substituting, and do any centimetre to metre conversion on that list rather than in your head.
Rearrange the equation before you put numbers in, not after. Substituting first and then trying to unpick the arithmetic is where most working goes wrong.
Give the answer to the same number of significant figures as the data and always attach a unit: $N\text{N}$ for force, $T\text{T}$ for flux density, $A\text{A}$ for current and $m\text{m}$ for length.
Use the phrase directly proportional only when a graph is a straight line through the origin, and quote the gradient as $B l$ when asked to find $B$ from one.

Common Mistake

Do not substitute a length in centimetres or millimetres. Convert to metres first.
Do not use the total length of the wire when only part of it lies in the field. Use the length inside the field.
Do not give the force in joules or watts. Force is measured in newtons, $N\text{N}$ .
Do not apply this equation when the conductor is not at right angles to the field, since it holds only in that case.
Do not treat $B$ as a force. It measures the strength of the field, and the force also depends on the current and the length.

Self review

State what magnetic flux density measures and give its unit in both accepted forms.
Write down the equation for the force on a conductor at right angles to a magnetic field and give the unit of each quantity.
Calculate the force on a $m0.15\ \text{m}$ length of wire carrying $A6.0\ \text{A}$ at right angles to a field of $T0.20\ \text{T}$ .
Explain what happens to the force when the current is trebled and the length is halved.
State which length must be used for $l$ when a long wire passes through a short magnet.
Describe how a graph of force against current can be used to find the magnetic flux density.

12.3.3 Force on a conductor

Magnetic flux density

The force equation

Calculating the force on a wire

Rearranging to find flux density

Using only the length inside the field

Measuring the force on a current-carrying conductor

Setting out a force calculation

Review 12.3.3 Force on a conductor by teaching Genie

12.3.3 Force on a conductor

Magnetic flux density

The force equation

Calculating the force on a wire

Rearranging to find flux density

Using only the length inside the field

Measuring the force on a current-carrying conductor

Setting out a force calculation

Review 12.3.3 Force on a conductor by teaching Genie

12.3.3 Force on a conductor

Magnetic flux density

The force equation

Calculating the force on a wire

Rearranging to find flux density

Using only the length inside the field

Measuring the force on a current-carrying conductor

Setting out a force calculation

Review 12.3.3 Force on a conductor by teaching Genie

12.3.3 Force on a conductor Revision Guide

12.3.3 Force on a conductor

Magnetic flux density

The force equation

Calculating the force on a wire

Rearranging to find flux density

Using only the length inside the field

Measuring the force on a current-carrying conductor

Setting out a force calculation

Review 12.3.3 Force on a conductor by teaching Genie

12.3.3 Force on a conductor Revision Guide