The size of the force

Definition

Magnetic flux density

A measure of the strength of a magnetic field, measured in tesla (T).

Definition

Tesla

The unit of magnetic flux density, where a field of one tesla exerts a force of one newton on each metre of wire carrying a current of one ampere at right angles to the field.

Definition

Directly proportional

A relationship between two quantities in which multiplying one by a factor multiplies the other by the same factor.

The size of the force on a current-carrying conductor in a magnetic field depends on three quantities.
The magnetic flux density, $B$ , measures how strong the magnetic field is, and a stronger field gives a bigger force.
The current, $I$ , in the conductor sets the strength of the wire's own magnetic field, so a bigger current gives a bigger force.
The length, $L$ , of conductor inside the field matters as well, because every part of the wire in the field is pushed, so a longer length gives a bigger force.
When the conductor is at right angles to the field, the force is given by $F = B I L$ , or $density×current×length\text{force}=\text{magnetic flux density}\times\text{current}\times\text{length}$ .
In this equation $F$ is the force in newtons ( $N\text{N}$ ), $B$ is the magnetic flux density in tesla ( $T\text{T}$ ), $I$ is the current in amperes ( $A\text{A}$ ) and $L$ is the length in metres ( $m\text{m}$ ).
The force is directly proportional to each of the three quantities, so doubling any one of them while the other two stay the same doubles the force.
The equation applies only when the conductor is perpendicular to the magnetic field, which is the arrangement that gives the largest force.
A conductor lying parallel to the field feels no force at all, whatever the values of $B$ , $I$ and $L$ .
The length $L$ is only the part of the conductor that lies inside the magnetic field, not the whole length of wire in the circuit.
The equation gives the size of the force, while Fleming's left-hand rule gives its direction.

Rearranging and converting units

The equation can be rearranged to make any one of the four quantities the subject.
To find the magnetic flux density, divide the force by the current multiplied by the length: $B=FILB=\dfrac{F}{IL}$ .
To find the current, use $I=FBLI=\dfrac{F}{BL}$ .
To find the length of conductor in the field, use $L=FBIL=\dfrac{F}{BI}$ .
The rearranged form $B=FILB=\dfrac{F}{IL}$ shows what one tesla means: a field of $T1\ \text{T}$ exerts a force of $N1\ \text{N}$ on each $m1\ \text{m}$ of wire carrying $A1\ \text{A}$ at right angles to it.
One tesla is a very strong field, and the magnets used in school laboratories produce fields well below $T1\ \text{T}$ , so the forces measured in class are small fractions of a newton.
Lengths given in centimetres or millimetres must be converted to metres before substituting, so $m5.0\ \text{cm}=0.050\ \text{m}$ and $m25\ \text{mm}=0.025\ \text{m}$ .
Currents given in milliamperes must be converted to amperes, so $A250\ \text{mA}=0.25\ \text{A}$ .
Flux densities given in millitesla must be converted to tesla, so $T40\ \text{mT}=0.040\ \text{T}$ .
Very small forces are often written in standard form, so $N0.0045\ \text{N}$ is written as $N4.5\times10^{-3}\ \text{N}$ .

Practical

Measuring the force on a wire

Aim: to investigate how the force on a current-carrying wire in a magnetic field depends on the current in the wire.
Apparatus: a top-pan balance reading to $g0.01\ \text{g}$ , two flat magnets on a U-shaped steel yoke, a stiff straight copper rod, two clamp stands with bosses and clamps, a low-voltage d.c. power supply, a variable resistor, an ammeter, a switch, connecting leads and a ruler.
Method:
- Place the magnets on the yoke with opposite poles facing each other, stand the yoke on the balance and press the tare button so the reading is zero.
- Clamp the copper rod horizontally so it passes through the gap between the poles at right angles to the field, without touching the magnets, the yoke or the balance pan.
- Measure the width of the magnet faces with the ruler, because this is the length of rod inside the field.
- Connect the rod in series with the ammeter, the variable resistor, the switch and the power supply.
- Close the switch, adjust the variable resistor until the ammeter reads $A0.50\ \text{A}$ , and record the balance reading once it has settled.
- Open the switch between readings so the rod does not heat up, and check that the balance returns to zero.
- Increase the current in steps of $A0.50\ \text{A}$ up to $A3.0\ \text{A}$ , recording the balance reading at each step.
- Repeat the whole set of readings twice more and calculate a mean balance reading for each current.
Processing:
- The balance reading is a change in mass caused by the force between the rod and the magnet, so convert it to kilograms and multiply by the gravitational field strength, $N/kgg=10\ \text{N/kg}$ , to find the force.
- Plot force on the $y$ -axis against current on the $x$ -axis and draw a line of best fit.
Variables: the independent variable is the current, the dependent variable is the force, and the control variables are the length of rod in the field, the magnets used and the right angle between the rod and the field.
Results:
- The points lie close to a straight line through the origin, which shows that the force is directly proportional to the current.
- The gradient of the line equals $B L$ , so dividing the gradient by the measured length gives the magnetic flux density between the poles.
- Reversing the current turns a positive balance reading into a negative reading of the same size, because the force on the magnet reverses with the current.
Watch out: a rod that touches the yoke or the balance pushes on it directly and spoils the reading, and a draught or a knock on the bench makes a sensitive balance drift, so tare it again before each set of readings.
Safety: the rod and leads become warm at the largest currents, so switch off between readings and let the rod cool before touching it.

Example

Force on a wire

A wire carries a current of $A2.5\ \text{A}$ at right angles to a magnetic field of flux density $T0.15\ \text{T}$ .
The length of wire inside the field is $cm8.0\ \text{cm}$ , which is $m0.080\ \text{m}$ .
Substitute the values into the equation:

F=BIL=0.15\times2.5\times0.080

F=0.030\ \text{N}

Example

Finding the flux density

A $cm5.0\ \text{cm}$ length of wire carrying $A3.2\ \text{A}$ at right angles to a magnetic field feels a force of $N0.024\ \text{N}$ .
Convert the length into metres: $m5.0\ \text{cm}=0.050\ \text{m}$ .
Rearrange the equation to make $B$ the subject and substitute:

B=\frac{F}{IL}=\frac{0.024}{3.2\times0.050}=\frac{0.024}{0.16}

B=0.15\ \text{T}

Example

Finding the current

A $mm25\ \text{mm}$ length of wire lies at right angles to a $mT60\ \text{mT}$ field and feels a force of $N9.0\times10^{-3}\ \text{N}$ .
Convert the units: $T60\ \text{mT}=0.060\ \text{T}$ and $m25\ \text{mm}=0.025\ \text{m}$ .
Rearrange the equation to make $I$ the subject and substitute:

I=\frac{F}{BL}=\frac{9.0\times10^{-3}}{0.060\times0.025}

I=6.0\ \text{A}

Common Mistake

Do not use the total length of wire in the circuit, because $L$ is only the length that lies inside the magnetic field.
Do not apply $F = B I L$ to a wire that is parallel to the field, because the force on it is zero.

Exam technique

Calculating the force on a conductor

The equation $F = B I L$ is one you select and apply rather than recall, so the marks come from choosing it, rearranging it and handling the units correctly.
Write the equation, then the substituted values, then the answer with its unit, because each stage can earn a mark even if a later step goes wrong.
Convert centimetres, millimetres, milliamperes and millitesla before substituting, since a missed conversion changes the answer by a factor of $100$ or $1000$ .
When a question asks what happens to the force if one quantity changes, use proportionality, so tripling the current triples the force and halving the length halves it.

Self review

State the equation that links force, magnetic flux density, current and length.
What is the unit of magnetic flux density?
Calculate the force on $m0.20\ \text{m}$ of wire carrying $A4.0\ \text{A}$ at right angles to a $T0.050\ \text{T}$ field.
What happens to the force if the current is halved and the length in the field is doubled?
What is the force on a wire that lies parallel to the magnetic field lines?

The size of the force

Definition

Magnetic flux density

A measure of the strength of a magnetic field, measured in tesla (T).

Definition

Tesla

The unit of magnetic flux density, where a field of one tesla exerts a force of one newton on each metre of wire carrying a current of one ampere at right angles to the field.

Definition

Directly proportional

A relationship between two quantities in which multiplying one by a factor multiplies the other by the same factor.

The size of the force on a current-carrying conductor in a magnetic field depends on three quantities.
The magnetic flux density, $B$ , measures how strong the magnetic field is, and a stronger field gives a bigger force.
The current, $I$ , in the conductor sets the strength of the wire's own magnetic field, so a bigger current gives a bigger force.
The length, $L$ , of conductor inside the field matters as well, because every part of the wire in the field is pushed, so a longer length gives a bigger force.
When the conductor is at right angles to the field, the force is given by $F = B I L$ , or $density×current×length\text{force}=\text{magnetic flux density}\times\text{current}\times\text{length}$ .
In this equation $F$ is the force in newtons ( $N\text{N}$ ), $B$ is the magnetic flux density in tesla ( $T\text{T}$ ), $I$ is the current in amperes ( $A\text{A}$ ) and $L$ is the length in metres ( $m\text{m}$ ).
The force is directly proportional to each of the three quantities, so doubling any one of them while the other two stay the same doubles the force.
The equation applies only when the conductor is perpendicular to the magnetic field, which is the arrangement that gives the largest force.
A conductor lying parallel to the field feels no force at all, whatever the values of $B$ , $I$ and $L$ .
The length $L$ is only the part of the conductor that lies inside the magnetic field, not the whole length of wire in the circuit.
The equation gives the size of the force, while Fleming's left-hand rule gives its direction.

Rearranging and converting units

The equation can be rearranged to make any one of the four quantities the subject.
To find the magnetic flux density, divide the force by the current multiplied by the length: $B=FILB=\dfrac{F}{IL}$ .
To find the current, use $I=FBLI=\dfrac{F}{BL}$ .
To find the length of conductor in the field, use $L=FBIL=\dfrac{F}{BI}$ .
The rearranged form $B=FILB=\dfrac{F}{IL}$ shows what one tesla means: a field of $T1\ \text{T}$ exerts a force of $N1\ \text{N}$ on each $m1\ \text{m}$ of wire carrying $A1\ \text{A}$ at right angles to it.
One tesla is a very strong field, and the magnets used in school laboratories produce fields well below $T1\ \text{T}$ , so the forces measured in class are small fractions of a newton.
Lengths given in centimetres or millimetres must be converted to metres before substituting, so $m5.0\ \text{cm}=0.050\ \text{m}$ and $m25\ \text{mm}=0.025\ \text{m}$ .
Currents given in milliamperes must be converted to amperes, so $A250\ \text{mA}=0.25\ \text{A}$ .
Flux densities given in millitesla must be converted to tesla, so $T40\ \text{mT}=0.040\ \text{T}$ .
Very small forces are often written in standard form, so $N0.0045\ \text{N}$ is written as $N4.5\times10^{-3}\ \text{N}$ .

Practical

Measuring the force on a wire

Aim: to investigate how the force on a current-carrying wire in a magnetic field depends on the current in the wire.
Apparatus: a top-pan balance reading to $g0.01\ \text{g}$ , two flat magnets on a U-shaped steel yoke, a stiff straight copper rod, two clamp stands with bosses and clamps, a low-voltage d.c. power supply, a variable resistor, an ammeter, a switch, connecting leads and a ruler.
Method:
- Place the magnets on the yoke with opposite poles facing each other, stand the yoke on the balance and press the tare button so the reading is zero.
- Clamp the copper rod horizontally so it passes through the gap between the poles at right angles to the field, without touching the magnets, the yoke or the balance pan.
- Measure the width of the magnet faces with the ruler, because this is the length of rod inside the field.
- Connect the rod in series with the ammeter, the variable resistor, the switch and the power supply.
- Close the switch, adjust the variable resistor until the ammeter reads $A0.50\ \text{A}$ , and record the balance reading once it has settled.
- Open the switch between readings so the rod does not heat up, and check that the balance returns to zero.
- Increase the current in steps of $A0.50\ \text{A}$ up to $A3.0\ \text{A}$ , recording the balance reading at each step.
- Repeat the whole set of readings twice more and calculate a mean balance reading for each current.
Processing:
- The balance reading is a change in mass caused by the force between the rod and the magnet, so convert it to kilograms and multiply by the gravitational field strength, $N/kgg=10\ \text{N/kg}$ , to find the force.
- Plot force on the $y$ -axis against current on the $x$ -axis and draw a line of best fit.
Variables: the independent variable is the current, the dependent variable is the force, and the control variables are the length of rod in the field, the magnets used and the right angle between the rod and the field.
Results:
- The points lie close to a straight line through the origin, which shows that the force is directly proportional to the current.
- The gradient of the line equals $B L$ , so dividing the gradient by the measured length gives the magnetic flux density between the poles.
- Reversing the current turns a positive balance reading into a negative reading of the same size, because the force on the magnet reverses with the current.
Watch out: a rod that touches the yoke or the balance pushes on it directly and spoils the reading, and a draught or a knock on the bench makes a sensitive balance drift, so tare it again before each set of readings.
Safety: the rod and leads become warm at the largest currents, so switch off between readings and let the rod cool before touching it.

Example

Force on a wire

A wire carries a current of $A2.5\ \text{A}$ at right angles to a magnetic field of flux density $T0.15\ \text{T}$ .
The length of wire inside the field is $cm8.0\ \text{cm}$ , which is $m0.080\ \text{m}$ .
Substitute the values into the equation:

F=BIL=0.15\times2.5\times0.080

F=0.030\ \text{N}

Example

Finding the flux density

A $cm5.0\ \text{cm}$ length of wire carrying $A3.2\ \text{A}$ at right angles to a magnetic field feels a force of $N0.024\ \text{N}$ .
Convert the length into metres: $m5.0\ \text{cm}=0.050\ \text{m}$ .
Rearrange the equation to make $B$ the subject and substitute:

B=\frac{F}{IL}=\frac{0.024}{3.2\times0.050}=\frac{0.024}{0.16}

B=0.15\ \text{T}

Example

Finding the current

A $mm25\ \text{mm}$ length of wire lies at right angles to a $mT60\ \text{mT}$ field and feels a force of $N9.0\times10^{-3}\ \text{N}$ .
Convert the units: $T60\ \text{mT}=0.060\ \text{T}$ and $m25\ \text{mm}=0.025\ \text{m}$ .
Rearrange the equation to make $I$ the subject and substitute:

I=\frac{F}{BL}=\frac{9.0\times10^{-3}}{0.060\times0.025}

I=6.0\ \text{A}

Common Mistake

Do not use the total length of wire in the circuit, because $L$ is only the length that lies inside the magnetic field.
Do not apply $F = B I L$ to a wire that is parallel to the field, because the force on it is zero.

Exam technique

Calculating the force on a conductor

The equation $F = B I L$ is one you select and apply rather than recall, so the marks come from choosing it, rearranging it and handling the units correctly.
Write the equation, then the substituted values, then the answer with its unit, because each stage can earn a mark even if a later step goes wrong.
Convert centimetres, millimetres, milliamperes and millitesla before substituting, since a missed conversion changes the answer by a factor of $100$ or $1000$ .
When a question asks what happens to the force if one quantity changes, use proportionality, so tripling the current triples the force and halving the length halves it.

Self review

State the equation that links force, magnetic flux density, current and length.
What is the unit of magnetic flux density?
Calculate the force on $m0.20\ \text{m}$ of wire carrying $A4.0\ \text{A}$ at right angles to a $T0.050\ \text{T}$ field.
What happens to the force if the current is halved and the length in the field is doubled?
What is the force on a wire that lies parallel to the magnetic field lines?

4.2.3 The force on a current-carrying conductor

The size of the force

Rearranging and converting units

Measuring the force on a wire

Force on a wire

Finding the flux density

Finding the current

Calculating the force on a conductor

Review 4.2.3 The force on a current-carrying conductor by teaching Genie

4.2.3 The force on a current-carrying conductor

The size of the force

Rearranging and converting units

Measuring the force on a wire

Force on a wire

Finding the flux density

Finding the current

Calculating the force on a conductor

Review 4.2.3 The force on a current-carrying conductor by teaching Genie

4.2.3 The force on a current-carrying conductor

The size of the force

Rearranging and converting units

Measuring the force on a wire

Force on a wire

Finding the flux density

Finding the current

Calculating the force on a conductor

Review 4.2.3 The force on a current-carrying conductor by teaching Genie

4.2.3 The force on a current-carrying conductor Revision Guide

4.2.3 The force on a current-carrying conductor

The size of the force

Rearranging and converting units

Measuring the force on a wire

Force on a wire

Finding the flux density

Finding the current

Calculating the force on a conductor

Review 4.2.3 The force on a current-carrying conductor by teaching Genie

4.2.3 The force on a current-carrying conductor Revision Guide