Calculating magnetic force

From the motor effect to a measurable force

A current-carrying conductor produces its own magnetic field. If the conductor is placed in an external magnetic field, the two fields interact and the conductor experiences a force. This is the motor effect. In this lesson, the concern is the size of that force rather than its direction.

The force becomes greater when the external magnetic field is stronger, when the current is larger, or when a greater length of the conductor lies within the field. Magnetic field strength is represented here by magnetic flux density, $B$, measured in tesla (T).

The force equation

For a straight conductor carrying a current at right angles to the magnetic field:

$$F=BIL$$

where:

• $F$ is force in newtons (N)
• $B$ is magnetic flux density in tesla (T)
• $I$ is current in amperes (A)
• $L$ is the within the magnetic field in metres (m).

The right-angle condition matters. This equation gives the full force when the current and field are perpendicular. If the conductor is parallel to the field, the force is zero; the simple equation $F=BIL$ must not be applied as though the full force acted.

OCR identifies this calculation as Higher Tier content.

Choosing the correct length

$L$ does not automatically mean the total length of the wire. It is the length of the current-carrying conductor that is actually within the relevant magnetic field. If a 0.50 m wire passes through a magnetic field for only 0.12 m, use $L=0.12\,\text{m}$.

Lengths must be expressed in metres. Useful conversions include:

$$1\,\text{cm}=0.01\,\text{m}, \qquad 1\,\text{mm}=0.001\,\text{m}$$

Calculating the force

First check that the conductor and field are at right angles. Then identify $B$, $I$ and the length within the field, convert the length to metres, and substitute into $F=BIL$.

For example, a conductor at right angles to a $0.20\,\text{T}$ field carries a current of $3.0\,\text{A}$, with $0.15\,\text{m}$ of conductor in the field. The force is

$$F=0.20\times3.0\times0.15=0.090\,\text{N}.$$

This means the magnetic field exerts a force of $0.090\,\text{N}$ on that section of conductor. The equation finds the magnitude of the force; Fleming’s left-hand rule, studied separately, determines its direction.

Rearranging the equation

The same relationship can be used when a different quantity is unknown:

$$B=\frac{F}{IL}, \qquad I=\frac{F}{BL}, \qquad L=\frac{F}{BI}.$$

Choose the rearrangement containing the unknown alone on one side, then substitute quantities in base units. For instance, if $F$, $B$ and $L$ are known, dividing the force by $BL$ gives the current in amperes.

Reading proportional relationships

Because $F=BIL$, force is directly proportional to each of $B$, $I$ and $L$ when the other quantities remain constant. Therefore:

• doubling $B$ doubles $F$;
• tripling $I$ triples $F$;
• halving $L$ halves $F$;
• doubling both $$ and makes four times as large.

These comparisons assume that the conductor remains at right angles to the field. They are also useful for checking calculations: increasing one factor should not produce a smaller force if everything else is unchanged.

Checking an answer

A reliable answer should state the equation, use tesla, amperes and metres, and give the force in newtons. It is also worth estimating the order of magnitude. If every factor is less than 1, for example, their product must also be less than 1; an answer of hundreds of newtons would signal a likely conversion or calculator error.

Equation and condition

For a current-carrying conductor at right angles to a magnetic field:

$F=BIL$

$F$ in N, in T, in A and in m. OCR assesses this as Higher Tier content.