Episode 412: The force on a conductor in a magnetic field
Lesson for 16-19
- Activity time 70 minutes
- Level Advanced
Having reminded your students that magnetic fields can be found near permanent magnets and in the presence of an electric current, the next step is to show how the
field can be quantified. Again, students should know that a conductor carrying a current in a magnetic field will experience a force and will probably remember that Fleming's Left Hand Rule can be used to find the direction of that force.
- Demonstrations: Leading to F = BIL (15 minutes)
- Discussion: Factors affecting the force (15 minutes)
- Discussion: Formal definitions (20 minutes)
- Student questions: BIL force calculations (20 minutes)
Demonstration: Leading to F = BIL
Several quick experimental reminders are possible.
These lead on to a further experiment in which the relationship F = BIL can be established.
Discussion: Factors affecting the force
The experiments above lead to the conclusion that the force F on the conductor is proportional to the length of wire in the field, L , the current I and the
strength of the field, represented by the flux density B . (There is also an
angle factor to consider, but we will leave this aside for now.)
Combining these we get F = BIL
(It can help students to refer to this force as the
Students will probably know that electric and gravitational fields are defined as the force on unit charge or mass. So by comparison, B = FIL
, and this gives a way of defining the
magnetic field strength. Physicists refer to this as the B-field or magnetic flux density which has units of N A-1 m-1 or tesla (T).
A field of 1 T is a very strong field. The field between the poles of the Magnadur magnets that are used in the above experiment is about 3 × 10-2 T while the Earth's magnetic field is about 1 × 10-5 T.
If your specification requires, you will need to develop the angle factor seen in the experiment into the mathematical formula:
F = BILsin( θ ).
For the mathematically inclined, it can be shown that the effective length of the wire in the field (i.e. that which is at right angles) is Lsin( θ ). If students find this difficult, then it can be argued that the maximum force occurs when field and current are at right angles,
θ = 90 °
(sin( θ ) = 1),
and that this falls to zero when field and current are parallel,
θ = 0 °
( sin( θ ) = 0 )
Discussion: Formal definitions
Some specifications require a formal definition of magnetic flux density and/or the tesla.
The strength of a magnetic field or magnetic flux density B can be measured by the force per unit current per unit length acting on a current-carrying conductor placed perpendicular to the lines of a uniform magnetic field.
The SI unit of magnetic flux density B is the tesla (T), equal to 1 N A-1 m-1. This is the magnetic flux density if a wire of length 1 m carrying a current of 1 A as a force of 1 N exerted on it in a direction perpendicular to both the flux and the current.
Study of the force between parallel conductors leading to the definition of the ampere may be required. Students may already have seen the effect in your initial experiments but this may need to be repeated here. The effect can be explained by considering the effect of the field produced by one conductor on the other and then reversing the argument.
(The most common alternative approach relies upon field lines only and describes a
catapult effect from regions where the field lines are tightly packed into regions where the lines are more widely spaced.)
The force between parallel conductors forms the basis of the definition of the unit of current, the ampere. A formal definition is not usually required but students should realize that in a current balance (such as was used above) measurement of force and length can be traced back to fundamental SI units (kg, m, s) leaving the current as the only
Some students are likely to be interested in the formal definition which is:
that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible cross-section, and placed 1m apart in a vacuum, would produce a force of 2 × 10-7 newton per metre of length.
Student questions: BIL force calculations
Although the electric motor could be discussed here, it is probably better to leave this until after electromagnetic induction has been covered so that the back emf can be included.