__
Magnetic Fields Tutorial 2 - Coils in Magnetic
Fields__

In the diagram below, we have a coil of wire, of n turns, suspended vertically in a magnetic field of strength B. It can rotate around the vertical axis.

A
current *I*
passes through the coil. We will look at the coil from
above. If we set the coil parallel to the field, we will get this:

The w term is the width of the coil. The n term refers to the number of turns on the coil.

The sides AC and BD are vertical, and will experience a force:

F = n(BIl)

The torque on the couple is given by:

G = Fw

The strange looking symbol, G, which looks a bit like a gallows, is “Gamma”, a Greek capital letter ‘G’. It is often used as the Physics code for torque. In some texts, it is written as t (tau).

Now the coil is at an angle q:

Because the force
F
is always at 90^{o} to
the magnetic field, the distance between the lines of action of the forces,
d,
reduces. Therefore the resultant torque reduces. In the equation
below we will call it
G '
("gamma-prime") to distinguish it from the original torque,
G,

G
′ = __Fw__ cos
q = BIlnw
cos q

Since length × width, lw, gives area, A, we can now write:

G ′ = BAnI cos q

When the coil is at 90^{o} to
the field,
cos
q
= 0. The magnitude of each of
the two forces are still F
N , but are in opposite directions, giving a **resultant force** of 0.
Therefore the torque is 0.

The bear-trap here is to think that the side AB is being acted on by the magnetic field. Remember that the force is acting on the vertical wires, AC and BD. We are looking down at the coil from above. The wires running between A and B (and C and D) are parallel with the magnetic field. Therefore they experience no force. |

A light coil of dimensions
3.0 × 5.0 cm with 500 turns is placed in a magnetic field of flux
density 14 mT. A steady current of 0.25 A flows through it. Calculate: |

The main application of this is with the **
electric motor**. The **universal motor** is found in a variety of
different household appliances. This one is from a vacuum cleaner.

Photograph by Marrrci from Wikimedia Commons. Captions translated from the original German.

Each
of the coils is connected to the outside circuit by a split-ring **commutator**
and spring-loaded carbon **brushes**.

Notice that the field magnets are curved. If
the field is radial, the angle that the coil makes with the magnetic field is
always constant at 0^{o}. In other words, the coil is always parallel
to the magnetic field lines through which it is turning. Therefore
F
remains constant,
and the torque remains constant.

This motor works
on** direct current**. You may well have made something similar in
class

The current is carried to the armature through carbon brushes and a split-ring commutator. The commutator acts as a change-over switch so that the current on the left hand side of the coil always moves towards us (and on the right hand side, away).

The armature
turns through 90 ^{o} and the torque is zero.

In this case, we see that the armature is at 90 degrees, so that the current is at 0 degrees to the field.

There is zero torque, and the motor stalls.

To prevent this, small DC motors have 3 poles so that at least 1 pole is providing a force. The motor keeps on turning. If we reverse the current, the direction of the rotation changes as well. In larger motors there are several coils. In larger motors, there are many pairs of poles. Only the poles that are connected to the brushes are providing a torque at any one time.

The power of the motor can be easily worked out. From A-level linear dynamics we learned that:

Power (W) = force (N) × speed (m s^{-1})

In Physics code:

P = Fv

As a motor rotates, many of the concepts of rotational dynamics apply. In this
case:

Power (W) = torque (N m) × angular velocity (rad s^{-1})

In Physics code:

P = tw

This equation is not on the AQA syllabus, but it may appear in a question, in which case it will be given.

A coil, which is part of an
electric motor armature, has dimensions 4.5 × 6.0 cm with 500 turns.
It is placed in a magnetic field of flux
density 140 mT. A steady current of 2.0 A flows through it. Calculate: (c) The voltage of the source, if the motor is 80 % efficient. |

In this simple **alternating current** motor
there is a permanent magnet that turns between two coils:

This motor has a cylindrical magnet (from an old pond pump) which is mounted on a shaft held by clamps mounted on clamp-stands. There are coils connected in parallel, connected to an a.c. supply. Changing the voltage does not affect the speed of the motor.

In most a.c. motors there is no permanent magnet.
Instead there is a magnetic field induced in the **rotor** by the alternating
current. Such a motor is called an **induction motor**. It does not work
on direct current. The picture below shows a small induction motor on a
bench drill.