CHAPTER 21 MAGNETIC FORCES AND FIELDS
Magnetic forces and magnetic fields are associated with
moving electric charge.
In the case of a permanent magnet made of iron, nickel,
cobalt or some combination of these and other elements,
the magnetic field comes from the alignment of electron
spin axes.
These elements are called ferromagnetic because they
can be magnetized. They have unpaired electrons that
can be aligned in regions called domains. When enough
of these domains within an object are aligned in the
same direction, the object becomes magnetic and has
north and south magnetic poles.
If the magnet is allowed to align itself with the earth's
magnetic field, the end that points to the earth's north
pole is called a north pole. The end that points south is
called a south pole.
This has caused some confusion.
The law of poles says like poles repell and unlike poles
attract. This can be demonstrated with two magnets of
similar strength.
Due to the way the poles are named, the earth's
magnetic south pole is near the earth's geographic north
pole. In about 100,000 years, the earth's magnetic poles
will reverse and our compasses will all point in the
opposite direction.
The direction of a magnetic field is represented by
magnetic field lines. They are drawn pointing from
North pole to South pole.
The relative distance between the field lines indicates
the magnetic field strength. Where they are close
together, it is stronger. Where they are parallel, it is
constant.
When a charge moves in a magnetic field, it experiences
a magnetic force. If it is at rest relative to the magnetic
field, it does not. This happens because a moving charge
generates a magnetic field that interacts with any other
magnetic field that is present. A charge at rest does not
generate a magnetic field and is not repelled by or
attracted to a magnet.
Also, a moving charge must have a component of
motion perpendicular to the direction of the magnetic
field causing the force. The magnetic field produced by
the moving charge forms a series of concentric circles
around the charge.
In the diagram below, the component of this magnetic
field that opposes the external field will cause an
attractive force to exist.(N attracts S) The component
oriented in the same direction will cause a repulsive
force to exist. In this diagram both of these forces result
in a net force directed upward. This force is always
perpendicular to the direction of the external magnetic
field and the velocity vector of the particle.
If the velocity vector is at some angle other than 90°
with respect to the magnetic field direction, we must
find the perpendicular component of the velocity by
using Vsinθ where θ is the angle between the velocity
vector and the direction of the magnetic field.
The equation for this relationship is:
B = F/(q0Vsinθ)
where B is the magnetic field strength in Teslas, F is the force
acting on the charged particle in Newtons, q0 is the charge, and
θ is the angle between the velocity vector and B.
One Tesla is one Newton per Coulomb meter per second. It is
usually described as one Newton per ampere meter.
The Tesla is a large unit of magnetic field strength and the
smaller unit, the Gauss is often used. The Gauss is
approximately equal to the strength of the earth's magnetic
field near the surface of the earth. 1 G = 10-4 T.
Example
A particle with a charge of 8.4 μC and a speed of 45 m/s enters
a uniform magnetic field whose magnitude is 0.30 T. For each
of the cases in the drawing, find the magnitude and direction of
the magnetic force acting on the particle.
Motion of a Charged Particle in a Magnetic Field
Motion of a Charge in Magnetic and Electric Fields
An important aspect of motion of a charged particle in a
magnetic field involves a comparison to motion in an electric
field.
In both cases, a positively charged particle enters a uniform
field with a velocity at 90° to the direction of the field.
In the case of the electric field the particle turns in the
direction of the field. Since the force exerted on the particle is
in the same direction as a component of the particle's velocity,
work is done on the particle, it gains kinetic energy and speeds
up.
In the case of the magnetic field, the force causes the particle to
turn in a direction that forms a 90° angle with both the velocity
and the magnetic field. Since there is no component of velocity
in the same direction as the force, no work is done and the
kinetic energy, therefore the speed remains constant.
When a charged particle enters a uniform magnetic field so
that its velocity is oriented at 90° to the direction of the field,
its path will become a circle.
The magnetic force will always remain perpendicular to the
velocity and is directed toward the center of the circle.
The radius of the circular path can be found using the equation
for centripetal force.
F = mv2/r
F = qBvsinθ
mv2/r = qBvsinθ
Since v is at right angles to B, θ = 90° and sinθ = 1.
mv = qBr
r = mv/qB
which gives us the equation for the radius of the path.
Example
A charged particle enters a uniform magnetic field and follows
the circular path shown in the drawing. Is the charge on the
particle positive or negative? If the particle's speed is 140 m/s,
the magnitude of the magnetic field is 0.48 T, the radius of the
path is 960 m, and the charge on the particle is 820 μC, find the
mass of the particle.
The mass spectrometer applies this relationship to separate
charged particles with different masses. Instead of the velocity
of the particle, the potential difference V which is applied
between the ion source and metal plate is known.
The ions pass through the metal plate, acquiring kinetic energy
equal to Vq as they enter a region with a constant magnetic
field, B where they follow a semicircular path to the detector.
Since Vq = ½mv2, we can replace v in the equation mv = qBr,
and derive the equation for the mass spectrometer which is:
m = (er2/2V)B2
which shows that the mass of an ion reaching the detector at 2r
is directly proportional to B2. Varying the field strength allows
ions of different masses to be detected.
Example
The ion source in a mass spectrometer produces singly and
doubly ionized species. The difference in mass between these
species is too small to be detected. Both species experience the
same potential difference and magnetic field. Find the ratio of
the radius of the path of particle 1(+e) to that of particle
2(+2e).
The Force on a Current in a Magnetic Field
When a charge moves through a magnetic field, the force it
experiences is proportional to the magnitude of the charge, its
speed, the magnetic field strength, and the sine of the angle
between the direction of the magnetic field and the direction of
the velocity of the particle.
The force on an electric current in a wire can be calculated
much the same way since a current consists of charges moving
through a wire.
F = qvBsinθ
F = (Δq/Δt)(vΔt)sinθ
Since Δq/Δt = I and vΔt = L(length of the wire), the equation
becomes:
F = ILBsinθ
Example
A square coil of wire containing a single turn is placed in a
uniform 0.25 T magnetic field. Each side has a length of 0.32 m
and the current in the coil is 12 A. Determine the magnitude of
the magnetic force on each of the four sides.
The Torque on a Current Carrying Coil
In the previos problem we calculated the force acting on the
four sides of a square coil of wire carrying a current in a
magnetic field. The coil was oriented so that the direction of
the magnetic field was at right angles to the direction of the
plane of the coil.
If we attach this loop to an axis with a set of bearings in the
same plane as the loop and through its center the forces
exerted on the loop will make it rotate until the normal to the
plane of the loop aligns with the external magnetic field.
If we multiply the force generated by the radius of the loop, we
get the torque generated by one force. Since this arrangement
acts like a couple, the torque is equal to the force generated
multiplied by the entiredistance across the coil.
τ = Fw
τ = ILBwsinφ
since F = ILB and φ = the angle between the force and the lever
arm connecting the axis of rotation to the wire in the coil. Also,
Lw = area of the coil, so
τ = IABsinφ
and if there are N coils instead of 1,
τ = NIABsinφ
The torque depends on the geometry of the coil, the current
through it, and the strength of the external magnetic field
causing the rotation.
NIA is called the magnetic moment of the coil and can be used
to compare torque generated by different coils placed in the
same magnetic field.
These equations apply to circular as well as square or
rectangular coils.
Example
The maximum torque experienced by a coil in a 0.75 T
magnetic field is 8.4 x 10-4 Nm. The coil is circular and consists
of 1 turn. If the current in the coil is 3.7 A, find the length of
wire from which the coil is made.
Electric motors work because of this torque generated by the
interaction of the magnetic field associated with the current in
a coil on the armature with the external magnetic field
produced by the field magnets.
Magnetic Fields Produced by Currents
Magnetic Field Produced by Long Straight Current-Carrying
Wire
Any time an electric current flows in a wire, it produces a
magnetic field.
In a long, straight wire, the direction of the magnetic field can
be determined using the right hand rule.
Point the thumb of your right hand in the direction of the
current in the wire and curl your fingers. Your fingers now
point in the direction of the magnetic field caused by the
current in the wire.
The strength of the field is directly proportional to the current
in the wire and inversely proportional to the distance from the
wire. The equation is:
B = μ0I/2πr
where μ0 is a proportionality constant called the
permeability of free space. Its value is 4π x 10-7 Tm/A.
Example
A long,straight wire carries a current of 48 amps. The strength
of the magnetic field produced by this current at a certain
point is 8.0 x 10-5 T. Find the shortest distance from the point
to the wire.
When two current carrying wires are near each other, each
one exerts a force on the other. This can be a repulsive force or
an attractive force depending on the current direction.
Consider wire 1 to be fixed. In diagram (a) wire 2 experiences
a force due to the magnetic field generated by wire 1 and the
current flowing through wire 2. Right hand rule # 2, when
applied to wire 1, tells us that the magnetic field at wire 2 acts
upward.
Right hand rule # 1 when applied to wire 2 tells us that the
force acting on wire 2 is to the right, perpendicular to both the
current direction and the magnetic field direction.
The magnitude of the force between two current carrying
wires can be calculated using the equations:
F2 = I2LBsinθ
Since this is the formula for the force acting on a wire
with a current of I2 of length L in a magnetic field of
strength B.
B = μ0I1/2πr
This equation is used to calculate the magnetic field
strength due to current I1 in the first wire. If we
substitute into the first equation for B we get:
F = I2L (μ0I1/2πr)sinθ
F = I1I2L(μ0/2πr)
where I1 and I2 are the currents through each wire, L is the
length of each wire, and r is the perpendicular distance
between them.
If the two currents are in the same direction, the force is
attractive. If they are in opposite directions, it is repulsive.
Example
Two rigid rods are parallel to each other and the ground. They
carry the same current in the same direction. The length of
each rod is 0.85 meters and the mass of each is 0.073 Kg. One
rod floats beneath the other rod at a distance of 0.0082 meters.
Find the current in each rod.
Magnetic Field Associated with a Loop of Wire
Magnetic Field Produced by a Current Loop
When a wire is bent into a circular loop, the magnetic field
lines generated resemble those of a bar magnet.
The direction of the magnetic field at the center of a loop can
be determined using right hand rule 2. The magnitude of the
magnetic field can be determined with the equation:
B = Nμ0I/2R
Example
Find the radius of a circular single loop of wire so that the
magnetic field at the center is 1.8 x 10-8 T when the loop carries
a current of 12 A.
Magnetic Field Produced by a Solenoid
A solenoid is a long coil of wire shaped like a helix
approximating a series of circles. The magnetic field inside a
solenoid is constant and can be calculated using the formula:
B = μ0nI
where n is the number of turns per meter of length and I is the
current. If the length of the solenoid is much greater than its
diameter the magnetic field outside the solenoid is nearly zero.
Solenoids are useful as electromagnets, electrically controlled
switches, in CRT image production and in magnetic resonance
imaging.
This MRI angiogram shows bleeding into the lower central
part of the brain.
Ampere's Law
For any current geometry that produces a magnetic field that
does not change in time,
ΣBparΔL = μ0I
This means that the sum of all the products of small distances
an the component of the magnetic field parallel to them is
equal to the product of the permeability of free space and the
current through the surface bounded by that total closed path.
P 678 Questions 2, 4, 5, 6, 7, 9, 10, 14, 15, 19
P 679 Problems 1, 3, 5, 6, 11, 12, 17, 18, 27, 29, 31, 36, 37, 39,
46, 47, 49, 53, 67, 71