Week 7
Lorentz Force
Ampere’s Law
Faraday’s Law
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the Lorentz force is the force on a point
charge due to electromagnetic fields.
It is given in terms of the electric field E and
magnetic flux density B:
F = q( E + v x B)
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If a charged particle moves
into a magnetic field, the
particle will take on a
curved trajectory:
Notice that the force and
the magnetic field are
perpendicular to each
other.
This means that the B field
does not do work on the
charged particles or
current.

The speed is just right, the forces will cancel
each other out and the charge will move in a
straight line
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The radius of the
trajectory is
proportional to the
mass-to-charge ratio
This allows us to
separate heavier ions
from lighter ones
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Since the force constantly
changes the direction of
the electron, the electron
will start moving in a
circular pattern preserving
its initial speed vo.
To derive an expression for
the radius of rotation r in
terms of Bo, set the
magnitude of F equal to the
centrifugal force ½ mer2
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
An electron with constant velocity v = vo x
enters in a magnetic field B = Bo z.
Calculate the initial magnetic force F exerted
on the electron.
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
In semiconductors, the current carriers can be
either electrons or electron holes.
Electron holes (or simply holes) have a
positive charge.

The forces on the charge carriers in a
conductor in the presence of a magnetic field
give rise to a voltage (Vab) across the width
of the conductor.

Determine which terminal of the
galvanometer is positive if the material is p
type.

This law relates the
magnetic flux density
B to its source, the
current I.

The line integral of the magnetic flux density B
over a closed contour is proportional to the net
current through the surface enclosed by the
contour:
Notice that
◦ the double integral is evaluated over the surface S
enclosed by the closed curve C.
◦ The line integral is evaluated around the closed curve C.
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The equation is correct in the special case
where the electric field is constant (i.e.
unchanging) in time.
Otherwise, the equation must be modified.
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The direction of the line differential and the
direction of the surface differential are resolved
using the right-hand rule:
When the index-finger of the right-hand points
along the direction of line integration, the
outstretched thumb points in the direction that
must be chosen for the vector area da, and
current passing in that same direction must be
counted as positive.

Using Ampere’s law, find B around a straight
wire carrying a current I. Assume the wire is
aligned with the z-axis.

The magnetic flux density Bφ in a cylindrical
region 0≤ r≤ a carrying a current Iz is given
by
Bφ =

μo Iz r
2π a2
Determine the surface current density Jz.

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A cable consisting of an inner
conductor, surrounded by a
tubular insulating layer
typically made from a flexible
material with a high dielectric
constant, all of which is then
surrounded by another
conductive layer (typically of
fine woven wire for flexibility,
or of a thin metallic foil), and
then finally covered again with
a thin insulating layer on the
outside.
The term coaxial comes from
the inner conductor and the
outer shield sharing the same
geometric axis.
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Coaxial cables are often used as
a transmission line or radio
frequency signals.
In a coaxial cable the
electromagnetic field carrying
the signal exists only in the
space between the inner and
outer conductors.
A coaxial cable provides
protection of signals from
external electromagnetic
interference, and effectively
guides signals with low
emission along the length of the
cable.
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
The curl of the magnetic flux density B is
proportional to the current density that
creates it:
Again, this equation only applies in the case
where the electric field is constant in time.

The absence of B fields around a coaxial
cable results in no interference in nearby
electrical equipment and wires.
◦ Show that if the current is the same magnitude in
each direction, the magnetic field B outside the
coaxial cable is zero.
◦ Find B for a<r<b.
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Taking a rectangular path
about which to evaluate
Ampere's Law such that the
length of the side parallel to
the solenoid field is L gives
a contribution (BL) inside
the coil.
The field is essentially
perpendicular to the sides
of the path, giving
negligible contribution.

Show that in this idealized case, Ampere's
Law gives B = μonI where n is the number of
turns N per length L.

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All of the loops of wire which
make up a toroid contribute to
the magnetic field in the same
direction inside the toroid.
The sense of the magnetic field
is that given by the right-hand
rule.
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A more detailed visualization of the field of
each loop can be obtained by examining the
field of a single current loop.
The current enclosed by the dashed line is
just the number of loops times the current in
each loop.

Show that the magnetic flux density B is given
by
Bφ =
μoNI
2π r

Determine the expression for B due to an
infinite plane sheet with uniform current
density J. Assume that the sheet is on the xy plane.

What if the electric field is
not constant in time?

Then we need to consider a
“displacement” current.

The force exerted on the rod is given by
F=ILxB

A straight wire carries a current I = 1 mA in
the –x direction. The wire feels a force of 1 N
per meter in the –z direction. Calculate the
magnetic flux density By.

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Parallel conductors carrying currents in same
direction will attract each other.
If the currents are in opposite directions the
conductors will repel each other.

The current I' is moving in the B field caused
by the current I, so it experiences a force
F = I' L x B.

Show that the magnetic force per unit length
that one wire exerts upon the other is
where

The ampere is defined to be the constant
current which will produce an attractive force
of 2 × 10–7 Newtons per meter of length
between two straight, parallel conductors of
infinite length and negligible circular cross
section placed one meter apart in free space.
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Read Sections 5-1, 5-4, and 5-8 in your
book.
Solve end-of-chapter problems 5-12, 5-15,
5-16, 5.20, 5.21.