Electric charges produce electric fields.

The electric field
produced by a point charge q at rest at the origin is
**E **= **F**/q = (k_{e}Q/r^{2}) (**r**/r).

The electric field of a charge distribution can be found using the principle of
superposition. If the charge distribution has a high degree of symmetry,
Gauss's law may be use to find the
electric field produced by the distribution.
Field lines can be
used to visualize the electric field.

If a charge q is placed into an electric field produced by other charges, it
will be acted on by a force **F **= q**E**. This force is parallel
or anti-parallel to the field, depending on the sign of the charge.

There are no magnetic charges.

Moving electric charges produce magnetic fields.

To produce a **magnetic field** **B**, a
current
density **j** is needed, i.e. **j **= ρ_{+}<**v**_{+}> + ρ_{-}<**v**_{-}>
cannot be zero.

The magnetic field produced by permanent magnets is due to "magnetization
current" densities inside ferromagnetic materials. This is
atomic-scale physics, and on that scale Classical Physics fails and
Quantum Mechanics is the model that is used. Many research
groups are currently investigating the magnetic properties of various
materials, both experimentally and theoretically, and many questions
still need to be answered.

Field lines can be used to visualize the
magnetic field.

Magnetic field lines have no beginning or end, they always form **closed
loops**. (Because there are no magnetic charges, there are no
sources or sinks.)

The field lines visualizing the magnetic field of a permanent bar magnet are shown on the right.

Link: Explore the magnetic field of a bar magnet (PhET)

We define the north and south pole of a bar magnet so that inside the bar magnet the magnetic field points from the south pole (S) to the north pole (N). Outside a bar magnet the field lines close the loop, from north to south. The density of the field lines is proportional to the strength of the magnetic field.

When a bar magnet is bend into the shape of a horseshoe, the magnetic field between the poles is nearly uniform and usually quite strong.

Magnetic fields exert forces on other moving charge.
The force a magnetic field exerts on a charge q moving with velocity **v** is called the magnetic
**Lorentz force**. It is given by
**F **= q

(The SI unit of B is Ns/(Cm) = T (Tesla))/p>

The force **F** is perpendicular to the direction of the magnetic
field **B**.

It also is perpendicular to the direction of the velocity **v**.
**F** is perpendicular to the plane that contains both **v** and **B**.

(Review the
vector or cross product!)

The magnitude of the Lorentz force **F** is F = qvB sinθ, where θ is
the smallest angle between the directions of the vectors **v** and
**B**.

If **v** and **B** are parallel or anti-parallel to each other,
then sinθ = 0 and **F **= 0. If **v** and
**B** are perpendicular to each other, then sinθ = 1 and **F** has its maximum
possible magnitude F = qvB.

If a charge q is moving with uniform velocity
**v** parallel to the direction of a uniform magnetic field **B**, it experiences no
force. It continues to move with uniform velocity **v** along a
straight line parallel to the field.

To find the direction of the
Lorentz force, use the **right-hand rule**. Let the fingers of your
right hand point in the direction of **v**. Orient the palm of
your hand, so that as you curl your fingers, you can sweep them over to
point into the direction of **B**. Your thumb points in the
direction of the vector product** v **×** B**. If q is
positive then this is the direction of **F**. If q is negative,
your thumb points opposite to the direction of **F**.

On the surface of a pulsar, or neutron star, the magnetic field may be as
strong as 10^{8 }T. Consider the electron in a hydrogen atom on
the surface of the neutron star. The average distance between the electron
and the proton is 0.53*10^{-10 }m. The average speed of the
electron is 2.2*10^{6 }m/s. Compare the magnitude of the electric force that the
electric field of the proton
exerts on the electron with the maximum magnitude of the magnetic force that the magnetic field of the
neutron star exerts on the electron. Is it reasonable to expect that the
hydrogen atom will be strongly deformed by the magnetic field?

Solution:

The electron in a hydrogen atom is at a distance r equal to 0.53*10^{-10
}m from the proton.

The magnitude of the electric force acting on the electron is
equal to

F_{el }= k_{e}e^{2}/r^{2
}= (9*10^{9}* (1.6*10^{-19})^{2}/(0.53*10^{-10})^{2})
N = 8.2*10^{-8 }N.

The maximum magnitude of the magnetic force acting on the electron when its velocity
**v** is perpendicular to **B** is

F_{mag }= evB = 1.6*10^{-19}*2.2*10^{6}*10^{8
}N = 3.5*10^{-5 }N.

The maximum magnitude of the magnetic force on the electron is more than 1000 times stronger than
the magnitude of the electric force. We expect that hydrogen atoms will be strongly deformed
or destroyed on the surface of a neutron star.

Consider a charged particle with mass m and charge q
which at t = 0 has a velocity **v **perpendicular to
**B**. This particle experiences a force with magnitude F = qvB perpendicular
to its velocity. A force perpendicular to the velocity results in
centripetal acceleration
a = F/m = v^{2}/r. The
particle will move along a circular path. The radius of the circle
is

r = mv^{2}/F = mv^{2}/(qvB) =
mv/(qB),

and the circle lies in a plane perpendicular to
**B**. The
diagram on the right shows the paths followed by two charges, one
positive (red) and one negative (blue), in a magnetic field that points
into the page.

Since the magnetic force is perpendicular to the
velocity **v **= ∆**r**/∆t, it is, at any time, perpendicular to
the displacement ∆**r**. The work done by the magnetic force is
therefore zero, ∆W = **F∙**∆**r **= 0.

The magnetic force does no work.

The magnetic force changes the direction of the velocity, but it does not change the speed or the kinetic energy of the particle.

Assume a particle at t = 0 is moving
with a velocity **v** which has a component **v**_{perpendicular}
perpendicular and a component **v**_{parallel} parallel to
the magnetic field.

The path of the particle will be a spiral.

There is no acceleration parallel to **B**, but in the
plane perpendicular to **B** the centripetal acceleration
is a = qv_{perpendicular}B/m, and the particle moves in a circle.
The superposition of these two motions results in a spiral path.

A
particle with mass M and charge q > 0 moves in a uniform magnetic field
**B** and also in the field of another charge Q < 0 located at the
origin. At t = 0 the particle is at x = z = 0, y = a, and its velocity
is v_{0}**i**. For what **B** will the trajectory of the
particle be a circle of radius a centered at the origin?

Solution:

The electomagnetic force, **F** = q(**E** +** v
**×**
B**) provides the centripetal force required for a circular orbit.

Consider the geometry shown in the figure on the
right.

Let **B**
= B**k**, i.e. let **B **point out of the page.

Then the electric force and the magnetic force point in the same direction,
towards the center of the circle.

Given a circular orbit of radius a and a speed v_{0}, we
need

Mv_{0}^{2}/a = k_{e}q|Q|/a^{2} + qv_{0}B,
B = Mv_{0}/qa - k_{e}|Q|/v_{0}a^{2}.

If B is negative, then
**B** points into the page. Then the magnetic force points
radially outward, away from the center of the circle.

In a current carrying wire electrons move with an average velocity, called
the drift velocity **v**_{d}. If the wire is placed into a
magnetic field **B**, a force will act on the wire.
Consider a straight section of wire of length dL. The number of
moving electrons in this section is n_{-}A dL, where n_{-}
is the electron density and A is the cross-sectional area of the wire. The
electrons move with the drift velocity **v**_{d}. The force
d**F** on the section of wire is the sum of the forces on all the moving electrons,

**dF **= -qn_{-}A dL**v**_{d }×** B
**= **j**A dL ×** B **= I**L **×** B**.

Here we have used that -qn_{-}**v**_{d }= ρ_{-}**v**_{d
}= **j**, and that jA = I for a wire. Since I is not a vector and
we have to preserve the directional aspects of the vector product, we assign the
direction of the current density to d**L**, which points in the
direction as **j**.

The force on a longer wire if **F **= ∫d**F **= ∫ I d**L**×** B**.
For a long straight wire of length l in a uniform field B we have

**F **= I**L **×** B**.

You can again use the right-hand rule to find the direction of the force.
Let the fingers of your right hand point in the direction of the current flow.
Orient the palm of your hand, so that as you curl your fingers, you can sweep
them over to point into the direction of **B**. Your thumb points in
the direction of the vector product** F**.

A wire carries a steady current of 2.4 A. A
straight section of the wire is 0.75 m long and lies along the x-axis
within a uniform field **B **= 1.6 T in the z-direction. If the
current is in the positive x-direction, what is the magnetic force on
the section of wire.

Solution:

The force on the wire is given by **F
**= I**L **×** B**.

The direction of
**L **×** B** is the negative y-direction.
Since **L** and **B** are perpendicular to each other,

F = ILB
= (2.4 A)(0.75 m)(1.6 T) = 2.88 N.

The force on the section of wire is
**F **= -2.88 N in the
negative y-direction.

A wire having a mass per unit length of 0.5 g/cm carries a 2 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

Solution:

Orient your coordinate system so that the x-axis points towards the south
and the z-axis points upward. We want the direction of the magnetic
force **F **= I**L**×**B** on the wire
segment to be upward and the magnitude to be equal to mg. To get the
maximum F for the minimum B, **B** hat to point into the y-direction. Then F = ILB.

To lift the wire we need B = mg/(IL) = ρLg/(IL) = ρg/I = (0.05 kg/m)(9.8
m/s^{2})/(2 A) = 0.245 T.

We need **B **= 0.245 T** j**.
**B** points eastward.

Link: The Lorentz force on a wire (Youtube)