## The magnetic force

### Review:

Electric charges produce electric fields.
The electric field produced by a point charge q at rest at the origin is E = F/q = (keQ/r2) (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 = qE.  This force is parallel or anti-parallel to the field, depending on the sign of the charge.

### The magnetic force on a moving 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
= qv × B.

(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.

#### Problem:

On the surface of a pulsar, or neutron star, the magnetic field may be as strong as 108 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*106 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
Fel = kee2/r2 = (9*109* (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
Fmag = evB = 1.6*10-19*2.2*106*108 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 = v2/r.  The particle will move along a circular path.  The radius of the circle is

r = mv2/F = mv2/(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 vperpendicular perpendicular and a component vparallel 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 = qvperpendicularB/m, and the particle moves in a circle.  The superposition of these two motions results in a spiral path.

#### Problem:

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 v0i.  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 = Bk, 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 v0, we need
Mv02/a = keq|Q|/a2 + qv0B, B = Mv0/qa - ke|Q|/v0a2.
If B is negative, then B points into the page.  Then the magnetic force points radially outward, away from the center of the circle.

### The magnetic force on a current-carrying wire

In a current carrying wire electrons move with an average velocity, called the drift velocity vd.  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 vd. The force dF on the section of wire is the sum of the forces on all the moving electrons,

dF = -qn-A dLvd × B = jA dL × B = IL × B.

Here we have used that -qn-vd = ρ-vd = 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 dL, which points in the direction as j.
The force on a longer wire if F = ∫dF = ∫ I dL× B.  For a long straight wire of length l in a uniform field B we have

F =  IL × 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.

#### Problem:

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 = IL × 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.

#### Problem:

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 = IL×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/s2)/(2 A) = 0.245 T.
We need B = 0.245 T jB points eastward.