Electric charges produce electric fields.  The electric field produced by a point charge q at rest at the origin is

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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.

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 = r₊<v₊> + r_-<v_-> cannot be zero.  The magnetic field produced by permanent magnets is due to magnetization current densities inside ferromagnetic materials.  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.  The field lines visualizing the magnetic field of a permanent bar magnet are shown below.

Inside the bar magnet the magnetic field points from the south pole (S) to the north pole (N).  Outside the 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 Lorentz force.  It is given by

F = qv´B.

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

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.

To find the direction of the 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.  The magnitude of F is F = qvBsinq, where q 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 v´B = 0 and F = 0, since sinq = 0.  If v and B are perpendicular to each other, then sinq = 1 and F has its maximum possible magnitude F = qvB.

Click here to review  the vector product.

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.

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 below 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 = dr/dt, it is, at any time, perpendicular to the displacement dr.  The work done by the magnetic force is therefore zero, dW = F×dr = 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 and a component v_|| 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_^B/m, and the particle moves in a circle.  The superposition of these two motions results in a spiral path.

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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 magnetic force will act on the wire.  Consider a small section of wire of length dl.  The number of moving electrons in this section is n_-Adl, 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 dF on the section of wire is the sum of the forces on all the moving electrons,

dF = -qn_-Adlv_d´B = jAdl´B = Idl´B.

Here we have used that -qn_-v_d = r_-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 dl, which points in the direction as j.

The force on a longer wire if F = òdF = òIdl´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.

Problems:

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Torque on a current loop

A current I is flowing in a square loop.  The sides of the square have length L. Let the direction of the normal to the loop be defined by the right-hand rule.  Curl the fingers of your right hand in the direction of the current flow.  Your thumb points into the direction of the normal.  Assume that the square loop is placed into a magnetic field and that the normal to the area is perpendicular to B.

The force on each side is given by F = IL´B.  The current I is the same for each side, but the vector L is different for each side.  The force on sides 1 and 3 is zero, since L´B is zero.  The force on side 2 has magnitude F = ILB and points out of the page.  The force on side 4 has magnitude F = ILB and points into the page.  The total force is zero.  In a uniform magnetic field the net force on a current loop is zero.  But each of the non-zero forces has a lever arm about the center of the loop, and therefore exerts a torque t = (L/2)F about the center of the loop.  (The torque exerted by a force F about a point P is t = r´F, where r is the vector pointing from P to the point where the force is applied.)  The total torque is t = LF = IL²B, lying in the plane of the page and pointing upward.  In a uniform magnetic field the total torque on a current loop is in general not zero.  The torque tries to align the normal to the area with the magnetic field.

For a current loop we define the magnetic moment m as

m = IAn.
(magnetic moment = current times area)

The magnetic moment is a vector.  The direction of the magnetic moment is the direction of the area of the loop as defined by the right-hand rule.  The torque on a current loop can be written as t = m´B.  This equation gives the magnitude as well as the direction of the torque.  The torque tries to align m and B.

A magnetic dipole is a current loop whose area goes to zero, or, for practical purposes, a current loop whose dimensions are very small compared to all the other dimensions important in the problem.

Problem:

The simplest DC electric motor consists of a wire-wound rotor mounted on an axle between the pole faces of a permanent magnet.

The magnetic field between the pole faces is nearly uniform, pointing from the north pole towards the south pole of the magnet.  When a current flows through the wire loop of the rotor, a torque tries to align the magnetic moment of the loop with the magnetic field.  The torque produces angular acceleration.  When m and B are aligned the torque is zero, but the loop has angular velocity and angular momentum.  It therefore overshoots the aligned position.  A split ring causes the current to reverse direction in the loop, just as it passes the aligned position.  This reverses the direction of the magnetic moment m.  The torque again tries to align m and B by accelerating the loop through another 180^o degree turn.  Then the current reverses its direction again, etc.  The split-ring configuration is a crucial feature of a DC motor.  It is called a split-ring commutator.

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