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

Moving electric charges produce magnetic fields.  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 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:

• Reasoning:
  The electron in a hydrogen atom is at a distance r = 0.53*10^-10 m from the proton.
  The magnitude of the electric force acting on the electron is equal to F_el = k_ee²/r².
  The maximum magnitude of the magnetic force acting on the electron when its velocity v is perpendicular to B is F_mag = evB.
• Details of the calculation:
  F_el = k_ee²/r² = (9*10⁹* (1.6*10^-19)²/(0.53*10^-10)²) N = 8.2*10^-8 N.
  F_mag = 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 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.

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:

• Reasoning:
  The electomagnetic force, F = q(E + v × B) provides the centripetal force required for a circular orbit.
  The magnitude of that force is F = mv²/r.
  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.
• Details of the calculation:
  Given a circular orbit of radius a and a speed v₀, we need
  Mv₀²/a = k_eq|Q|/a² + qv₀B, B = Mv₀/qa - k_e|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.

An interactive 3D animation of the motion of a charged particle in a uniform magnetic field.

Click here!

The magnetic force on a current-carrying wire

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 L.  The number of moving electrons in this section is n_-AL, 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 F on the section of wire is the sum of the forces on all the moving electrons,

F = -qn_-ALv_d × B = jAL × B = IL × 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 L, which points in the direction as j.
The force on a straight section of wire in a uniform field B is

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:

• Reasoning:
  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, the magnitude F = ILB.
• Details of the calculation:
  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:

• Reasoning:
  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.
• Details of the calculation:
  To lift the wire we need B = F/(IL) = 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 j.  B points eastward.

External link:  The Lorentz force on a wire  (Youtube)

Embedded Question 1

A demo:  We pass a current through a wire a section of which passes between the poles of two magnets, as shown below.
Make a prediction. When we connect the power supply, which way the wire will move?

Discuss this with your fellow students in the discussion forum!
Review with them the Lorentz force and the right-hand rule.

In this video clip  a hand crank generator is used to produce a voltage across a thin rod which can move in a magnetic field produced by a set of magnets.  The north pole of the magnets points up.

When the crank is turned in the direction of the arrow shown, the red lead is positive, and a current flows from the left to the right through the rod.  If the crank is turned the other way, a a current flows from the right to the left through the rod.

You can verify that the direction of the force F on the wire is the direction of IL × B.

The force between two parallel wires

Parallel wires carrying currents will exert forces on each other.  Each wire produces a magnetic field, which influences the other wire.  When the currents in both wires flow in the same direction, then the force is attractive.  When the currents flow in opposite directions, then the force is repulsive.

In the diagram on the right, both wires carry a current in the same direction.  The magnetic field produced by wire 1 at the location of wire 2 is B = μ₀I₁/(2πd) pointing into the page.  The force on a section of wire 2 of length L is is F = I₂L × B.  F  has magnitude F = I₂LB = (μ₀I₁I₂L)/(2πd) and points towards wire 1.  The force per unit length is (F/L) = (μ₀I₁I₂)/(2πd) towards wire 1.  Similarly the force per unit length on wire 1 due to the field produced by wire 2 is (F/L) = (μ₀I₁I₂)/(2πd) towards wire 2.  (Newton's third law)

Remember:
Two parallel wires carrying current in the same direction  attract each other.
Two parallel wires carrying current in opposite direction  repel each other.

External link:  Two wires  (Youtube)