Consider the motion of a wire through a magnetic field. If the wire moves to the right and the magnetic field is into the paper, then we see, via (v x B), that a force will be generated of size F = q(v x B) on the particles in the wire.

If this is a normal conducting wire, electrons are the charge carriers and a downward force is generated. This causes the electrons to move to the bottom end of the wire which makes top end of the wire positively charged and the lower end of the wire negatively charged. The field due to this charge configuration points to the bottom of the wire and thus the top is at higher potential than is the bottom of the wire. The motion thus induced an EMF.

Note that the induced electron flow (downward) leads to another q(v x B) force this time pointing to the left, opposing the motion. So, if no external push is supplied the motion rapidly comes to a halt. This is due to an application of Lenz's Law.

If I push the wire so that it moves at a steady rate, a steady EMF will be induced. How large is the EMF? Let's consider the work done on a positive charge and then use the work to deduce the potential difference between the top and bottom of the wire.

• The force is q(v x B). If the wire has a length of l, then the work required to move a positive q is Work = qvBl.
• The potential difference is then Voltage = Work/q = vBl.
• The induced EMF is thus vBl.

Note that the wire is not a closed circuit ===> charges build up at both ends of the wire. The charges in fact build up until they generate an electric field which is strong enough to balance the magnetic force, i.e.,

F(E) + F(B) = 0 = E + q(v x B) = 0 ===> E = -q(v x B)

When this happens, the current flow across the wire ceases. In steady state, we have since, V = EMF = El (pg 631),

El = vBl ===> E = vB

Examples

Imagine a U-shaped conducting rail which has a lamp hooked (of resistance 100 ohms) hooked in series upon which we place a conducting rod of length l = 10 cm which is free to slide. Suppose that there is a B-field of strength 1 mT directed into the paper.

• Push the wire at a speed of v = 10 cm /s, what is the induced EMF and what is the force needed to maintain a constant speed?
  • EMF = vBl = 0.1 m/s x 1 T x 0.1 m = 0.01 Volts

    The upper end of the wire is at higher potential than is the bottom of the wire. This drives a current I around the circuit in the CCW-direction.

  • The current through the circuit is V = EMF = 0.01 Volts = I x 100 ohms ===> I = 0.0001 A

  • The Force is then F = l (I x B) = 0.10 m x 0.0001 A x 1 T = 10^(-5) N

  Comment--we have just finished talking about induced EMF's and come up with two equations:

  • EMF = - delta(PHI)/delta(t)
  • EMF = vBl

  How are these to be reconciled? Look at the U-shaped rail problem again. Following Faraday, imagine that we can think of the process as cutting field lines. Since B is defined as the density of field lines per unit area, the number of field lines a wire cuts is given by,

  • # = B x (rate area is covered) = B x (v x l) = BVl

  Now, note that if I consider the wire to be a point A at time t = 0 s and then to be at point B after delta(t) = 1 s. The change in the area of the loop is simply

  • delta(Area) = v x 1 s = l = v x delta(t) x l===> v = (1/l) x delta(Area)/delta(t)

  Now, plugging in this expression for v

  • # = B x delta(Area)/delta(t) = delta(Phi)/delta(t)

  Although this argument was made assuming constant B, it is in fact general.

  The two formulations and ways of thinking about the problem turn out to be nearly equivalent.