Example 20-1 Changing Magnetic Flux I: A Sliding Bar in a Magnetic Field
A copper bar of length L slides at a constant speed v along stationary, U-shaped copper rails (Figure 20-5). A uniform magnetic field of magnitude B is directed perpendicular to the plane of the bar and rails. The moving bar and stationary rails form
a closed circuit, and an emf is produced in this circuit because the wire is moving
in a magnetic field. Determine the emf in the circuit (a) by using the expression for
the magnetic force on a charge in the moving wire and (b) by using Faraday’s law
of induction, Equation 20-2.
Figure 20-5 ​A sliding copper bar What emf is generated in the bar as it slides in the presence of
a magnetic field Bs?
Set Up
For a battery, the magnitude of the emf equals
the change in electric potential (potential
energy per charge) for a charge that traverses
the battery. That is, it’s equal to the work per
charge that the battery does on charges that
travel from one terminal to the other. We’ll use
the same idea in part (a) to calculate the emf in
terms of the work done by the magnetic force
on a charged particle that travels the length of
the moving bar. In part (b) we’ll find the emf by
instead using Equation 20-2. While the magnetic field doesn’t change, the area of the loop
outlined by the moving bar and the rails does
change, and so the magnetic flux through this
loop changes.
Solve
(a) Find the magnetic force on a charged particle moving along with the copper bar.
Copper
rails
v
L
A uniform magnetic field B
points perpendicular to the
plane of the rails.
Magnetic force on a moving
charged particle:
F =  q vB sin u
increase in area
∆A = L∆d = Lv ∆t
i
(19-1)
Work done by a constant force
that points in the same direction
as the straight-line displacement:
(6-1)
W = Fd
Magnetic flux:
B = AB # = AB cos u
Copper bar slides
on the rails at
constant velocity v.
B
L
i
q
+
B
v
F
i
∆d = v ∆t
(20-1)
Faraday’s law of induction:
e = `
B
`
t
(20-2)
For a positive charge q moving with the bar, the direction of the
s.
v is perpendicular to the direction of the magnetic field B
velocity s
So u = 90° in Equation 19-1, and the magnetic force Fs on such a
charge has magnitude
F = qvB sin 90° = qvB (1) = qvB
s,
The direction of Fs is perpendicular to the directions of both s
v and B
and so is directed along the length of the moving bar.
Use the magnetic force on a charged particle to
find the emf produced in the bar.
The magnetic force Fs on a charge q causes it to move along
the length L of the bar. Since Fs is in the same direction as the
displacement of the charge, the work done on the charge as it
travels this length is
W = FL = qvBL
The magnitude of the emf in the bar equals the work done per charge:
e =
qvBL
W
=
= vBL
q
q
(b) Find the emf using Faraday’s law of
induction.
s points perpendicular to the plane of the loop
The magnetic field B
outlined by the moving copper bar and the copper rails. If the area of
this loop is A and we take the positive x direction to point out of the
s), then u = 0 in
plane of the above figure (in the same direction as B
Equation 20-1. The magnetic flux through the loop is then
B = AB cos 0 = AB(1) = AB
The magnetic field is constant, but the area A changes with time
because the bar moves. The speed v of the bar is just the distance d
that the bar moves divided by the time t that it takes to move that
distance, so
v =
d
t
and
d = v t
During time t the area A of the loop outlined by the moving bar
and rails increases by an amount A = L d = Lv t. Therefore the
change in magnetic flux through the loop during this time is
B = (A)B = (Lv t)B = vBL t
From Equation 20-2, the magnitude of the emf in the loop is
Reflect
e = `
B
vBL t
` = `
` = vBL
t
t
We find the same expression for the emf in both parts (a) and (b), as we must. We certainly haven’t proved that
Equation 20-2 is correct in all cases where an emf appears due to a wire moving in a magnetic field (a motional emf)
or due to a changing magnetic field (an induced emf). But we do have added confidence that Equation 20-2 is valid,
and a host of experiments backs up this conclusion.