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Handout 9: Electromagnetic induction, Faraday’s law
Electromagnetic induction
Consider Fig. 1 where a bar magnet is inserted
into a coil. As the magnet is moving the towards the coil,
there is a current flowing in the coil as seen from the
galvanometer. This current is called “induced current”
and the corresponding emf required to cause this
current is called an “induced emf”. The induced emf
occurs only when the bar magnet is moving towards or
away from the coil. When the bar magnet is at rest, there
is no induced emf.
Figure 1: Electromagnetic induction
We will see later that the induced emf is
associated with the rate of change of magnetic flux
through the coil. When the magnet is at rest, there is no
change in magnetic flux through the coil and therefore
the induced emf is not developed.
Magnetic flux
In Fig. 2, a uniform magnetic field 𝐵 makes an
angle 𝜃 to the normal to the surface of area 𝐴. The
perpendicular component of the field 𝐵 cos 𝜃. The
magnetic flux through the surface is defined as
Φ = 𝐵 cos 𝜃 𝐴
or
Φ = 𝐁∙𝐀.
If 𝐁 is not uniform throughout the surface, the general
expression for the magnetic flux must be used:
Φ =
𝐁 ∙ 𝑑𝐀,
where 𝑑𝐀 is the small area element of the surface. The
unit of magnetic flux is Tm2 or weber (Wb).
Example 1 A solenoid is 8.0 cm long, 1.4 cm in diameter,
and has 340 turns. When the current through the
solenoid is 68.0 mA, what is the magnetic flux through
one turn of the solenoid?
Figure 2: Magnetic flux through a
surface
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Faraday’s law
Faraday’s law states that the induced emf in a
closed loop is equal to the negative of the rate of change
of the magnetic flux through the loop. In symbols,
Faraday’s law is
ℰ = −
𝑑Φ
,
𝑑𝑡
where ℰ is the induced emf.
Lenz’s law
Lenz’s law is a method for determining the
direction of an induced current or emf. It states that the
direction of the induced current is such as to oppose
the change causing it. The demonstration of Lenz’s law
can be seen in Fig. 3. In Fig. 3(a), the N-pole of a bar
magnet moves away from the ring. The field through the
ring points to the right and the flux through the ring
decreases as the magnet moves away. By Lenz’s law, the
induced current in the ring has to flow in such the
direction to prevent the change of the flux. In Fig. 3(b),
the N-pole of the magnet is moved towards the ring. The
induced current in the ring flows, opposing the change
of the magnetic flux through the ring.
Example 3 In a physics laboratory experiment, a coil
with 200 turns enclosing an area of 12 cm2 is rotated in
0.04 s from a position where its plane is perpendicular
the earth’s magnetic field to one where its plane is
parallel to the field. The earth’s magnetic field is 60 𝜇T.
What is the average emf induced?
Example 4 A coil of radius 𝑟 = 4.0 cm containing
𝑁 = 500 turns is placed in a uniform field 𝐵 tesla. At
time 𝑡 seconds, 𝐵 = 1.2 × 10−2 𝑡 + 3.0 × 10−5 𝑡 4 . The
coil is connected to a 600-Ω resistor and its plane is
perpendicular to the field. The resistance of the coil can
be neglected.
a) Find the induced emf in the coil at time 𝑡 seconds.
b) Find the current in the resistor at time 𝑡 = 5.0 s
(a)
(b)
Figure 3: The direction of induced current 𝐼
in the ring. (a) the N-pole of the bar magnet
moves away from the ring and (b) the Npole moves towards the ring.
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Example 5 A square loop (0.15 m × 0.08 m) of wire is
moved at constant speed 𝑣 = 2.0 ms-1 into a region of a
transverse uniform magnetic field 𝐵 = 0.70 T.
a) Find the magnitude of induced emf in the loop.
What is the direction of the induced current?
b) When the loop is completely inside region of
magnetic field, what is the induced emf?
c) When the loop is leaving the region of magnetic
field, how is the answer in a) changed?
Example 6 A coil of wire with area 𝐴, containing 𝑁 turns
is rotating at angular speed 𝜔 about its symmetry axis
through the plane of the coil. The coil is placed in a
uniform magnetic field 𝐵. Determine the magnitude of
the induced emf in the coil as a function of time 𝑡.
Motional emf
Figure 4 shows a conducting rod of length ℓ is
moving at speed 𝑣 perpendicular to the magnetic field 𝐵.
The magnetic force 𝐹𝐵 = 𝑞𝑣𝐵 pushes the electrons
towards the lower end. The lower end becomes
negatively charged and the upper end positively
charged. The electric field 𝐸 is developed in the rod. The
electric force 𝐹𝐸 = 𝑞𝐸 on the electrons is upward. The
accumulation of electrons continues until the upward
electric force cancels the downward magnetic force.
Figure 4: A conducting rod moving in
a uniform magnetic field.
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Then 𝑞𝐸 = 𝑞𝑣𝐵 or 𝐸 = 𝑣𝐵 and the charges are in
equilibrium.
The potential difference 𝑉 is developed between
the ends of the rod. From 𝑉 = 𝐸ℓ, we obtain 𝑉 = 𝑣𝐵ℓ.
This is called the motion emf, denoted by
ℰ = 𝑣𝐵ℓ.
When motion of the rod “cuts” the magnetic field line,
motional emf is induced between the ends of the rod. If
the two ends of the rod are connected to a lamp (Fig. 5),
the lamp will shine.
Example 7 A conducting rod with length ℓ = 85.0 cm is
placed in a uniform magnetic field 𝐵 = 0.95 T which is
perpendicular to the length of the rod. The rod cuts the
magnetic field at a constant speed 𝑣 = 0.25 ms-1.
a) Find the induced emf.
b) The ends of the rod are connected to a lamp. The
total resistance is 0.75 Ω. Find the induced
current.
*Example 8 A bar moves along a straight line at
constant speed 𝑣 without friction on parallel conducting
rails as shown on the left diagram. The rails separated
by distance ℓ are connected by resistance 𝑅 at the ends.
A uniform magnetic field with magnitude 𝐵 is directed
into the page. The motional emf is developed and the
equivalent circuit is shown on the right diagram.
a) Show that the force needed to move the bar at the
constant speed is given by
𝐵 2 ℓ2 𝑣
𝐹 =
.
𝑅
b) Show that the rate of work done on the bar is
equal to the power lost in the resistor. Comment
on the result.
Figure 5: A moving conducting wire
provides emf lighting the lamp.
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Induced electric fields
By changing magnetic flux through a circular
wire, the emf is induced in the wire and hence the
induced current flows. This is shown in Fig. 5. There is
work done 𝑊 on charge 𝑞 around the loop:
𝑊 =
𝐅 ∙ 𝑑𝓵 = 𝑞
𝐄 ∙ 𝑑𝓵.
Work done per unit charge is emf. Hence,
ℰ =
𝐄 ∙ 𝑑𝓵.
The emf is line integral (along stationary path) of
electric field around a closed loop. Thus, we can rewrite
Faraday’s law as
𝐄 ∙ 𝑑𝓵 = −
𝑑Φ
.
𝑑𝑡
In Fig. 5, the electric field 𝐄 is tangential to the circle and
the magnitude is the same every point on the circle.
Therefore, 𝐄 ∙ 𝑑𝓵 = 𝐸 𝑑ℓ = 𝐸 2𝜋𝑟 , where 𝑟 is the
radius of the circle.
*Example 9 Magnetic field 𝐵 is confined in a circular
region of radius 𝑅. The magnetic field is changing at rate
𝑑𝐵 𝑑𝑡 > 0. As a result, an electric field 𝐸 at distance 𝑟
from the center is induced, circling around the region of
magnetic field.
a) Determine the direction of the induced electric
field.
b) Show that the induced electric field is given by
𝑅 2 𝑑𝐵
,
𝑟>𝑅
2𝑟 𝑑𝑡
𝐸(𝑟) =
𝑟 𝑑𝐵
,
𝑟 < 𝑅.
2 𝑑𝑡
Figure 6: Induced electric field