FARADAY'S LAW
No. of
lectures
allocated
3
Actual No.
of lectures
dates :
9/5/09-14
/5/09
31.1
Faraday's Law of Induction
In the previous chapter we learned that electric
current produces magnetic field. After this
important discovery, scientists wondered: if electric
current produces magnetic field, is it possible that
magnetic field can produce an electric current?
Consider loop of wire that is connected to a
galvanometer and a magnet is moved in the vicinity
of the loop. It was observed that if both the loop
and the magnet held stationary relative to each
other there will be no deflection in the
galvanometer. If the loop is held stationary while
the magnet is moved toward the loop, the
N
(a)
N
S
(b)
S
N
S
(c)
Figure 31.1 (a) A bar magnet is stationary relative to the loop and there is no
deflection in the galvanometer. (b) The bar magnet is moving toward the loop and
the galvanometer deflects in one direction. (c) The bar magnet is moving away
from the loop and the galvanometer deflects in the opposite direction.
galvanometer will deflect in one direction. If the
magnet is moved away from the loop, the
galvanometer will deflect in the opposite direction.
The same observations is occurred when the loop is
moved while the magnet is held stationary. From
these observations one concludes that a current is
produced in the loop as long as there is relative
motion between the loop and the magnet. Such a
current is called the induced current, and it source
is called the induced emf. The phenomenon itself
(the production of electric current from changing
magnetic field) is called the electromagnetic
induction.
Michael Faraday studied these observations
quantitatively and put them in a mathematical
formula considered as one of the fundamental laws
in electromagnetic theory. He found that the
induced emf is proportional to the rate of change of
the magnetic flux, i.e.,
ε =− N
dΦ m
dt
31.1
Where N is the number of turns in the loop. For a
uniform magnetic field the magnetic flux becomes
Φ m = BAcosθ
With θ is the angle between the magnetic field and
the area ( The direction of the area of a plane is
normal to the plane). The SI unit of the magnetic
flux is Weber (Wb) with 1 Wb equals to 1 T.m2.
Faraday's law now reads
ε =− N
d ( BAcosθ )
dt
31.2
From this expression we conclude that an induced
emf can be created if either B, A, θ, or a
combination of them vary with time.
The minus sign in Faraday's law is a
consequence of the law of conservation of energy.
Example 31.1 A coil consists of 200 turns of
wire. Each turn is a square of side 18 cm, and a
uniform m. field perpendicular to the plane of the
oil is turned on. If ∆B = 0.5 T in 0.8 s , what is the
induced emf in the coil.
Solution Knowing that θ = 0 , and the area A is
constant we have
ε = − NAcos θ
d ( B)
dB
= − NA
dt
dt
ε = 200(0.18) 2
0.5
= 4 .1 V
0.8
Example 31.2 A loop of wire of area A is placed
in a m. field perpendicular to the plane of the loop.
The magnitude of B varies with time according
to B = Bmax e −αt , what is the induced emf in the loop.
Solution
we have
Again θ = 0 , and the area A is constant
d ( B)
dB
ε = − NAcos θ
= − NA
dt
dt
ε = αABmax e −αt
31.2
MOTIONAL EMF
To understand how the induced emf is
originated we now study in details the nature of the
induced emf. Consider a rod of length l moving
with speed v in a uniform magnetic field B directed
into the page, as shown in Figure 31.2. The free
electrons inside the rod will experience a magnetic
force Fm = evB which is directed downward. The
electrons then will accumulate at the lower end of
the rod leaving a net positive at its upper end. As a
result of this charge separation, a net electric field
E will be set up inside the rod. Therefore, free
electrons now will be affected by an upward
electric force Fe = eE in addition to the magnetic
field. Charges continue to build up at the ends of
the rod until the two forces balanced. At this point
motion of charges ceases leaving the rod with two
opposite polarities at its end, that is an emf is
produced across the rod. To calculate this emf we
have, from the equilibrium condition
Fm = Fe or evB= eE
so E = vB
×
×
B
×
×
×
×
×
×
×
×
×
×
v×
×
×l ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
(a)
×
×
×
×
×
×
×
×
×
×
×
×
B
l×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
x×
×
×
v
(b)
Figure 31.2 (a) A conducting rod moving in a uniform magnetic field into the page.
The magnetic force makes electrons to accumulate at the lower end of the rod, leaving
the upper end with positive charges. (b) A conducting rod slides along conducting rails
in a uniform magnetic field into the page. A current will be induced in the loop.
Since the electric field in uniform inside the rod,
the potential difference across the rod is related to
this electric field according to
ε =V = El = vlB
31.3
This potential difference is maintained across the
ends of the rod as long as it is moving in the field
and is called the motional emf. If the rod is a part
of closed loop, as shown in Figure 31.2(b), a
current will flow in the loop from the positive end
to the negative end (counterclockwise).
Let us now prove Equation 31.3 using
Faraday's law. Consider again Figure 14.2(b) where
the rod is sliding along conducting rails in the
magnetic field such that it forms a closed loop.
The rate of change of the magnetic flux through the
loop is now proportional to the change in the area
of the loop. As the area of the loop at some instant
is A= lx , the magnetic flux through the loop is
Φ m = Blx
Where x, the width of the loop is changing with
time as the rod moves. Using Faraday's law, we
find that the induced emf in the loop is
dΦ m
d
dx
(
)
ε =−
=−
Blx = − Bl
dt
dt
dt
But (dx dt ) represents the speed of the rod, so we
obtain
ε = − vlB
31.4
Which is the same result of Equation 31.3 except of
the minus sign.
Example 31.3 A
conducting × × × × × ×
v
bar of length l rotates with a
constant angular speed ω about × × × × × ×
r
l
one end. A uniform m. B is × × × × × ×
perpendicular to the plane of the × × × × × ×
rotation. What is the potential
difference induced between the ends of the bar.
Solution It is clear that v is not constant along the
length of the bar
⇒ We have to divide the bar
into small elements each of length dr. Noe the emf
across one of these e elements is
dV = vBdr = ωBrdr
Integrate to find the emf across the bar ⇒
L
V = ωB ∫ rdr = 12 Bωl 2
0
31.3
Lenz's Law
It tells us that the induced current must be in a
direction such that it produces a magnetic field
to oppose the change in the magnetic flux. Lenz's
law can be explained by the following two rules:
(1) If the magnetic flux through the loop is
increasing, the direction of the induced current is
such that it produces a magnetic field opposite to
the source magnetic field, (2) If the magnetic flux
through the loop is decreasing, the direction of the
induced current is such that it produces a magnetic
filed in the same direction as that of the source
magnetic field.
Example 9.3 A square loop of
side L and resistance R moves with
constant speed v through a region
of width 3L in which a uniform
magnetic field B directed out of the
page as shown. Plot the flux and
the induced emf in the loop as a
function of x, the position of the
right side of the loop.
Solution The magnetic flux is zero
before the loop enters the field. As
the loop is entering the field,
Φm= Blx , that is, the flux increases
linearly with x, reaches its
maximum value, Bl2, when the
loop is entirely in the field.
Finally, as the loop is leaving the
field Φm= Bl(4L-x), that is the flux
decreases linearly with x, reaches
to zero when the loop is entirely
outside the field.
3L
x
Φ
ε
F
Now
ε =−
dΦ m
dΦ m dx
dΦ m
=−
=−
v
dt
dx dt
dx
Noting that dΦm/dx is the slope of the curve in the first graph,
(Φm vs. x). While the loop is entering the filed the flux is
increasing and, according to Lenz’s rule, the magnetic field set
up by the induced current is into the page (opposite to the
original field). Hence the induced emf is clockwise. While the
loop is leaving the field, Φm is decreasing and the magnetic
field set up by the induced current in, this case, is out of the
page (similar to the original field). This means that the induced
emf is counterclockwise.
To find the force on the loop it clear that while the loop is
entering the only side that cause the net force is the right side.
Now
r r
r
F = Il R × B = ILB(− j × k ) = ILB (− iˆ )
While the loop leaving the left side is the only side that cause
the force. Again
r r
r
F = Il L × B = ILB(− j × k ) = ILB (− iˆ )
9.3 INDUCED ELECTRIC FIELD
A changing magnetic flux creates an induced
emf and thus an induced current in a conducting
loop. Therefore, an electric field must be present
along the loop. This field, which is created by
changing magnetic flux, is called induced electric
field and given by.
ε in = ∫ E ⋅ ds
31.5
Using Equation 31.1, Faraday’s law can be
rewritten as
∫ E ⋅ ds = −
dΦ m
dt
31.6
It should be noted that this result is also valid for
any hypothetical closed path.
The induced electric field given by Equation 31.6 is
quite different from the electrostatic field
(produced by static charges). The formal one is a
non-conservative field produced by a changing
magnetic flux. Hence no electric potential can be
associated by the induced electric field. The
potential difference between two points i and f is
f
Vf − Vi = − ∫ E ⋅ ds
i
which would be zero for a closed path, contrary to
Equation 31.6
The direction of the induced e.f. is determined
by Lenz's rule
Example 31.8 A long solenoid of radius R has n
turns of wire per unit length and carries a current
given by I = I max cos ωt , with Bo and ω are constant
and the time t is in seconds. Calculate the induced
electric field inside and outside the
R
Solenoid.
R
r
Solution The m. flux through the
area enclosed by the closed loop is
( )
Φ m = BAcos θ = µ o nI π r 2 = µ o nπ r 2 I max cos ωt
Now applying
∫ E ⋅ ds = −
dΦ m
= µ o nπ r 2ωI max sin ωt
dt
By symmetry, the magnitude of E is constant
around the path and tangent to it
⇒
E (2πr ) = µ o nπ r 2ωI max sin ωt
From which we find that
E=
µ o n ωI max
r sin ωt
2
To calculate the electric field
outside the sphere, the closed path
has now a radius r>R. Since the
magnetic field is confined only to
the region r<R , the magnetic flux
through the path is
(
)
Φ m = µ o nI π R 2 = µ o nπ R 2 I max cos ωt
The electric field is now
E (2πr ) = µ o nπ R 2ωI max sin ωt
From which we find that
µ o n ωR 2 I max
E=
sin ωt
2r
R
R
r