Transcript It is sometimes difficult to find the polarity of an

It is sometimes difficult
to find the polarity of an
induced emf.
The net magnetic field
penetrating a coil of
wire results from two
factors.
One is the original magnetic
field that produced the
changing flux that leads to
the induced emf. The other
comes from the fact that the
induced current produces its
own magnetic field.
The field
produced by an
induced current is
called an induced
magnetic field.
Lenz’s Law states that the
induced emf resulting from a
changing magnetic flux has a
polarity that leads to an
induced current whose
direction is such that the
induced magnetic field
opposes the original flux
change.
This three step strategy
helps find the direction of
the induced emf:
1. Determine whether
the magnetic flux that
penetrates a coil is
increasing or decreasing.
2. Find what the
direction of the induced
magnetic field must be
to oppose the change in
flux by adding to or
subtracting from the
original field.
3. After finding the direction
of the induced magnetic field,
use RHR-2 to find the
direction of the induced
current. Then the polarity can
be found (conventional flow
from + to -).
Ex. 8 - A permanent magnet
approaches a loop of wire.
The external circuit attached
to the loop consists of the
resistance R. Find the
direction of the induced
current and the polarity
of the induced emf.
Ex. 9 - There is a constant magnetic field in
a rectangular region of space. This field is
directed perpendicularly away from us.
Outside this region there is no magnetic
field. A copper ring, perpendicular to the
direction of the field, slides through the
region. The ring is completely outside the
region, partway through, completely inside,
partway out, and completely out at different
times of its motion. For each of the five
positions: is there an induced current?,
and what is its direction?
An ac generator consists of
a coil of wire spinning in a
magnetic field. The equation
for the emf produced is:
ε = N(2BLv sin θ). This can
be written using angular speed
as: ε = ε0 sin wt, where
w = 2πf and E0 = NABw.
The polarity of the
coil alternates as it
spins, so this electric
generator is called an
alternating current
(ac) generator.
Ex. 10 - The coil of an
ac generator rotates at a
frequency of f = 60.0 Hz
and develops an emf of
120 V (rms). The coil has an
area of A = 3.0 x 10-3 m2. and
consists of N = 500 turns. Find
the magnitude of the magnetic
field in which the coil rotates.
Ex. 11 - A generator is mounted on a
bicycle to power a headlight. A small
wheel on the shaft of the generator is
pressed against the bike tire and turns the
armature 44 times for each revolution of
the tire. The tire has a radius of 0.33 m.
The armature has 75 turns, each with an
area of 2.6 x 10-3 m2, and rotates in a
0.10-T magnetic field. When the peak emf
being generated is 6.0 V, what is the
linear speed of the bike?
The devices connected to a
generator are called the “load”.
If no devices are connected, the
only force from the generator is that
needed to mechanically move the
loop. However, if devices are
connected, the induced current
produces a magnetic force F called
a countertorque.
This countertorque
opposes the motion
and requires more
force to be added to the
generator. The greater
the current, the more
force must be applied.
When an electric motor is
operating, two sources of emf
are present: (1) the applied
emf V that drives the motor,
and (2) the emf ε induced by
the generator-like action of
the rotating coil.
This induced emf ε acts
to oppose the applied
emf V as Lenz’s law
predicts. It is called the
back emf or the counter
emf of the motor.
V and ε have opposite
polarities, so the net
emf in the circuit is
V-ε. The current I
drawn by the motor is
V-ε /R.
Ex. 12 - The coil of an ac motor
has a resistance of R = 4.1 Ω.
The motor is plugged into an outlet
where V = 120.0 volts (rms), and the
coil develops a back emf of
ε = 118.0 volts (rms) when rotating at
normal speed. The motor is turning a
wheel. Find (a) the current when the
motor first starts up and (b) the current
when the motor is operating at normal
speed.
A changing current (ac, for
example) in a coil of wire can
produce an emf in a nearby coil.
The initial coil is called the
primary coil and is an
electromagnet that produces
a magnetic field. This field
creates a magnetic flux in the
secondary coil.
This production of current
in another circuit is called
mutual induction. The net
magnetic flux in the
secondary coil is N2F2,
and is proportional to
the change in current ∆I1
in the primary coil.
Using a proportionality
constant M (mutual
inductance), an equation
is created: N2F2 = MI1.
This can be substituted
into Faraday’s law to form:
ε2 = -M∆I1/∆t.
ε2 = -M∆I1/∆t
This is the emf
produced in a
secondary coil due to
mutual induction.
The unit of mutual
inductance is V•s/A, which
is called a henry (H).
It is usually found
experimentally for coil
geometry and core
material, if present.
An ac current in a coil
produces a changing
magnetic field which,
in turn, creates a
changing flux through
the coil. This is selfinduction.
Using a constant L,
called the selfinductance of the coil,
we find that:
ε = -L ∆I/∆t.
This is the emf due to
self-inductance.
Ex. 13 - A long solenoid of length
l = 8.0 x 10-2 m and crosssectional area A = 5.0 x 10-5 m2
contains n = 6500 turns per meter.
(a) Find the self-inductance of the
solenoid, assuming the core is air.
(b) Determine the emf induced in
the solenoid when the current
increases from 0 to 1.5 A in a
time of 0.20 s.
A transformer uses mutual
induction and self-induction to
increase or decrease the ac
voltage. In a high-quality
transformer: εS/εP = NS/NP. This
is because the rate of change of
flux is equal in both coils and
cancels out.
High-quality transformers
have negligible resistance,
so the emfs are nearly
equal to the voltages. The
equation εS/εP = NS/NP
becomes: VS/VP = NS/NP.
This is the transformer
equation.
The average power in
both coils is constant,
and P = IV. Therefore:
IP/IS = VS/VP =
NS/NP
A transformer that steps
up the voltage
simultaneously steps
down the current, and
vice-versa. However, the
average power remains
constant.
Ex. 14 - A step-down transformer has
330 turns in the primary coil
and 25 turns in the secondary coil.
It is connected to a 120-V wall socket,
and there is a current of
0.83 A in the primary coil. Find
(a) the voltage across the secondary
coil, (b) the current in the secondary
coil, and (c) the average electrical
power delivered to the circuits
connected to the secondary coil.