Unit Nine Summary
Grade 12 Advanced
Ministry Of Education Curriculum
Electromagnetic
induction
Prepared By Mr.Mohannad
Mohannad
Sami Karajah
Sami
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9.1: Faraday’s Experiments
In1830s Michael Faraday and independently Joseph Henry. Prove by their
experiments that a changing magnetic field could generate a potential
difference in a conductor, strong enough to produce an electric current.
https://www.youtube.com/watch?v=3rvGtHcd_7s
1- Define the magnetic flux.
Magnetic flux is the number of magnetic field lines passing
through a certain area.
2- According to faraday’s experiments, how an induced current can
produce in a wire loop?
By changing the magnetic flux through the loop.
3- How can we change the magnetic flux through the loop?
By one of these ways:
➢ Changing the magnitude of the magnetic field.
➢ Changing the area of the loop.
➢ changing the angle that loop makes with respect to the magnetic field.
4- According to the teacher explanation and the video that you watch,
determine the direction of the induced current in the wire loop in
each of the following cases.
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Note: By using two conducting loops as shown in the figure If a constant
current is flowing through loop 1, no current is induced in loop 2. If the
current in loop 1 is increased, a current is induced in loop 2 in the opposite
direction If the current in loop 1 is decreased, a current is induced in loop 2
in the same direction
Concept Check 9.1
The four figures show a bar magnet and a low-voltage light bulb connected
to the ends of a conducting loop. The plane of the loop is perpendicular to
the dotted line. In case 1, the loop is stationary, and the magnet is moving
away from the loop. In case 2, the magnet is stationary, and the loop is
moving toward the magnet. In case 3, both the magnet and loop are
stationary, but the area of the loop is increasing. In case 4, the magnet is
stationary, and the loop is rotating about its center. In which of these
situations will the light bulb be burning?
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9.2: Faraday’s law of induction
1- What is the mathematical definition of the magnetic flux?
Magnetic flux is defined as the surface integral of the magnetic field
passing through a differential element of area
⃗ . 𝑑𝐴
𝛷𝐵 = ∬ 𝐵
➢ Note: Integration of the magnetic flux over a closed surface yield zero,
this result is often termed Gauss’s Law for Magnetic Fields
2- Write the formula of the magnetic flux in case of a flat loop of area
A in a constant magnetic field.
𝛷𝐵 = 𝐵. 𝐴. cos 𝜃
➢ Note: 𝜃 is the angle between the surface normal vector to the plane
of the loop and the magnetic field lines.
✓ If the magnetic field is perpendicular to the plane of the loop, 𝜃 = 0°
✓ If the magnetic field is parallel to the plane of the loop, 𝜃 = 90
3- What is the unit of the magnetic flux?
The unit of magnetic flux 𝛷𝐵 is 𝑇𝑚2. This unit has received a special name,
the weber (𝑊𝑏)
1𝑊𝑏 = 1 𝑇. 𝑚2
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➢ Note: According to faraday’s Law, an induced potential difference
produces in a loop when the magnetic flux change by the time through
it, ∆𝑉𝑖𝑛𝑑 which called also the induced electromotive force emf
∆𝑉𝑖𝑛𝑑 = −
𝑑𝛷𝐵
𝑑𝑡
Induction in a Flat loop inside a Magnetic Field
➢ The induced potential difference in a flat loop inside a uniform
magnetic field is:
∆𝑉𝑖𝑛𝑑 = −𝐴 cos 𝜃
𝑑𝐵
𝑑𝐴
− 𝐵 cos 𝜃
+ 𝜔𝐴𝐵 sin 𝜃
𝑑𝑡
𝑑𝑡
The case
The magnitude of the magnetic field changes with time
The area of the loop changes with time
⃗ changes with time
The angle between 𝐴, 𝐵
The formula of ∆𝑉𝑖𝑛𝑑
𝑑𝐵
−𝐴 cos 𝜃
𝑑𝑡
𝑑𝐴
−𝐵 cos 𝜃
𝑑𝑡
𝜔𝐴𝐵 sin 𝜃
If we are dealing with 𝑁 flat loops so the area will be 𝑁. 𝐴
EXAMPLE 9.1
A current of 600 mA is flowing in an ideal solenoid, resulting in a magnetic
field of 0.025 T. Then the current increases with time, t, according to:
𝑖(𝑡) = 𝑖0 (1 + 2.4𝑡 2 )
If a circular coil of radius 3.4 cm with N = 200 windings is located inside the
solenoid with its normal vector parallel to the magnetic field as shown in
the figure, what is the induced potential difference in the coil at t = 2.0 s?
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EXAMPLE 9.2
A rectangular wire loop of width w = 3.1 cm and depth d0 = 4.8 cm is pulled
out of the gap between two permanent magnets. A magnetic field of
magnitude B = 0.073 T is present throughout the gap as shown in the figure.
If the loop is removed at a constant speed of 1.6 cm/s, what is the induced
voltage in the loop as a function of time?
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Concept Check 9.2
A power supply is connected to loop 1 and an ammeter as shown in the
figure. Loop 2 is close to loop 1 and is connected to a voltmeter. A graph
of the current i through loop 1 as a function of time, t, is also shown in the
figure. Which graph best describes the induced potential difference, ∆V ind,
in loop 2 as a function of time, t?
Concept Check 9.3
A long wire carries a current, i, as shown in the figure. A square loop moves
in the same plane as the wire as indicated. In which cases will the loop have
an induced current?
A.
B.
C.
D.
Cases 1 and 2
Cases 1 and 3
Cases 2 and 3
All the loops will have
an induced current.
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Exercises
9.1 A solenoid with 200 turns and a cross-sectional area of 60 cm2 has a
magnetic field of 0.60 T along its axis. If the field is confined within the
solenoid and changes at a rate of 0.20 T/s, the magnitude of the induced
potential difference in the solenoid will be:
A.
B.
C.
D.
0.0020 V.
0.02 V.
0.24 V.
0.001 V.
9.3 Which of the following will induce a current in a loop of wire in a
uniform magnetic field?
A.
B.
C.
D.
Decreasing the strength of the field.
Rotating the loop about an axis parallel to the field.
Moving the loop within the field.
All of the above.
9.5 A conducting ring is moving from left to right through a uniform
magnetic field, as shown in the figure. In which regions is there an induced
current in the ring?
A.
B.
C.
D.
Regions B and D
Regions B, C, and D
Region C
Regions A through E
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9.28 A circular coil of wire with 20 turns and a radius of
40.0 cm is lying flat on a horizontal table as shown in the
figure. There is a uniform magnetic field extending over
the entire table with a magnitude of 5.00 T and directed
to the north and downward, making an angle of 25.8°
with the horizontal. What is the magnitude of the
magnetic flux through the coil?
9.29 Suppose a magnet with an initial field of 1.20 T is
quenched in 20.0 s, and the final field is approximately zero.
Under these conditions, what is the average induced potential
difference around a conducting loop of radius 1.00 cm
oriented perpendicular to the field?
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9.30 An 8-turn coil has square loops measuring 0.200 m along a side and a
resistance of 3.00 Ω. It is placed in a magnetic field that makes an angle of
40.0° with the plane of each loop. The magnitude of this field varies with time
according to B = 1.50 t3, what is the induced current in the coil at t = 2.00 s.
9.32 A respiration monitor has a flexible loop of copper wire, which wraps
about the chest. As the wearer breathes, the radius of the loop of wire
increases and decreases. When a person in the Earth’s magnetic field
(assume 0.426×10–4 T) inhales, what is the average current in the loop, if it
has a resistance of 30.0 Ω and increases in radius from 20.0 cm to 25.0 cm
over 1.00 s? Assume that the magnetic field is perpendicular to the plane of
the loop.
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9.3: Lenz’s Law
1- What is Lenz’s Law to determine the direction of the induced current
in a loop?
An induced current will have a direction such that the magnetic field due
⃗ 𝑖𝑛𝑑 ) opposes the change in the magnetic flux
to the induced current (𝐵
that induces the current.
2- According to Lenz’s Law and your teacher explanation determine the
direction of the induced current in the following solenoids. (Draw on
the figure)
3- According to Lenz’s Law determine the direction of the induced
magnetic field and current in the following cases. (Draw on the
figure)
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4- A square conducting loop with very small resistance is moved at
constant speed from a region with no magnetic field through a region
of constant magnetic field and then into a region with no magnetic
field, as shown in the figure. As the loop enters the magnetic field,
what is the direction of the induced current? As the loop leaves the
magnetic field, what is the direction of the induced current?
Induced Potential Difference on a Wire Moving in a Magnetic Field
1- Watch this video to understand the idea
https://www.youtube.com/watch?v=GR44Ajut3hU
2- What is the formula of the induced potential difference in a wire
moving in a magnetic field?
∆𝑉𝑖𝑛𝑑 = 𝐸𝑙
∆𝑉𝑖𝑛𝑑 = 𝑣𝑙𝐵
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Concept Check 9.4
⃗ through a uniform magnetic
A metal bar is moving with constant velocity 𝒗
field pointing into the page, as shown in the figure, which of the following
most accurately represents the charge distribution on the surface of the
metal bar
EXAMPLE 9.3
In 1996, the Space Shuttle Columbia deployed a tethered satellite on a wire
out to a distance of 20 km as shown in the figure. The wire was oriented
perpendicular to the Earth’s magnetic field at that point, and the
magnitude of the field was B = 5.1×10–5 T. Columbia was traveling at a
speed of 7.6 km/s.
What was the potential difference induced between the ends of the wire?
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EXAMPLE 9.4
A conducting rod is pulled horizontally by a constant force of magnitude
F = 5.00 N, along a set of conducting rails separated by a distance
a = 0.500 m as shown in the figure. The two rails are connected, and no
friction occurs between the rod and the rails. A uniform magnetic field with
magnitude B = 0.500 T is directed into the page. The rod moves at constant
speed, v =5.00 m/s.
What is the magnitude of the induced
potential difference in the loop created by the
connected rails and the moving rod?
Exercises
9.9 Calculate the potential difference induced between the tips of the
wings of a Boeing 747-400 with a wingspan of 64.67 m in level flight at a
speed of 913 km/h. Assume that the downward component of the Earth’s
magnetic field is B = 5.00×10–5 T.
A.
B.
C.
D.
0.821 V
2.95 V
10.4 V
225 V
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9.62 A wire of length 𝒍= 10.0 cm is moving with constant velocity in the xyplane; the wire is parallel to the y-axis and moving along the x-axis. If a
magnetic field of magnitude 1.00 T is pointing along the positive
z-axis,
what must the velocity of the wire be in order to induce a potential
difference of 2.00 V across it?
7.75 A conducting rod of length 50.0 cm slides over two parallel metal bars
placed in a magnetic field with a magnitude of 1000. G, as shown in the
figure. The ends of the rods are connected by two resistors, R1 = 100Ω and
R2 = 200 Ω the conducting rod moves with a constant speed of 8.00 m/s.
A. What are the currents flowing through the two resistors?
B. What power is delivered to the resistors?
C. What force is needed to keep the rod moving with constant velocity?
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Eddy Currents
1- Define eddy currents.
Eddy currents are loops of electrical current, induce in the solid metal
plates when the magnetic flux change through this plate.
Note: eddy currents can be useful and are employed in certain practical
applications, such as the brakes of train cars, because it can convert the
mechanical energy of a moving metal plate into heat energy. The eddy
currents disperse heat in the metal because of its finite resistance.
2- Where can eddy currents be undesirable?
Eddy currents are often undesirable. For example: in some electrical
devices such as electrical motors, eddy current resist the motion.
3- How can reduce the undesirable effects of eddy currents?
By segmenting or laminating electrical devices that must operate in an
environment of changing magnetic field
An experiment shows the effect of eddy currents in a moving metal plate
Two pendulums, one consisting of an arm and a solid metal plate and a second consisting
of an arm and a slotted metal plate. The five frames are in time sequence from left to right,
with the two pendulums starting their motion together in the second frame from the left.
The pendulum with the solid plate stops in the gap, while the pendulum with the slotted
plate passes through the gap.
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Metal detector
1- What is the scientific principle of the metal detector working?
A metal detector works by using electromagnetic induction, often called
pulse induction.
2- What are the components of the metal detector?
A metal detector has a transmitter coil and a
receiver coil. An alternating current is applied
to the transmitter coil, which then produces
an alternating magnetic field. As the magnetic
field of the transmitter coil increases and
decreases, it induces a current in the receiving
coil that tends to counteract the change in the
magnetic flux produced by the transmitter
coil. The induced current in the receiver coil is
measured when nothing but air is between the
coils.
3- How does metal detector work?
The effect of the metal plate when it passes through the metal detector
is to decrease the observed current in the receiver coil. That happens
because the changing magnetic field from the transmitter coil induces
eddy currents in the metal plate, and this will decrease the induced
current in the receiver coil.
4- Name some applications of the metal detector.
➢ Airport metal detector
➢ Metal detectors to control traffic.
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9.4: Generators and motors
1- Electric motors and electric generators are everyday applications of
magnetic induction.
2- Electric generator: a device that produces electric current from
mechanical motion.
3- Electric motor: a device that produces mechanical motion from
electric current.
4- A simple generator consists of a loop forced to rotate in a fixed
magnetic field.
5- In a direct-current generator (DC), the rotating loop is connected to an
external circuit through a split commutator ring.
As the loop turns, the connection is reversed twice per revolution, so
the induced potential difference always has the same sign.
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6- In alternating-current generator (AC), each end of the loop is
connected to the external circuit through its own solid slip ring. Thus,
this generator produces an induced potential difference that varies
from positive to negative and back.
7- An alternating current is a current that varies in time between positive
and negative values, with the variation often showing a sinusoidal
form.
Self-test Opportunity 9.4/Pg:239
A generator is operated by rotating a coil of 𝑵 turns in a constant magnetic
field of magnitude 𝑩 at a frequency 𝒇. The resistance of the coil is 𝑹, and
the cross-sectional area of the coil is 𝑨. Decide whether each of the
following statements is true or false.
A. The average induced potential difference doubles if the frequency, 𝒇,
is doubled.
B. The average induced potential difference doubles if the resistance,
𝑹, is doubled.
C. The average induced potential difference doubles if the magnetic
field’s magnitude, 𝑩, is doubled.
D. The average induced potential difference doubles if the area, 𝑨, is
doubled.
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Regenerative braking
Hybrid cars are propelled by a combination of gasoline power and electrical
power. One attractive feature of a hybrid vehicle is that it is capable of
regenerative braking. When the brakes are used to slow or stop a nonhybrid
vehicle, the kinetic energy of the vehicle is turned into heat in the brake pads.
This heat dissipates into the
environment, and energy is lost. In a
hybrid car, the brakes are connected
to the electric motor (Figure 29.19),
which functions as a generator,
charging the car’s battery. Thus, the
kinetic energy of the car is partially recovered during braking, and this energy
can later be used to propel the car, contributing to its efficiency, and greatly
increasing its gas mileage in stop-and-go driving
Exercises
9.43 A simple generator consists of a loop rotating inside a constant
magnetic field (see Figure 29.17). If the loop is rotating with frequency f, the
magnetic flux is given by 𝜱(t) = BA cos (2πft). If B = 1.00 T and A = 1.00 m2,
what must the value of f be for the maximum induced potential difference
to be 110. V?
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9.44 A motor has a single loop inside a magnetic field of magnitude 0.87 T.
If the area of the loop is 300. cm2, find the maximum angular speed possible
for this motor when connected to a source of emf providing 170 V.
9.45 Your friend decides to produce electrical power by turning a coil of
1.00×105 circular loops of wire around an axis parallel to a diameter in the
Earth’s magnetic field, which has a local magnitude of 0.300 G. The loops have
a radius of 25.0 cm.
a) If your friend turns the coil at a frequency of 150.0 Hz, what peak current
will flow in a resistor, R = 1500. Ω, connected to the coil?
b) The average current flowing in the coil will be 0.7071 times the peak current.
What will be the average power obtained from this device?
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9.5: Induced Electric Field
➢ Notes:
1- Changing magnetic flux in a wire loop induces an electric field.
2- The induced electric field B vector is directed tangentially to the loop.
3- The magnitude of the induced electric field can be calculated from this
equation, which can be applied to any closed path in a changing
magnetic field, even if no conductor exists in the path.
න 𝐸⃗ . 𝑑𝑠 = −
𝑑𝛷𝐵
𝑑𝑡
External Problem
The magnetic field inside a wire loop with radius of 15.0 cm, change its
magnitude according to B=2t2 find the magnitude of the induced electric field
at t=2 s.
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9.6: Inductance of a Solenoid
1- Define the flux linkage.
The flux linkage is the product of the number of windings and the
magnetic flux in one loop, or 𝑁𝛷𝐵.
➢ Note: The flux linkage is proportional to the current, so we can write
the following equation:
𝑁𝛷𝐵 = 𝐿𝑖
➢ Note: 𝐿 is called the inductance, its unit in Henry (𝐻)
1𝑇𝑚2
1𝐻 =
1𝐴
2- Name the factors that effect on the inductance.
The inductance of a solenoid depends only on the geometry of the
device.
✓ The length of the solenoid
✓ The cross-sectional area of the solenoid.
✓ The Number of turns of the solenoid.
𝐿=
𝐿
𝑛
𝑙
𝐴
𝑁𝛷𝐵
= 𝜇0𝑛2 𝑙𝐴
𝑖
The physical quantity
The solenoid inductance
Number of turns per unit length
Solenoid length
Solenoid cross sectional area
The unit
𝐻
𝑡𝑢𝑟𝑛𝑠/𝑚
𝑚
𝑚2
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External conceptual questions
1- Which of the following identical solenoids has the highest
inductance?
ABCD-
Solenoid 1
Solenoid 2
Solenoid 3
All of them has the same inductance.
2- If the length of a solenoid doubled, how will its inductance change?
ABCD-
It will not change.
I will be double.
It will be half.
It will be quadruple.
External Problems
1- A solenoid with 50 cm length and a diameter of 10 cm has 700 turns.
Calculate the inductance of the solenoid.
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2- What is the cross-sectional area of 20 cm solenoid with inductance
of 3.4 𝒎𝑯, and 300 turns?
9.7: Self-inductance and Mutual induction
https://www.youtube.com/watch?v=VFPwgjGHqFQ
1- Self- induction:
Changing current in a coil change the magnetic
flux in the coil and as a result of that, an
induced potential difference produces and
opposes the changes of the current. This
phenomenon is called self-induction.
∆𝑉𝑖𝑛𝑑 = −
𝑑(𝑁𝛷𝐵 )
𝑑𝑖
= −𝐿
𝑑𝑡
𝑑𝑡
2- Mutual induction:
If two coils are close to each other, the
changing current in the first coil produces a
changing magnetic flux in the second coil and
induces a potential difference in it. This
phenomenon is called mutual induction.
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3- Define the mutual inductance 𝑴:
Let’s consider two adjacent coils with their central axes aligned. Coil 1 has
N1 turns, and coil 2 has N2 turns. The current in coil 1 produces a magnetic
⃗ 1 . The flux linkage in coil 2 resulting from the magnetic field in coil
field, 𝐵
1 is N2𝜱1→2. The mutual inductance, M1→2, of coil 2 due to coil 1 is defined
as:
𝑀1→2 =
𝑁2 𝛷1→2
𝑖1
If we reverse the roles of the two coils
𝑀2→1 =
𝑁1 𝛷2→1
𝑖2
➢ Important note: The mutual inductance is equal in the two cases so:
𝑀1→2 = 𝑀2→1 = 𝑀
𝑀 unit is Henry 𝐻
4- How to calculate the induced potential difference according to the
mutual induction in each of the previous cases?
∆𝑉𝑖𝑛𝑑,1 = −𝑀
𝑑𝑖2
𝑑𝑡
∆𝑉𝑖𝑛𝑑,2 = −𝑀
𝑑𝑖1
𝑑𝑡
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SOLVED PROBLEM 9.2
A long solenoid with circular cross section of radius r 1 = 2.80 cm and n = 290
turns/cm is inside and coaxial with a short coil with circular cross section of
radius r2 = 4.90 cm and N = 31 turns as shown in the figure. The current in
the solenoid is increased at a constant rate from zero to i= 2.20 A over a
time interval of 48.0 ms.
What is the potential difference induced in the short coil while the current is
changing?
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Exercises
9.7 Which of the following statements regarding self-induction is correct?
A. Self-induction occurs only when a direct current is flowing through a
circuit.
B. Self-induction occurs only when an alternating current is flowing
through a circuit.
C. Self-induction occurs when either a direct current or an alternating
current is flowing through a circuit.
D. Self-induction occurs when either a direct current or an alternating
current is flowing through a circuit as long as the current is varying.
9.46 The figure shows the current through a 10.0-mH inductor over a time
interval of 8.00 ms. Draw a graph showing the self-induced potential
difference, ∆𝑽𝒊𝒏𝒅,𝑳 for the inductor over the same interval.
External: A current flow through a solenoid with 50 cm length and a
diameter of 10 cm has 700 turns, if the current changes from 12 A to 27 A
in 125 ms. What is the magnitude and the direction of the induced emf in
the solenoid?
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9.48 A short coil with radius R = 10.0 cm contains N = 30.0 turns and
surrounds a long solenoid with radius r = 8.00 cm containing n = 60 turns
per centimeter. The current in the short coil is increased at a constant rate
from zero to i = 2.00 A in a time of t = 12.0 s. Calculate the induced potential
difference in the long solenoid while the current is increasing in the short
coil.
9.63 The magnetic field inside the solenoid in the figure changes at
the
rate of 1.50 T/s. A conducting coil with 2000 turns surrounds
the solenoid, as shown. The radius of the solenoid is 4.00 cm, and
the radius of the coil is 7.00 cm. What is the potential difference
induced in the coil?
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External Problems
1- A cupper coil of 30 turns and 5 cm2 cross-sectional area, rotates
during 0.25 s within a uniform magnetic field from a vertical position
to a horizontal one relative to the magnetic field direction. If the
induced potential difference in the coil is 2 V what is the magnitude
of the magnetic field?
2- The graph below shows the magnetic flux through a conducting loop
as a function of time. What is the induced potential difference in the
loop during the first 20 seconds?
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9.8: RL circuit
In the circuit with both the resistor and the inductor, the increasing current
flowing through the inductor creates a self-induced potential difference that
tends to oppose the increase in current. As time passes, the change in
current decreases, and the opposing self-induced potential difference also
decreases. After a long time, the current becomes steady at the value V emf/R.
We can use Kirchhoff’s Loop Rule to analyze this circuit:
𝑉𝑒𝑚𝑓 − 𝑖𝑅 − 𝐿
𝑑𝑖
𝑑𝑡
=0
The solution to this differential equation
𝑖(𝑡) =
𝑉𝑒𝑚𝑓
(1 − 𝑒 −𝑡/(𝐿/𝑅) )
𝑅
𝑖(𝑡) =
𝑖(𝑡) = 𝑖0 (1 − 𝑒 −𝑡/(𝐿/𝑅) )
𝑉𝑒𝑚𝑓
(1 − 𝑒 −𝑡/𝜏 )
𝑅
𝑖 (𝑡) = 𝑖0 (1 − 𝑒 −𝑡/𝜏 )
The quantity 𝐿/𝑅 is the time constant of the 𝑅𝐿 circuit:
𝜏=
𝐿
𝑅
For 𝑡 = 0, the current is zero.
For 𝑡 → ∞, the current is given by
𝑖 = 𝑖0 = 𝑉𝑒𝑚𝑓 /𝑅
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Now if the source of emf had been connected and is suddenly removed
We can use Kirchhoff’s Loop Rule to analyze this circuit:
𝑖𝑅 + 𝐿
𝑑𝑖
=0
𝑑𝑡
The resistor causes a potential drop, and the inductor has a self-induced
potential difference that tends to oppose the decrease in current. The
solution of equation is
𝑖 (𝑡 ) =
𝑉𝑒𝑚𝑓 −𝑡/(𝐿/𝑅)
𝑒
𝑅
𝑖(𝑡) =
𝑖(𝑡) = 𝑖0 𝑒 −𝑡/(𝐿/𝑅)
𝑉𝑒𝑚𝑓 −𝑡/𝜏
𝑒
𝑅
𝑖(𝑡) = 𝑖0 𝑒 −𝑡/𝜏
For 𝑡 → ∞, the current is zero.
For 𝑡 = 0, the current is given by
𝑖 = 𝑖0 = 𝑉𝑒𝑚𝑓 /𝑅
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Mohannad Sami
Concept Check 9.7
Consider the RL circuit shown in the figure. When the switch is closed, the
current in the circuit increases exponentially to the value 𝒊 = 𝑽𝒆𝒎𝒇 /𝑹. If
the inductor in this circuit is replaced with an inductor having three times
the number of turns per unit length, the time required to reach a current
of magnitude 𝟎. 𝟗𝒊
A. Increase
B. Decrease
C. Stay the same.
Exercises
9.49 Consider an RL circuit with resistance R = 1.00 MΩ and inductance
L = 1.00 H, which is powered by a 10.0-V battery.
a) What is the time constant of the circuit?
b) If the switch is closed at time t = 0, what is the current just after that
time? After 2.00 µs? When a long time has passed?
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Mohannad Sami
9.50 In the circuit in the figure, R = 120. Ω, L = 3.00 H, and Vemf = 40.0 V.
After the switch is closed, how long will it take the current in the inductor
to reach 300. mA?
9.51 The current is increasing at a rate of 3.6 A/s in an 𝑹𝑳 circuit with
R = 3.25 Ω and L = 440 𝒎𝑯. What is the potential difference across the
circuit at the moment when the current in the circuit is 3.0 A?
9.52 In the circuit in the figure, a battery supplies V emf = 18 V and
R1 = 6.0 Ω, R2 = 6.0 Ω, and L = 5.0 H.
Calculate each of the following immediately after the switch is closed:
A.
B.
C.
D.
E.
F.
G.
the current flowing out of the battery
the current through R1
the current through R2
the potential difference across R1
the potential difference across R2
the potential difference across L
the rate of current change across R1
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Mohannad Sami
Problem 9.52 solution
9.53 In the circuit in the figure, a battery supplies V emf = 18 V and
R1 = 6.0 Ω, R2 = 6.0 Ω, and L = 5.0 H.
Calculate each of the following a long time after the switch is closed:
A.
B.
C.
D.
E.
F.
G.
the current flowing out of the battery
the current through R1
the current through R2
the potential difference across R1
the potential difference across R2
the potential difference across L
the rate of current change across R1
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Mohannad Sami