SHS
General Physics 2 Activity Sheet
Quarter 3 – MELC 54, 55,
58 & 59
Week 8
Magnetic Field, Magnetic Flux, and
Magnetic Force
REGION VI – WESTERN VISAYAS
General Physics 2
Activity Sheet No. 14 - Magnetic Field, Magnetic Flux, and Magnetic Force
First Edition, 2021
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Name of Learner: ___________________________________________________
Grade and Section: ____________________________Date: __________________
GENERAL PHYSICS 2 ACTIVITY SHEET No. 14
Magnetic Field, Magnetic Flux, and Magnetic Force
I.
Learning Competencies with Codes
• Differentiate electric interactions from magnetic interactions.
(STEM_GP12EMIIIh-54)
• Evaluate the total magnetic flux through an open surface.
(STEM_GP12EMIIIh-55)
• Describe the motion of a charged particle in a magnetic field in terms of
its speed, acceleration, cyclotron radius, cyclotron frequency, and kinetic
energy. (STEM_GP12EMIIIh-58)
• Evaluate the magnetic force on an arbitrary wire segment placed in a
uniform magnetic field. (STEM_GP12EMIIIh-59)
II.
Background Information for Learners
Physics is really both fascinating and puzzling. The nature of light is a
good example for this --- light is electrical in nature, and magnetic too!
The link between electricity and magnetism was primarily established by
Hans Christian Oersted when he discovered that electricity can produce
magnetism. Moreover, scientists also found out that electricity can be
generated from magnetism. For instance, Alessandro Volta invented the voltaic
pile which is the world’s first source of direct current. Michael Faraday also
discovered that a changing magnetic field is the key to getting electricity from
magnetism.
In this learning activity, you will get to know about the force due to
magnetic fields, as well as the sources of magnetic fields.
ELECTRIC INTERACTIONS VS. MAGNETIC INTERACTIONS
Aside from being
the
only
inhabitable
planet, there are so many
interesting things about
Earth. The Earth is a
giant, natural magnet,
and its magnetic field
serves as its “shield”.
Earth’s magnetic field is a
lot like that of a giant bar
magnet, and this protects
the Earth from the direct Figure
1.
Earth-Sun
System.
(Retrieved
from
hit of solar flares. As https://svs.gsfc.nasa.gov/hyperwall/index/download/events/2013shown in Figure 1, the agu/lika/201312_agu_lika_0001_sun_earth_overview.hwshow.html)
magnetosphere which is
a protective cavity of the Earth, is created as the Earth’s magnetic field carves
out a hollow in the solar wind.
What is even more interesting is the fact that the Earth has two kinds of
poles: (1) magnetic poles and (2) geographic poles.
Figure 2. The geographic and magnetic poles of the
Earth.
(Retrieved
from
https://www.netclipart.com/isee/iRbhwiT_earthmagnetic-field-drawing/)
The geographic north pole is also
known as the true north. This is the
location on Earth where all the lines of
longitude intersect. The geographic
south pole is directly opposite on the
other side of the planet.
Remember the law of charges? The
same law applies to magnets: Like poles
repel; unlike poles attract.
Now, carefully read the next line,
“Earth’s magnetic poles are upside
down!” (See Figure 2)
Yes, you read it right! This is another
interesting fact, isn’t it? The Earth’s
geographic north pole is in the Arctic
ocean, while its magnetic pole is located
somewhere in Antarctica. On the other
hand, the Earth’s geographic south can
be found somewhere in Antarctica, while
its magnetic south pole is somewhere in
Northern Canada.
Before you delve deeper into
magnetic field, let us first identify the
differences
between
electric
and
magnetic interactions.
Exercise 1. May the Electric and Magnetic Forces Be With You
Analyze the following figures and answer the questions that follow.
Figure 3. Electric Field. (Retrieved from
https://www.ck12.org/book/ck-12-physicalscience-for-middle-school/section/23.1/)
Figure 4. Magnetic Field. (Retrieved from
https://www.istockphoto.com/photos/electromag
netic-field)
5
Guide Questions:
1) What are the similarities of electric and magnetic fields?
2) What are the differences of electric and magnetic fields?
In the previous discussions, you learned that a field refers to property of a
region of space that can exert a force to objects found in that region of space.
More so, you learned that region of space that exerts a force to charged objects
in that region of space is known as the electric field. Now, since you are talking
about magnet, you will be dealing more about magnetic field. Every magnet, as
well as the current-carrying wire is surrounded by a magnetic field. Magnetic
field is the area in which magnetic force is exerted. This means that if ever
another magnet or current-carrying wire is introduced into this region will
experience a magnetic force. Other differences of electric and magnetic fields
are summarized in the table below.
Basis for
Comparison
Definition
Existence of Pole
Symbol
Equation
SI Unit
Instrument Used for
Measuring the
Intensity
Orientation in an
Electromagnetic
Field
Electric Field
Magnetic Field
A force (vector) field that fills
the space around every
electric charge or group of
charges.
A force (vector) field that
fills the space around
every magnet or currentcarrying wire.
This is induced by a single
charge (positive or negative
charge) and is directly
proportional to the flux.
This is induced by a north
and south pole of the
magnet, and depends on
the number of field lines
produces by the magnet
Dipole
⃗
𝐵
∅
⃗ =
𝐵
𝐴
Tesla
Monopole
𝐸⃗
𝐸⃗ =
𝐹
𝑞
Volt/meter
or
Newton/coulomb
Electrometer
Perpendicular
magnetic field
Magnetometer
to
the Perpendicular
electric field
to
an
Like charges repel, unlike Like poles repel, unlike
charges attract
poles attract
Interaction
(Electric
Force) (Magnetic Force)
Stationary/moving
A moving charge whose
charges are both affected
velocity has a component
that is perpendicular to
the magnetic field vector
interacts with
magnetic field
6
Emerge
from
positive
charges and terminate in
negative charges
Do not form a loop
Exists in two dimensions
Point out of the north
Field Lines
pole and curve around
toward the south pole
Loop of Field Lines
Form a loop
Exists
in
three
Dimension
dimensions
Can do work (the speed and Cannot do work (speed
Work
direction
of
particles of
particles
remain
changes)
constant)
Table 1. Differences Between Electric Field and Magnetic Field (Source:
Circuit Globe, 2021)
MAGNETIC FLUX THROUGH AN OPEN SURFACE
Magnetic flux is just the same with the
electric flux in the sense that it is the number
A
of field lines in a specific area. Moreover, the
magnetic flux is a scalar quantity that is
defined as the dot product of the magnetic field
and area vector (Santisteban-Cook and
Baguio, 2018). Mathematically, magnetic flux
⃗ ∙ 𝑨 = 𝑩𝑨𝒄𝒐𝒔𝜽, where
is expressed as Ф𝑩 = ⃗𝑩
 Ф𝑩 is the magnetic flux in the unit
Tesla•meter squared (T•m2) which is
Figure 5. Magnetic flux is the product of the area
also equal to weber (Wb)
and the component of the magnetic field
⃗⃗ is the magnetic field intensity in the perpendicular to the area. (Retrieved from
 𝑩
https://commons.princeton.edu/59-tigerunit Tesla (T)
 A is the area vector which has a cub/electrical/electromagnetism/)
direction that is perpendicular to the surface and is measured in the unit
meter squared (m2)
 Θ is the angle between the area and magnetic field vectors.
Additionally, the total magnetic flux through the surface is the sum of
the contributions from the individual area elements. The magnetic flux is
affected by the angle θ between the area and the magnetic field vectors. Since
the magnetic flux is a dot product, it is therefore maximum when the angle is
either zero and zero when the angle is 90°.
Now, apply the equation on the following sample problems.
⃗ and θ are 0.02T
1) The dimension of a rectangular loop is 0.50m and 0.60m. 𝐵
and 45° respectively. Determine the magnetic flux through the surface.
Given: A = 0.50 m x 0.60 m
Req’d: Ф𝐵 = ?
⃗𝐵 = 0.02 T
θ = 45°
⃗ ∙𝐴=𝐵
⃗ 𝐴𝑐𝑜𝑠𝜃
Eq’n: Ф𝐵 = 𝐵
Sol’n: Determine the area of the rectangular loop.
𝐴 = 𝑙𝑤
𝐴 = (0.50 𝑚)(0.60 𝑚)
𝐴 = 0.30 𝑚2
7
Solve for the magnetic flux.
⃗ 𝐴𝑐𝑜𝑠𝜃
Ф𝐵 = 𝐵
Ф𝐵 = (0.02 𝑇)(0.30 𝑚2 )𝑐𝑜𝑠45°
Ф𝑩 = 𝟒. 𝟐𝟒 × 𝟏𝟎−𝟑 𝑾𝒃
2) A square loop of wire 10.0 cm on a side is in a 1.25-T magnetic field B. What
are the maximum and minimum values of flux that can pass through the
loop?
Given: s = 10 cm
Req’d: Ф𝐵𝑚𝑎𝑥 = ?
⃗𝐵 = 1.25 T
Ф𝐵𝑚𝑖𝑛 = ?
⃗
⃗
Eq’n: Ф𝐵 = 𝐵 ∙ 𝐴 = 𝐵 𝐴𝑐𝑜𝑠𝜃
1𝑚
Sol’n: First, convert cm to m → (10 𝑐𝑚) (100 𝑐𝑚) = 0.1 𝑚
Then, determine the area
→ 𝐴 = 𝑠 2 → 𝐴 = (0.1 𝑚)2 → A = 0.01 m2
⃗ 𝐴𝑐𝑜𝑠𝜃. It is maximum for θ = 0°,
The flux is given by the equation Ф𝐵 = 𝐵
⃗ . Hence,
which occurs when the plane of the loop is perpendicular to 𝐵
2
Ф𝐵𝑚𝑎𝑥 = (1.25 𝑇)(0.01 𝑚 )𝑐𝑜𝑠0°
Ф𝑩𝒎𝒂𝒙 = 𝟎. 𝟎𝟏𝟐𝟓 𝑾𝒃
The minimum value occurs when θ = 90° and the plane of the loop is aligned
⃗ . Thus,
with 𝐵
Ф𝐵𝑚𝑖𝑛 = (1.25 𝑇)(0.01 𝑚2 )𝑐𝑜𝑠90°
Since cos90° = 0, therefore, Ф𝑩𝒎𝒊𝒏 = 𝟎.
3) A rectangular loop of wire (2.5cm X 4.0cm) is placed
in between the poles of a horseshoe magnet such that
the upward magnetic field lines make a 30.0˚ angle
with the loop surface as shown in the figure on the
right. The magnetic field has a strength of 1.2 Tesla.
Calculate the magnetic flux through the loop.
Given: A = 2.5cm X 4.0cm
Req’d: Ф𝐵 = ?
Figure 6. Horseshoe magnet
with a rectangular loop of
⃗𝐵 = 1.2 T
wire.
(Retrieved
from
⃗ 𝐴𝑐𝑜𝑠𝜃
Eq’n: Ф𝐵 = 𝐵
http://www.pstcc.edu/depar
Sol’n: First, determine the value of θ. The angle the tments/natural_behavioral_s
magnetic field lines make with the surface of the loop ciences/Web%20Physics/Cha
pter22.htm)
⃗ makes with the
is 30.0°. However, θ is the angle the 𝐵
line normal to the surface. Hence, 𝜃 = 90.0° − 30.0° = 60.0°. Then, solve for
the area.
𝐴 = 𝑙𝑤
𝐴 = (0.025 𝑚)(0.04 𝑚)
𝐴 = 1 × 10−3 𝑚2
Now, solve for the magnetic flux.
⃗ 𝐴𝑐𝑜𝑠𝜃
Ф𝐵 = 𝐵
Ф𝐵 = (1.2 𝑇)(1 × 10−3 𝑚2 )𝑐𝑜𝑠60°
Ф𝑩 = 𝟔 × 𝟏𝟎−𝟒 𝑾𝒃
8
For sure, you can now on your own solve word problems involving the
concept of magnetic flux. Are you ready for the next exercise? I know you will
do well on this.
Exercise 2. Let’s Do the Math!
Young scientist, it’s time to showcase your mathematical skills! Solve the
following word problems and show your complete process (Given-RequiredEquation-Solution). Box all your final answers.
1) Thea is driving down the expressway on her way to the office in a town where
the horizontal component of Earth’s magnetic field is 3.5 x 10-5 T to the north.
The driver’s side window of Thea’s car has an area of 0.42 m2.
(a) What is the magnitude of the flux through the window if the car is moving
south?
(b) What is the magnitude of the flux through the window if the car is moving
west?
2) A magnetic field of magnitude 5.9 x 10-5 T is directed 72° below the horizontal
and passes through a horizontal region 130 cm by 82 cm. Calculate the flux
from the vertical.
3) A house has a floor of dimensions 22 m by 18 m. The local magnetic field
due to Earth has a horizontal component 2.6 x 10-5 T and a downward vertical
component 4.2 x10-5 T. Calculate the magnetic flux.
MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD
By now, you know that a charged particle moving in a magnetic field is
acted upon by a magnetic force. Furthermore, the
motion of the charged particle is determined by
S
Newton’s laws. This time you will learn about
circular motion of the charged particle, as well as the
other motion brought about by the charged particle
P
entering a magnetic field.
Exercise 3. Visualize and Explain
To help you visualize the motion of a charged
particle in a magnetic field, consider Figure 7. As
shown in the figure, a particle with positive charge
q is at point O, moving with velocity 𝑣 in a uniform
⃗ directed into the plane of the figure.
magnetic field 𝐵
9
O
Figure 7. A charged particle moves in a plane
⃗.
perpendicular to a uniform magnetic field 𝐵
Specifically, this shows the orbit of a charged
particle in a uniform magnetic field (Retrieved
from
https://web.njit.edu/~levyr/Physics_121/chapt
er29.ppt)
Guide Questions:
⃗ with magnitude 𝐹 = 𝑞𝑣𝐵,
1) For the magnetic force to be 𝐹 = 𝑞𝑣 × 𝐵
⃗ with respect to each
what should be the orientation of vectors 𝑣 and 𝐵
other?
2) Based on Fig. 7, what is the orientation of the force 𝐹 with respect to
𝑣? Why is it that the 𝐹 is always in this orientation with respect to 𝑣?
3) Describe the path and the speed of a charge moving at right angles
to a uniform magnetic field.
4) Describe the direction of the magnetic force in Figure 7. What is the
effect of this magnetic force?
If a charged particle is under the sole action of magnetic field, then its
motion is always with constant speed. With this, both 𝐹 and 𝑣 have constant
magnitudes. As shown in Fig. 7, at points where the directions of force and
velocity have changed, such as at points P and S, their magnitudes remained
constant. Hence, the particle moves under the influence of a constantmagnitude force that is always perpendicular to the velocity of the particle. With
this, the magnetic force can never do work on the particle.
𝑊 = 𝐹 ⋅ ∆𝑟 = 0 since 𝐹𝐵 ⊥ ∆𝑟
The magnetic field cannot change the kinetic energy of a moving particle,
but can change its direction of motion.
𝑊𝑛𝑒𝑡 = ∆𝐾𝐸 = 0
Exercise 4. Think and Share
Do you know that there is a way to confine electrons and make them
move in a circular motion without spatial restrictions of their paths? Get to know
more about this by analyzing the illustration below and answering the questions
that follow.
Figure 8. Electrons confined to move within a circular path
(Retrieved from Teaching Guide for SHS GP2, n.d.)
10
Guide Questions:
1) What kind of force in classical physics is most closely related to the
motion of electrons in the picture above, if the electrons are
assumed as classical mass particles?
2) What is the equation for a centripetal force expressed in terms of
mass and centripetal acceleration?
3) What is the property of the electrons that determines the radius of
their circular motion (the pink part on the illustration) based on the
equation for question #2?
Recall that in General Physics 1, you learned that object moving in a
circular motion has an acceleration 𝑎𝑐 , velocity 𝑣 = 𝑟𝜔, and 𝜔 = 2𝜋𝑓. For a
charged particle moving in a circular motion in a region of uniform magnetic
field, it experiences the centripetal force that is perpendicular to its velocity.
It can also be observed in Figure 7 that particle’s path is a circle, traced
2
out with constant speed v. The centripetal acceleration is 𝑣 ⁄𝑟 and only the
magnetic force acts, so from Newton’s second law,
𝐹𝐵 = 𝐹𝑐 = 𝑚𝑎𝑐
𝑣2
𝐹𝑐 = 𝑚
𝑟
where m is the mass of the particle. Noting that sinθ = 1, we see that
𝐹 = |𝑞|𝑣𝐵. The Lorentz magnetic force supplies the centripetal force, so the
following terms are equal
𝑣2
|𝑞|𝑣𝐵 = 𝑚
𝑟
Hence, to solve for the radius r of the circular orbit in a magnetic field, we can
use the equation,
𝒓=
𝒎𝒗
|𝒒|𝑩
This radius is known as the cyclotron radius or gyroradius. A cyclotron radius
is the radius of the charged particle’s circular motion in the presence of a
uniform magnetic field.
The angular speed 𝜔 of the particle can be found from the equation 𝑣 =
𝑚𝑣
𝑟𝜔. Combining this with 𝑟 = |𝑞|𝐵, we get
|𝑞|𝐵 |𝑞|𝐵
𝑣
𝜔= =𝑣
=
𝑟
𝑚𝑣
𝑚
Frequency, which is the number of revolutions per unit time is
quantitatively expressed as 𝑓 = 𝜔⁄2𝜋. This frequency is known as the
cyclotron frequency and is independent of the radius of the path. In a particle
accelerator called a cyclotron, particles moving in nearly circular paths are
given a boost twice each revolution, increasing their energy and their orbital
radii but not their angular speed or frequency. (Young and Freedman, 2012).
Exercise 5. Think Critically
11
My dear young scientist, I know you have learned a lot from this
discussion. This time you will be showcasing your analytical as well as critical
thinking skills as you discuss the following.
1) At a given instant, an electron and a proton are moving with the same
velocity in a constant magnetic field. Compare the magnetic forces on these
particles. Compare their accelerations.
MAGNETIC FORCE OF A CURRENT-CARRYING WIRE
Can you still recall the right-hand rule? To refresh your memory in the
said rule, study the attached handout on the next page regarding right hand
rule for this will be an essential thing for discussion on the magnetic force of a
current-carrying wire.
Source: https://www.csusm.edu/stemsc/handouts/project2_handouts/phys202_right_hand_rule.pdf
12
Do you know what makes an electric motor
work? Within the motor are conductors that carry
currents (that is, whose charges are in motion), as
well as magnets that exert forces on the moving
charges. Hence there is a magnetic force along the
length of each current-carrying conductor, and these
forces make the motor turn.
You can compute the force on a currentcarrying conductor starting with the magnetic force
⃗ on a single moving charge. Figure 9
𝐹 = 𝑞𝑣 × 𝐵
shows the following information:
Figure 9. Forces on a moving
 a straight segment of a conducting wire, with positive charge in a current-carrying
(Retrieved from Young
length l and cross-sectional area A; the conductor.
and Freedman, 2012).
current is from bottom to top
⃗ perpendicular to the plane of the
 the wire is in a uniform magnetic field 𝐵
diagram and directed into the plane
⃗
 the drift velocity 𝑣𝑑 is upward, perpendicular to 𝐵
⃗ , directed to the left
 the average force on each charge is 𝐹 = 𝑞𝑣𝑑 × 𝐵
 since 𝑣𝑑 and 𝐵 are perpendicular, the magnitude of the force is 𝐹 = 𝑞𝑣𝑑 𝐵
You can derive an expression for the total force on all the moving charges
in a length l of conductor with cross-sectional area A. The number of charges
per unit volume is n; a segment of conductor with length l has volume Al and
contains a number of charges equal to nAl. The total force 𝐹 on all the moving
charges in this segment has magnitude
𝐹 = (𝑛𝐴𝑙)(𝑞𝑣𝑑 𝐵) = (𝑛𝑞𝑣𝑑 𝐴)(𝑙𝐵)
You have learned that the current density 𝐽 = 𝑛𝑞𝑣𝑑 . The product JA is the total
current I, so the equation can be written as 𝐹 = 𝐼𝑙𝐵.
⃗ perpendicular to the wire (and to the drift
Only the component of 𝐵
velocities of the charges) exerts a force; this component is 𝐵⊥ = 𝐵 sin 𝜙. The
magnetic force on the wire is then
𝑭 = 𝑰𝒍𝑩⊥ = 𝑰𝒍𝑩 𝐬𝐢𝐧 𝝓.
The force is always perpendicular to both the conductor and the field, with
the direction determined by the same right-hand rule we used for a moving
positive charge. Hence this force can be expressed as a vector product, just
like the force on a single moving charge. We represent the segment of wire with
a vector 𝑙 along the wire in the direction of the current; then the force 𝐹 on this
segment is
⃗ = 𝑰𝒍 × 𝑩
⃗⃗
𝑭
⃗ , length 𝑙, and force 𝐹 vectors for a straight wire carrying a current I. (Retrieved from Young
Figure 10. Magnetic field 𝐵
and Freedman, 2012).
13
Now, let us apply the equation on the following sample problems.
1) A straight horizontal copper rod carries a current of 50.0 A from west to east
in a region between the poles of a large electromagnet. In this region there
is a horizontal magnetic field toward the northeast (that is, 45° north of east)
with magnitude 1.20 T.
(a) Find the magnitude and direction of the force on a 1.00-m section of rod.
(b) While keeping the rod horizontal, how should it be oriented to maximize
the magnitude of the force? What is the force magnitude in this case?
This is a straight wire segment in a uniform magnetic field. Our target
variables are the force 𝐹 on the segment and the angle 𝜙 for which the force
magnitude F is greatest. We find the magnitude of the magnetic force using
𝐹 = 𝐼𝑙𝐵⊥ = 𝐼𝑙𝐵 sin 𝜙 and the direction from the right-hand rule.
(a) Given: I = 50.0 A
Req’d: 𝐹 = ?
𝑙 = 1.00 m
A = 1.20 T
𝜙 = 45°
Eq’n: 𝐹 = 𝐼𝑙𝐵 sin 𝜙
Sol’n:
𝐹 = (50.0 𝐴)(1.00 𝑚)(1.20 𝑇) sin 45°
𝑭 = 𝟒𝟐. 𝟒𝟑 𝑵
The direction of the force is perpendicular to the plane of the current and the
field, both of which lie in the horizontal plane. Thus the force must be vertical;
the right-hand rule shows that it is vertically upward (out of the plane of the
figure).
⃗ are
(b) From 𝐹 = 𝐼𝑙𝐵 sin 𝜙, F is maximum for 𝜙 = 90° so that 𝑙 and 𝐵
⃗ upward, we rotate the rod clockwise by from
perpendicular. To keep 𝐹 = 𝐼𝑙 × 𝐵
its orientation, so that the current runs toward the southeast. Then
Given: I = 50.0 A
Req’d: 𝐹 = ?
𝑙 = 1.00 m
A = 1.20 T
𝜙 = 90°
Eq’n: 𝐹 = 𝐼𝑙𝐵 sin 𝜙 = 𝐼𝑙𝐵
Sol’n:
𝐹 = (50.0 𝐴)(1.00 𝑚)(1.20 𝑇) sin 90°
𝑭 = 𝟔𝟎 𝑵
2) A wire 36 m long carries a current of 22 A from west to east. Find the
magnetic force on the wire if the magnetic field of Earth at this location is
directed from south to north and has a magnitude of 5.0 x 10-5 T.
Given: B = 5.0 x 10-5 T
Req’d: 𝐹 = ?
I = 22 A
𝑙 = 36 m
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n:
𝐹 = (22 𝐴)(36 𝑚)(5.0 × 10−5 𝑇)
𝑭 = 𝟎. 𝟎𝟒 𝑵
14
Using your right hand, point your thumb to the east and your fingers north.
Your palm points up, so the force is up.
3) A 2.5 N magnetic force acts on a 475 m wire that is perpendicular to a 0.50
T magnetic field. What is the current in the wire?
Given: F = 2.5 N
Req’d: 𝐼 = ?
𝑙 = 475 m
B = 0.50 T
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n:
𝐹
𝐹 = 𝐼𝑙𝐵 → 𝐼 = 𝑙𝐵
2.5 𝑁
𝐼=
(475 𝑚)(0.50 𝑇)
𝑰 = 𝟎. 𝟎𝟏 𝑨
Exercise 6. Let’s Do the Math!
Young scientist, it’s your turn! Solve the following word problems and
show your complete process (Given-Required-Equation-Solution). Box all your
final answers.
1) Calculate the force on the wire shown in the following figure given B = 1.75
T, 𝑙 = 5.00 cm, and I = 15 A.
Retrieved from https://courses.lumenlearning.com/physics/chapter/22-7-magnetic-force-on-a-currentcarrying-conductor/#:~:text=The%20force%20on%20a%20current,is%20given%20by%20RHR%2D1
2) A 0.45 m long wire is carrying a 1.75 A current [W] while the wire is
perpendicular to a 0.35 T magnetic field directed to the north. What is the
size and direction of the force acting on the wire?
3) The magnetic force acting on a wire that is perpendicular to a 0.25 T uniform
magnetic field is 3.4 N. If the current in the wire is 2.75 A, what is the length
of the wire that is inside the magnetic field?
4) A copper wire 25 cm long carries a current of 6.0 A and weighs 0.35 N. A
certain magnetic is strong enough to balance the force of gravity that acts
on the wire. What is the strength of the magnetic field?
15
III.
Reflection
Before you proceed to your next journey, you need to complete the
following prompts:
Stop: I’m totally confused of
______________________________________________________________
Proceed with Caution: I still need some clarification on
______________________________________________________________
Go: I’m ready to move on because I have learned lots of things such as
______________________________________________________________
16
Exercise 1:
1) The following are the similarities of electric and magnetic fields:
 Electric fields are produced by two kinds of charges, positive and negative.
Magnetic fields are associated with two magnetic poles, north and south,
although they are also produced by charges (but moving charges).
 Like charge or poles repel; unlike charges or poles attract.
 Electric field points in the direction of the force experienced by a positive
charge. Magnetic field points in the direction of the force experienced by a
north pole.
 Forces in these fields are considered as action-at-a-distance forces.
17
2) The following are the differences of electric and magnetic fields:
 Positive and negative charges can exist separately. North and south poles
always come together. Single magnetic poles, known as magnetic
monopoles, have been proposed theoretically, but a magnetic monopole has
never been observed.
 Electric field lines have definite starting and ending points. Magnetic field
lines are continuous loops. Outside a magnet the field is directed from the
north pole to the south pole. Inside a magnet the field runs from south to
north.
Exercise 2:
1)
⃗ = 3.5 x 10-5 T
(a) Given: 𝐵
Req’d: Ф𝐵 = ?
A = 0.42 m2
⃗ 𝐴𝑐𝑜𝑠𝜃
Eq’n: Ф𝐵 = 𝐵
Sol’n:
⃗ 𝐴𝑐𝑜𝑠𝜃
Ф𝐵 = 𝐵
Ф𝐵 = (3.5 × 10−5 𝑇)(0.42𝑚2 ) cos 90°
Ф𝑩 = 𝟎
⃗ = 3.5 x 10-5 T
(b) Given: 𝐵
Req’d: Ф𝐵 = ?
A = 0.42 m2
⃗ 𝐴𝑐𝑜𝑠𝜃
Eq’n: Ф𝐵 = 𝐵
Sol’n:
= (3.5 × 10−5 𝑇)(0.42𝑚2 )
Ф𝑩 = 𝟏. 𝟒𝟕 × 𝟏𝟎−𝟓 𝐖𝐛
Ф𝐵
⃗ = 5.9 x 10-5 T
2) Given: 𝐵
Req’d: Ф𝐵 = ?
A = 1.3 m x 0.82 m = 1.066 m2
θ = 90° - 72° = 18°
⃗ 𝐴 cos 𝜃
Eq’n: Ф𝐵 = 𝐵
Sol’n:
⃗ 𝐴 cos 𝜃
Ф𝐵 = 𝐵
Ф𝐵 = (5.9 × 10−5 𝑇)(1.066 𝑚2 ) cos 18°
Ф𝑩 = 𝟓. 𝟗𝟖 × 𝟏𝟎−𝟓 𝐖𝐛
IV.
Answer Key
18
⃗ ℎ = 2.6 x 10-5 T
3) Given: 𝐵
Req’d: Ф𝐵 = ?
⃗ 𝑣 = 4.2 x 10-5 T
𝐵
A = 22 m x 18 m = 396 m2
⃗𝐴
Eq’n: Ф𝐵 = 𝐵
Sol’n: The horizontal component of the magnetic field is parallel to the floor, so it
does not contribute to the flux. Use the equation to calculate the flux using
the vertical component.
⃗𝐴
Ф𝐵 = 𝐵
Ф𝐵 = (4.2 × 10−5 𝑇)(396 𝑚2 )
Ф𝑩 = 𝟎. 𝟎𝟏𝟔𝟔𝟑𝟐 𝐖𝐛
Exercise 3:
⃗ should be perpendicular with each other.
1) Vectors 𝑣 and 𝐵
2) The force is always perpendicular to 𝑣 so it cannot change the magnitude of the
velocity, only its direction.
3) A charge moving at right angles to a uniform magnetic field moves in a circle at
constant speed because 𝐹 and 𝑣 are always perpendicular to each other.
4) The force is always directed toward the center of the circular path. This magnetic
force causes a centripetal acceleration, changing the direction of the velocity of
the particle.
Exercise 4:
1) Centripetal force
2) 𝐹𝑐 = 𝑚𝑎𝑐 = 𝑚
3) Mass
𝑣2
𝑟
Exercise 5:
1) The magnitude of the proton and electron magnetic forces are the same since
they have the same amount of charge. The direction of these forces however is
opposite of each other. The accelerations are opposite in direction and the
electron has a larger acceleration than the proton due to its smaller mass.
Exercise 6:
1) Given: 𝐵 = 1.75 T
Req’d: 𝐹 = ?
𝑙 = 5.00 cm
I = 15 A
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n: Convert 5.00 cm to m → 5.00 cm 0.05 m
𝐹 = (15 𝐴)(0.05 𝑚)(1.75 𝑇)
𝑭 = 𝟏. 𝟑𝟏 𝑵
This large magnetic field creates a significant force on a small length of wire.
19
2) Given: 𝐵 = 0.35 T
Req’d: 𝐹 = ?
𝑙 = 0.45 m
I = 1.75 A
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n:
𝐹 = (1.75 𝐴)(0.45 𝑚)(0.35 𝑇)
𝑭 = 𝟎. 𝟐𝟖 𝑵, downward
Req’d: 𝑙 = ?
3) Given: 𝐵 = 0.25 T
𝐼 = 2.75 A
F = 3.4 N
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n:
𝐹
𝐹 = 𝐼𝑙𝐵 → 𝑙 = 𝐼𝐵
3.4 𝑁
𝑙=
(2.75 𝐴)(0.25 𝑇)
𝒍 = 𝟒. 𝟗𝟓 𝒎
4) Given: 𝑙 = 25 cm
Req’d: 𝐵 = ?
𝐼 = 6.0 A
F = 0.35 N
Eq’n: 𝐹 = 𝐼𝑙𝐵
Sol’n: Convert 25 cm to m → 0.25 m
𝐹
𝐹 = 𝐼𝑙𝐵 → 𝐵 = 𝐼𝑙
0.35 𝑁
𝐵=
(6.0 𝐴)(0.25 𝑚)
𝑩 = 𝟎. 𝟐𝟑 𝑻
V.
References
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