MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 14: MAGNETISM and E/M INDUCTION
MAGNETISM CLASS NOTES
ELECTROMAGNETIC FORCE
Electricity and magnetism are both part of the Electromagnetic Force (EMF). For example, magnetic
fields affect moving charges and moving charges produce magnetic fields. The ultimate source of all
magnetic fields is electric current, the motion of charged particles.
MAGNITUDE OF MAGNETIC FIELDS
The magnetic force is maximum when a charge moves perpendicular to the magnetic field. Particles
with charge q, moving with a velocity v in a magnetic field B, experience a magnetic force F:
FB = qvB(sin
MAGNITUDE OF MAGNETIC FIELDS
The SI unit of magnetic field intensity is the tesla:
[T] = [N.s / C.m]
THE RIGHT-HAND RULE
Start with the equation FB = qvB(sin. Point the thumb along the direction of the force. The four
fingers wrap around the force vector first in the direction of the velocity, and then in the direction of the
magnetic field lines.
THE RIGHT-HAND RULE
Example 1. A positive charge and a negative charge enter a region where the magnetic field is uniform
and into the page (between the poles of a magnet with a square cross section). Describe the deflections
of each charge.
Example 1. Diagram
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
1A. The positive charge deflects toward the bottom of the page and the negative charge towards the top
of the page. While in the field, their paths are circular arcs.
RIGHT-HAND RULE FOR WIRES
Conducting wires form circular magnetic fields surrounding them. In these cases, point your thumb in
the direction of current. Allow your fingers to curl “around” the wire.
RIGHT-HAND RULE FOR WIRES
The “x” represent magnetic field lines going into the page and the “.” represent magnetic field lines
coming out of the page.
MAGNETIC FIELDS SURROUNDING WIRES
The magnitude of the magnetic field around a conducting wire can be determined:
B = oI / 2r
o = vacuum permeability
o = 4 x 10-7 (T.m) / A
Example 2. An electron moves through a region of crossed electric and magnetic fields. The electric
field E = 1000 V/m and is directed straight down. The magnetic field B = 0.40 T and is directed to the
left. For what velocity v of the electron into the paper (perpendicular to the plane) will the electric force
exactly cancel the magnetic force?
Example 2. Diagram
2A.
(1) FE – FB = 0
(2) FE = FB
(3) FE = Eq
(4) FB = qvB(sin
FqvB
(6) Eq = qvB

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MR. SURRETTE
VAN NUYS HIGH SCHOOL

(continued…)
v = E / B
(8) v = (1000 V/m) / (0.40 T)
(9) v = 2500 m/s
FB AND FE
In some cases, electric and magnetic fields of equal strength “crisscross” and cancel out. Under these
circumstances, charged particles are not deflected and the following can be used to determine their
speed:
v=E/B
MAGNETIC FIELD: SPEED AND RADIUS
When a charged particle enters the region of a uniform magnetic field, the particle will move in a
circular path of constant radius. The circular path will have a radius which is proportional to the linear
momentum (p = mv) of the moving charge and will be in a plane which is perpendicular to the direction
of the magnetic field.
MAGNETIC FIELD: SPEED AND RADIUS
r = mv / qB
MAGNETIC FIELD: PERIOD OF
REVOLUTION
The period of revolution (the time required in seconds) for the circular path is independent of the value
of the radius:
T = 2r / v
T = 2m / qB
Example 3. A proton, mass 1.67 x 10-27 kg and charge 1.60 x 10-19 C, moves in a circular orbit
perpendicular to a uniform magnetic field of 0.75 T. Find the time for the proton to make one complete
circular orbit.
3A.
(1) T = (2m) / (qB)
(2) T = (2)(1.67x10-27 kg) / (1.6 x 10-19 C) (0.75 T)
(3) T = 8.74 x 10-8 s
(4) T = 87.4 ns
MAGNETIC TORQUE
When a closed conducting loop is placed in a uniform external magnetic field, a net torque will be
exerted on the loop. The magnitude of the torque will depend on the angle between the direction of the
magnetic field, the number of loops (n), and the direction of the normal (or perpendicular) to the plane
of the loop:
 = BnIA(sin
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 4. A circular loop carrying a current of 2.00 A is oriented in a magnetic field of 3.50 T. The
loop has an area of 0.12 m2 and is mounted on an axis, perpendicular to the magnetic field, which allows
the loop to rotate. What is the torque on the loop when its plane is oriented at a 37.0o angle to the field?
4A.
(1)  = BnIA(sin
(2)  = (3.50 T)(1)(2.00 A)(0.12 m2)(sin(90-37))o
(3)  = (3.50 T)(1)(2.00 A)(0.12 m2)(sin(53)o)
(4)  = 0.67 N.m
Example 5. A square loop (L = 0.200 m) consists of 50 closely-wrapped turns each of which carries a
current of 0.500 A. The loop is oriented as shown in a uniform magnetic field of 0.40 T directed in the
positive y-direction. What is the magnitude of the torque on the loop?
Example 5. Diagram
5A.
(1)  = BnIA(sin
(2)  = (0.40 T)(50)(0.500 A)(0.04 m2)(sin 60o)
(3)  = 0.35 N.m
WIRES IN A MAGNETIC FIELD
A magnetic field is exerted on a straight conductor placed in a magnetic field:
F = BIL(sin
Example 6. A current-carrying wire of length 0.300 m is positioned perpendicular to a uniform
magnetic field. If the current is 5.00 A and it is determined that there is a resultant force of 2.25 N on
the wire due to the interaction of the current and field, what is the magnetic field strength?
6A.
(1) F = BIL(sin
(2) F = BIL
(3) B = F / IL
(4) B = (2.25 N) / (5.00 A)(0.300 m)
(5) B = 1.5 T
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 7. A long straight wire passes through a region where the magnetic field is uniform and
directed out of the page. The current is directed to the top of the page.
Example 7. Diagram
7a. What is the direction of the magnetic force on the wire?
A. To the right.
7b.
A.
(1)
(2)
(3)
(4)
Calculate the magnitude of the force.
F = BIL(sin)
F = BIL
F = (0.40 T)(10 A)(0.20 m)
F = 0.80 N
PARALLEL CONDUCTORS
The magnitude of the magnetic force per unit length between parallel conductors is:
(F / L) = (oI1I2 / 2r)
o is the “permeability of free space”
o = 4 x 10-7 T.m/A
Example 8. Two parallel cables of a high voltage transmission line carry equal and opposite currents of
1500 A. The distance between the cables is 3.00 m. What is the magnetic force acting on a 40.0 meter
length of each cable?
8A.
(1) F / L = oI1I2 / 2r
(2) F = oI1I2L / 2r
(3) F = (4 x 10-7 T.m/A)(1500 A)(1500 A)(40.0 m) / (2)(3.00 m)
(4) F = 6 N
SOLENOIDS
Within a solenoid, the magnetic field is parallel to the axis of the solenoid and pointing in a sense
determined by applying the right-hand rule B to one of the coils:
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
SOLENOIDS
The magnetic field inside a solenoid which has n turns of conductor per unit length is:
B = onI
Example 9. A superconducting solenoid is to be designed to generate a magnetic field of 10.0 T. If the
solenoid winding has 2000 turns/meter, what is the required current?
9A.
(1) B = onI
(2) I = B / on
(3) I = (10 T) / (4 x 10-7 A.m/T)(2000)
(4) I = 3979 A
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 14: MAGNETISM and E/M INDUCTION
ELECTROMAGNETIC INDUCTION CLASS NOTES
ELECTROMAGNETIC INDUCTION
The discovery that electric current can produce magnetic fields led to the discovery that magnetic fields
can produce electric currents. The production of currents by a changing magnetic field through a
conducting loop is called induction.
MAGNETIC FLUX
The total magnetic flux through a plane area A placed in a uniform magnetic field depends on the angle
between the direction of the magnetic field and the direction perpendicular to the surface area:
 = BA(cos

GNETIC FLUX
Example 1. A square loop 2.00 m on a side is placed in a magnetic field of 0.300 T. If the field makes
an angle of 50.0o with the normal to the plane of the loop, determine the magnetic flux through the loop.
1A.
(1)  = BA(cos
(2) A = s2
(3) A = (2.00 m)2
(4) A = 4.00 m2
(5)  = (0.300 T)(4.00 m2)(cos 50o)
(6)  = 7.71 x 10-1 T.m2
FARADAY’S LAW
Changing the magnetic flux  through a loop of wire induces a current. Faraday’s law states that the
emf (electromagnetic force)  induced in a wire is proportional to the rate of change of the flux through
the loop:
 = - N / t
FARADAY’S LAW
 is indistinguishable from the voltage generated from a battery. N is the number of loops in the wire.
 is the change of flux in time t. The minus sign indicates the polarity of the induced emf, which
acts in the opposite direction of .
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 2. A 50-turn solenoid (r = 0.2 m) is placed in a magnetic field of 0.545 T. The field makes a
12o angle with the normal to the plane of the loop. The magnetic field then decreases to 0.100 T in 0.3 s.
Determine the average voltage induced during this time interval.
2A.
(1)  = BA cos 
(2) A = r2
(3) A = (0.2 m)2
(4) A = 0.126 m2
(5)  = (B)A cos 
(6) Bo = 0.545 T
(7) B = 0.100 T
(8) B = 0.545T – 0.100T
(9) B = 0.445 T
(10)  = (0.445 T)(0.126 m2)(cos 12o)
(11)  = 5.47 x 10-2 T.m2
(12)  = - N / t
(13)  = (50)(5.47 x 10-2 T.m2) / 0.3 s
(14)  = 9.12 V
SLIDE WIRES
Magnetic flux can be changed through a loop by changing the size of the loop. Imagine a slide wire
where L is the length of the wire that moves in contact with the U-shaped wire.
SLIDE WIRE
SLIDE WIRE EQUATION
To determine the induced emf of slide wires:
 = BLv
 = emf (voltage)
B = magnetic field
L = length of slide wire
v = velocity of slide wire
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 3. A 0.20 m wire is moved perpendicular to a 0.50 T magnetic field at a speed of 1.50 m/s.
What emf is induced across the ends of the wire?
3A.
(1)  = BLv
(2)  = (0.50 T)(0.20 m)(1.50 m/s)
(3)  = 0.15 V
ROTATING WIRE LOOPS
A wire loop can be rotated on its axis in a magnetic field. This induces an emf  and a constant electric
current. As a wire loop rotates, so does its magnetic flux  as per the equation:  = BA cos .
ROTATING WIRE LOOPS
For every half turn a wire loop makes in a magnetic field, the value of  changes by 180o. As a result,
the direction of its flux (and induced electric current) reverses every half rotation.
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