My lecture slides are posted at
http://www.physics.ohio-state.edu/~humanic/
Information for Physics 112 midterm, Wednesday, May 2
1) Format: 10 multiple choice questions (each worth 5 points) and two
show-work problems (each worth 25 points), giving 100 points total.
2) Closed book and closed notes.
3) Can make up your own equation sheet on a single normal sized piece of
paper (8.5” x 11"), both sides of sheet
-- handwritten
-- no worked out problems
-- only equations and physical constants allowed
-- signed equation sheet must be turned in with the exam
4) Covers the material in Chapters 18, 19, 20 – Electricity only
Chapter 21
Magnetic Forces and
Magnetic Fields
21.3 The Motion of a Charged Particle in a Magnetic Field
Conceptual Example 2 A Velocity Selector
A velocity selector is a device for measuring
the velocity of a charged particle. The device
operates by applying electric and magnetic
forces to the particle in such a way that these
forces balance.
How should an electric field be applied so that
the force it applies to the particle can balance
the magnetic force?
From RHR-1, FM points upward, so FE should
point downward so the forces balance, i.e.,
FM - FE =0 --> FM = FE
Since FM = qvB
FE = qE
Then, qvB = qE --> v = E/B , knowing E and B
gives v independent of q
21.3 The Motion of a Charged Particle in a Magnetic Field
Work done on a charged particle moving
through electric and magnetic fields.
For a constant force, F, --> WAB = FΔs = ΔKE
where Δs is the displacement along the
direction of F and ΔKE is the change in the KE.
The electrical force can do work on a charged
particle since it can displace the particle
in the direction of the force and thus change
its kinetic energy.
The magnetic force cannot do work on a
charged particle since it acts perpendicular
to the motion of the particle so that no
displacement occurs along the direction of
the force and thus its speed remains constant
and its kinetic energy does not change.
21.3 The Motion of a Charged Particle in a Magnetic Field
The circular trajectory.
Since the magnetic force always remains
perpendicular to the velocity, if a charged
particle moves perpendicular to a uniform
B-field its path will be circular. The magnitude
of the force remains constant and it is directed
toward the center of the circular path, i.e. it is
the centripetal force for the motion.
Centripetal
Force
FM = F c
Solve for r
v2
Fc = m
r
v2
qvB = m
r
mv
r=
qB
Example. Find the radius of curvature for a fast electron with speed
v = 2 x 107 m/s in a) the Earth’s magnetic field at the surface, and b) in a
10 T magnetic field.
a)  BEarth ~ 0.5 gauss = 0.5 x 10-4 T
r = mv/(qB) = (9.11 x 10-31)(2 x 107)/[(1.6 x 10-19)(0.5 x 10-4)]
= 2.3 m
b)  B = 10 T
r = mv/(qB) = (9.11 x 10-31)(2 x 107)/[(1.6 x 10-19)(10)]
= 1.1 x 10-5 m = 11 µm
21.5 The Force on a Current in a Magnetic Field
A magnet force can be exerted on
charged particle moving in a
magnetic field.
è A magnetic force can also be
exerted on a current of charged
particles moving in a magnetic field,
e.g. a current in a wire
The magnetic force on the
moving charges pushes the
wire to the right.
Use RHR-1 to find the direction
of the force on a wire by putting
the thumb in the direction of the
velocity of moving positive charges,
i.e. the thumb points in the direction of
the conventional current.
21.5 The Force on a Current in a Magnetic Field
Derivation of the force F on a wire of
length L with current I and angle θ
with the B field.
F = qvB sin θ
{
( Δq %
F =&
vΔt )B sin θ
#(
Δ
t$ L
'
I
F = ILB sin θ
21.5 The Force on a Current in a Magnetic Field
Example 5 The Force and Acceleration in a Loudspeaker
The voice coil of a speaker has a diameter of 0.025 m, contains 55 turns of
wire, and is placed in a 0.10-T magnetic field. The current in the voice coil is
2.0 A. (a) Determine the magnetic force that acts on the coil and the
cone. (b) The voice coil and cone have a combined mass of 0.020 kg. Find
their acceleration.
21.5 The Force on a Current in a Magnetic Field
(a)
(b)
F = ILB sin θ
= (2.0 A)[55π(0.025 m)](0.10 T)sin 90o
= 0.86 N
F
0.86 N
a= =
= 43 m s 2
m 0.020 kg
21.5 The Force on a Current in a Magnetic Field
Magnetohydrodynamic (MHD) propulsion for ships and submarines.
MHD propulsion uses the principle of forces on charged currents in magnetic
fields. As water is expelled from the rear of the ship by the magnetic force,
Newton’s 3rd law causes the ship to recoil forward with the same force.
è Has promise to be a low-noise, reliable and inexpensive system.
21.6 The Torque on a Current-Carrying Coil
First, consider the forces on a current-carrying loop in a magnetic field:
The two forces on the loop have equal magnitude but an application
of RHR-1 shows that they are opposite in direction.
21.6 The Torque on a Current-Carrying Coil
The loop tends to rotate such that its
normal becomes aligned with the magnetic
field.
21.6 The Torque on a Current-Carrying Coil
Derivation for the net torque on a current-carrying loop in a B field.
torque = (force) x (moment arm)
A = Lw
Net torque = τ = ILB(12 w sin φ )+ ILB(12 w sin φ )= IAB sin φ
τ = NIA B sin φ
{
For a coil of N loops (turns) è
Magnetic moment
of the coil
21.6 The Torque on a Current-Carrying Coil
Example 6 The Torque Exerted on a Current-Carrying Coil
A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns,
and contains a current of 0.045 A. The coil is placed in a uniform magnetic
field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil.
(b) Find the maximum torque that the magnetic field can exert on the
coil.
magnetic
moment
(a)

NIA = (100 )(0.045 A ) 2.0 ×10 − 4 m 2 = 9.0 ×10 − 4 A ⋅ m 2
(
)
magnetic
moment

−4
2

−4
(b) τ = NIA B sin φ = 9.0 × 10 A ⋅ m (0.15 T )sin 90 = 1.4 × 10 N ⋅ m
(
)
21.6 The Torque on a Current-Carrying Coil
The basic components of
a dc motor.
}
Split-ring commutator
21.6 The Torque on a Current-Carrying Coil
How a dc motor works.
When a current exists in the coil,
the coil experiences a torque.
Because of its inertia, the coil
continues to rotate when there
is no current.