My lecture slides are posted at
http://www.physics.ohio-state.edu/~humanic/
Information for Physics 1201 Midterm I
Wednesday, February 20
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) Equations and constants will be provided on the midterm
4) Covers the material in Chapters 16, 17, 18, 19, and 20
Chapter 20
Magnetic Forces and
Magnetic Fields
Magnetic Fields Produced by Currents
A solenoid is made up of many current loops which extend a finite
distance along the axis of the loops characterized by the number of
turns (number of loops) per unit length, n. The B-field in the interior
of a long solenoid depends only on n and the current in the loops, I, as
B = µ o nI
Use RHR
for direction
of B inside
Interior of a long solenoid
(B ~ 0
outside)
B-field lines similar to those of a bar
magnet with N and S poles shown.
Example: Calculation of the B-field in a solenoid
A 0.25 m long solenoid with 5000 turns has a current of
3.5 A applied to it. Find the B-field in the solenoid.
B = µo nI
5000 turns
4 turns
n=
= 2.0 ×10
m
0.25 m
B = ( 4π ×10
−7
) (2.0 ×10 ) (3.5) = 0.088 T = 880 G
4
Magnetic Materials
The intrinsic “spin” and orbital motion of electrons gives rise to the magnetic
properties of materials è electron spin and orbits act as tiny current loops.
In ferromagnetic materials groups of 1016 - 1019 neighboring atoms form
magnetic domains where the spins of electrons are naturally aligned with
each other; magnetic domain sizes are ~ 0.01 - 0.1 mm.
An external magnetic field can induce magnetism in ferromagnetic
materials by merging and aligning domains. Depending on the material,
the induced magnetism may or may not become permanent.
Putting iron in the center of a solenoid can create a strong electromagnet
with fields 100x - 1000x the applied fields (also, can turn fields on and off).
Ampere’s Law
AMPERE’S LAW FOR STATIC MAGNETIC FIELDS
For any current geometry that produces a magnetic field that does not change in time,
sum up all of the segment lengths Δl multiplied by the B|| è this equals a constant
multiplied by the current enclosed by the path.
In mathematical notation,
∑ B Δ = µ I
||
o
net current
passing through
surface bounded
by path
Note: this is a messy thing to use for anything except in situations where
the geometry of the problem has a lot of symmetry, e.g. the long straight wire
Ampere’s Law
Example: An Infinitely Long, Straight, Current-Carrying Wire
Use Ampere’s law to obtain the magnetic field è lots of symmetry!
∑ B Δ = µ I
||
o
Can choose a path
where B|| is constant
B (∑ Δ )= µ o I
Since path is a
circle of radius r
B 2π r = µ o I
On RHS, I is the only
current enclosed by path
µo I
B=
2π r
è We get our familiar equation!
Example: Ampere’s law applied to a solenoid
∑ B Δ = µ I
||
o
( B||Δ)ab + ( B||Δ)bc + ( B||Δ)cd + ( B||Δ)da = µ0 I enclosed
⇒
n turns/length
Since,
B ≈ 0 outside solenoid

B ⊥ Δ along bc and da
I
d
Use closed
path abcda

B
a
L
c
then,
b
( B||Δ)cd = µ0 I enclosed

Since B || Δ
and
along cd
I enclosed = nLI
∴ BL = µ 0 nLI ⇒ B = µ 0 nI
as seen earlier
The Force on a Current in a Magnetic Field
Example: 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.
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
The Torque on a Current-Carrying Coil
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 shows that they are opposite in direction.
The Torque on a Current-Carrying Coil
The loop tends to rotate such that its
normal becomes aligned with the magnetic
field.
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
L
For a coil of N loops (turns) è
τ = NIABsin φ = MBsin φ
M = NIA Magnetic moment
of the coil
The Torque on a Current-Carrying Coil
Example: 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.
(a)
M = NIA = (100 ) ( 0.045 A ) ( 2.0 ×10 −4 m 2 ) = 9.0 ×10 −4 A ⋅ m 2
(b)
τ = MBsin ϕ = ( 9.0 ×10 −4 A ⋅ m 2 ) ( 0.15 T ) sin 90 = 1.4 ×10 −4 N ⋅ m
The galvanometer revisited
How a galvanometer works. The coil of
wire and pointer rotate when there is a
current in the wire. The pointer stops when
the magnetic torque on the coil, τ, is canceled
by the opposing torque of the spring, τs .
τ M = NIABsin φ
τ S = kΔφ
∑τ = τ
M
Δφ is the angular displacement of the needle
k is the spring constant (Hooke’s Law)
−τS = 0
NIABsin φ
NIABsin φ − kΔφ = 0 ⇒ Δφ =
⇒ Δφ ∝ I
k
The Torque on a Current-Carrying Coil
The basic components of
a dc motor.
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Split-ring commutator
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.