Announcements
• WebAssign HW Set 5 due October 10
• Problems cover material from Chapters 18
• HW set 6 due on October 17 (Chapter 19)
•Prof. Kumar tea and cookies Tuesdays from 5 –
6… pm in room 2165
• Exam 1 statistics
average 12.78
stand. dev. 3.51
QUESTIONS? PLEASE ASK!
NUMBER SCORE . 1 2 3 4 5 6 7 8 910 . 1 2 3 4 5 6 7 8 910 . 1 2 3 4 5 6 7 8 910 . 1 2 3 4 5 6 7 8 910 .
0 0 :
:
0 1 :
:
0 2 :
:
0 3 :
:
4 4 :********
:********
2 5 :****
:****
11 6 :**********************
:**********************
17 7 :**********************************
:**********************************
21 8 :******************************************
:******************************************
21 9 :******************************************
:******************************************
40 10 :********************************************************************************
:********************************************************************************
45 11 :******************************************************************************************
:******************************************************************************************
36 12 :************************************************************************
:************************************************************************
40 13 :********************************************************************************
:********************************************************************************
40 14 :********************************************************************************
:********************************************************************************
47 15 :**********************************************************************************************
:**********************************************************************************************
35 16 :**********************************************************************
:**********************************************************************
33 17 :******************************************************************
:******************************************************************
20 18 :****************************************
:****************************************
12 19 :************************
:************************
5 20 :**********
:**********
From last time


Magnets and earth’s magnetic
field
F
Magnetic Fields: B º
qvsin q



Units are T = N/A.m
Use right hand rule to determine
direction of force
Force on a wire:
F = B I L sin θ
Continuing




Forces and torques
on current carrying
wires and loops.
On moving charges.
Magnetic Fields and
Amperes law.
Forces between two
current carrying
wires.
  IANB sin   B sin 
r  mv / qB
o I
B
2r
 B.l   o I
F  o I1 I 2

l
2d
Torque on a Current Loop

Torque  = B I A N sin 





Applies to any shape loop
N is the number of turns in the
coil
Torque has a maximum value
of NBIA (when  = 90°)
Torque is zero when the field is
parallel to the plane of the loop
Magnetic Moment  = IAN


 is a vector
Torque can be written as
 = B sin

Example Problem 19.31

A long piece of wire with a mass of
0.100 kg and a length of 4.00 m is
used to make a square coil with a
side of 0.100 m. The coil is hinged
along a horizontal side, carrying a
3.40 A current, and is placed in a
vertical magnetic field of 0.010 T.
(a) Determine the angle that plane
of the coil makes with the vertical
when the coil is in equilibrium. (b)
Find the torque acting on the coil
due to the magnetic force at
equilibrium
Electric Motor

electric motor - converts
electrical energy to
mechanical energy


The mechanical energy is
in the form of rotational
kinetic energy
An electric motor
consists of a rigid
current-carrying loop
that rotates when
placed in a magnetic
field
Electric Motor


Torque acting on the
loop will rotate the loop
to smaller values of θ
until the torque
becomes 0 at θ = 0°
If the loop turns past
this point and the
current remains in the
same direction, the
torque reverses and
turns the loop in the
opposite direction

Bad!!
Electric Motor


So, we need to be
clever…
To provide continuous
rotation in one direction,
the current in the loop
must periodically
reverse


In AC motors, this
reversal naturally occurs
In DC motors, a split-ring
commutator and brushes
are used

Actual motors would
contain many current
loops and commutators
Force on a Charged
Particle in a Magnetic Field



Consider a particle
moving in an external
magnetic field so that its
velocity is perpendicular
to the field
The force is always
directed toward the
center of the circular path
The magnetic force
causes a centripetal
acceleration, changing
the direction of the
velocity of the particle
Force on a Charged Particle

Equating the magnetic and centripetal
forces:
2
mv
F = qvB =
r

mv
Solving for r: r =
qB


r is proportional to the momentum of the
particle and inversely proportional to the
magnetic field
Sometimes called the cyclotron equation
Particle Moving in an External
Magnetic Field

If the particle’s
velocity is not
perpendicular to the
field, the path
followed by the
particle is a spiral

The spiral path is
called a helix
Example Problem 19.42

A cosmic ray proton in interstellar space
has an energy of 10 MeV and executes a
circular orbit having a radius equal to that
of Mercury’s orbit around the Sun (5.8 x
1010 m). What is the magnetic field in
that region of space?
Solution to 19.31
Solution to 19.42
Example Problem 19.42

A cosmic ray proton in interstellar space
has an energy of 10 MeV and executes a
circular orbit having a radius equal to that
of Mercury’s orbit around the Sun (5.8 x
1010 m). What is the magnetic field in
that region of space?
Magnetic Fields –
Long Straight Wire

A current-carrying wire
produces a magnetic field B


Right hand rule # 2 to
determine direction of B
Magnitude of B at a distance
r from a wire carrying
current of I is:
B =

mo I
2p r
µo = 4  x 10-7 T.m / A

µo is called the permeability
of free space
Ampère’s Law: General

relationship between I in
an arbitrarily shaped
wire and B produced by
the wire:
B|| Δℓ = µo I


Choose an arbitrary
closed path around the
current
Sum all the products of
B|| Δℓ around the closed
path
Ampère’s Law Applied to a
Long Straight Wire


Use a closed circular
path
The circumference of
the circle is 2  r
B =

mo I
2p r
This is identical to the
result previously
shown
Example Problem 19.54

Two long parallel wires separated
by a distance 2d carry equal
currents in the same direction.
The currents are out of the page
in the figure. (a) What is the
direction of the magnetic field at P
on the x-axis set up by the two
wires? (b) Find an expression for
the magnitude of the field at P. (c)
From (b), determine the field
midway between the two wires.
Magnetic Force Between
Two Parallel Conductors


The force on wire 1 is due
to the current in wire 1 and
the magnetic field produced
by wire 2
The force per unit length is:
F


=
mo I 1 I 2
2p d
Parallel conductors carrying
currents in the same
direction attract each other
Parallel conductors carrying
currents in the opposite
directions repel each other
Magnetic Field of a Current
Loop

The magnitude of the
magnetic field at the
center of a circular loop
with a radius R and
carrying current I is
B =

mo I
2R
With N loops in the coil,
this becomes:
B =N
mo I
2R
Magnetic Field of a Solenoid

Solenoid – long straight wire is
bent into a coil of several
closely spaced loops


B field lines inside the solenoid
are nearly parallel, uniformly
spaced, and close together


Electromagnet - acts like a magnet
only when it carries a current
B is nearly uniform and strong
The exterior field is
nonuniform, much weaker, and
in the opposite direction to the
field inside the solenoid
Magnetic Field in a
Solenoid, Magnitude


The magnitude of the
field inside a solenoid is
constant at all points far
from its ends
B = µo n I



n is the number of turns
per unit length
n=N/ℓ
The same result can be
obtained by applying
Ampère’s Law to the
solenoid
Example Problem 19.60

A certain superconducting magnet in the
form of a solenoid of length 0.5 m can
generate a magnetic field of 9.0T in its
core when the coils carry a current of 75
A. The windings, made of a niobiumtitanium alloy, must be cooled to 4.2K.
Find the number of turns in the solenoid.
Solution to 19.42
Solution to 19.54
Solution to 19.60