Sources of Magnetic Field
Young and Freedman
Chapter 28
Electric and magnetic fields
• We learnt in electrostatics that the electric field of a static
electric charge acts on another static electric charge to
give a force
– note: the charge actually generates a field regardless if it is
moving or not
• If the charge is moving, it also generates a magnetic field.
• Other moving charges will experience a force due to this
magnetic field.
!
E is dependent on the magnitude and position of the charge.
!
B is dependent on the magnitude, position and velocity of the
charge
Permeability of free space
!
The “strength” of the source of E is q
!
The “strength” of the source of B is qv
µ0
#7 N ! s 2
= 4$ " 10
C2
4$
!
E =
1 q
4!" 0 r 2
! µ0 qv sin !
B=
4"
r2
EXACTLY
µ0 Is deeply related to ! 0 and we will later discover that:
1
µ0! 0 = 2
c
Speed of light
For the time being consider µ0 to be an independent constant like ! 0
Magnetic field of a point charge
µ0 q vsin "
B=
2
4!
r
It is found that the magnetic field is perpendicular to
both the velocity of the charge and the vector rˆ pointing
from the charge (source point) to the field point
In vector notation (note the unit vector!)
! µ0 q v! " rˆ Magnetic field of a
point charge with
B=
constant velocity
4! r 2
Magnetic field of a current element
µ0 q vsin "
B=
2
4!
r
From before
(point source)
Instead of a single charge we now have a current element.
We use superposition to sum up all the magnetic fields from
the individual charges. The total charge is an element is:
dQ = nqAdl where n is the number of charges (q) per unit
volume, A is the cross sectional area.
µ0 n q vd Adl sin " µ0 Idl sin "
dB =
=
2
4!
r
4!
r2
Magnetic field of a current element
µ0 Idl sin "
dB =
2
4!
r
In vector form
!
! µ0 Idl " rˆ
dB =
2
4! r
!
! µ0 Idl " rˆ
B=
#
2
4!
r
Magnetic field of a current element
For a complete circuit, law of Biot
and Savart
a
Example Ex 28.24
! µ0
B=
b
4!
#
!
Idl " rˆ
r2
c
Calculate the magnitude and direction of the magnetic field at
point P.
Break the section into three parts, labeled
! a,b and c. !The
magnetic field due to a and c is 0 as r is parallel to dl .
P is at the origin of the !semicircle b, so at!all points on b
!
r is perpendicular to dl, so sinφ=1 and dl ! rˆ = dl.
! µ0 I 1
B=
4! r 2
! µ0 I
B=
4r
µ0 I 1
" dl = 4 ! r 2 !r
total length of b
Magnetic field of a straight current-carrying conductor
!
! µ0 Idl " rˆ
The dB’s are all in the
same direction (into the B = 4 ! # r 2
plane). r = (x 2 + y 2 )
sin ! = sin(" # ! ) =
µ0 I
B=
4!
x
x 2 + y2
a
xdy
# (x 2 + y2 )3/2
"a
µ0 I
2a
By integration we find: B =
4! x x 2 + a2
In the limit where a>>x, x 2 + a 2 ! a and by using axial symmetry
µ0 I
B=
2 !r
Magnetic field of a (infinite) straight current-carrying conductor
Field around an infinite straight conductor
with current I
• The magnitude of the
magnetic field around an
infinite straight wire is
µ0 I
B=
2 !r
• The direction is found
using a slightly different
version of the right hand
rule
• Point your thumb in direction of current, your fingers
show the direction of the field
Force between two infinite conductors
!
! !
F = IL ! B
The lower wire produces a field, which has the
magnitude B = µ0 I at the location of the upper wire.
2 !r
The force per unit length on the upper wire is dependent
on its current I’, and B:
µ0 II ' Two long parallel current-carrying conductors
F/L =
by the RH rule the force is attractive if the
2 !r Note:
currents are in the same direction and vice versa
Definition of the ampere
µ0 II '
F/L =
2 !r
One ampere is that unvarying current that, if
present in two parallel conductors of infinite length
and one meter apart in empty space, causes each
conductor to experience a force of exactly 2 x 10-7
Newtons per meter of length
Magnetic field of a circular current loop
!
! µ0 Idl " rˆ
B=
#
2
4!
r
Biot and Savart Law
µ0 I
dl
dB =
2
2
4 ! (x + a )
Due to rotational symmetry around the x-axis, the components
perpendicular to this axis for one element cancel out with those
for another element on the opposite side of the loop. However,
the x-components sum. (Note r2=x2+a2.)
µ0 I
dla
dBx = dB cos! =
4 " (x 2 + a 2 )3/2
Magnetic field of a circular current loop
µ0 I
dla
dBx =
4 ! (x 2 + a 2 )3/2
Integrate to find Bx
µ0 Ia
Bx =
4 ! (x 2 + a 2 )3/2
" dl
The integral over dl is just the circumference of the circle, 2πa
µ0 Ia 2
Bx =
2(x 2 + a 2 )3/2
On the axis of a circular loop
Now imagine we have N tightly wound coils, each contributes equally to the
total magnetic field, and the field is N times that for a single loop
µ0 NIa 2
Bx =
2(x 2 + a 2 )3/2
On the axis of N circular loops
Magnetic field of N circular current loops
µ0 Ia 2
Bx =
2(x 2 + a 2 )3/2
On the axis of a circular loop
Now imagine we have N tightly wound coils, each contributes
equally to the total magnetic field, and the field is N times that
for a single loop
µ0 NIa 2
Bx =
On the axis of N circular loops
2
2 3/2
2(x + a )
This is maximum when x=0, i.e. the center of the solenoid. At
this point Bx is
µ0 NI
Bx =
2a
At the center of N circular loops
Magnetic field lines
•
•
Electric field lines start at a +ve charge and end at a -ve charge
There are no free magnetic N and S poles. Magnetic field lines are
continuous loops.
Magnetic flux
• Magnetic flux through a surface,
ΦB, is completely analogous to
electric flux ΦE.
• To calculate ΦB we integrate
over elements of the flux dΦB
relating to elements of area dA
of our surface.
!
! B = # B" dA = # B cos $ dA = # B • dA
• Units of magnetic flux are called
weber (Wb)
1Wb = 1Tm = 1Nm / A
2
Gauss’s law for magnetism
• Remember, Gauss’s law for electricity showed us
that the total electric flux is related to the total
enclosed charge.
• However, there are no magnetic monopoles, always
dipoles. Like in electric dipole case, total flux through
a closed surface for a dipole is 0!
• For any closed surface you can draw, every magnetic
field line which enters the surface also exits.
!" !"
#! B • d A = 0
Biot and Savart law and Coulomb’s law
! µ0
B=
4!
#
!
Idl " rˆ
2
r
Except for a few special cases,
evaluation of the integral is very hard
Compare to Coulomb’s Law:
!
E=
1
4 !" 0
$
!
# (r )
rˆdv
2
r
! charge density
!dv = dq
Very difficult for anything other than the most simple cases,
so we used Gauss’s Law and everything became easier...
Q
! ! Qencl
E4 !r 2 =
"0
E • dA =
r Gaussian
surface
"0
Q
Qencl
E=
4 !" 0 r 2
!
Comparison: electric and magnetic fields
Electric
!
!
F = qE
!
E=
#!
closed
surface
1 q
r̂
2
4!" 0 r
!" !" Qencl
E • dA =
"0
Magnetic
!
! !
F = qv ! B
! µ0 q
B=
v " r̂
2
4! r
?
Not Gauss’s Law
!" !"
#! B • d A = 0
Closed paths and line integrals
•
•
We need a different formulation than the closed surface
integrals used in Gauss’s law.
Consider a “closed path” through a magnetic field.
•
•
Consider a small vector dl which is tangential to the path
!" "
Consider
B • dl
#!
IMPORTANT: Do NOT confuse “closed path” with a
current loop! The path is imaginary, just like Gaussian
surfaces.
First (simple) example: Uniform field
a
b
c
d
!" " b !" " c !" " d !" " a !" "
#! B • dl = ! B • dl + ! B • dl + ! B • dl + ! B • dl
a
b
c
# d &
= B ! dl + 0 + % "B ! dl ( + 0
$ c '
a
= BL " BL
=0
b
d
Line integral around a long straight wire
µ0 I
B=
2 !r
Field at distance r from a
long straight wire
• Consider a long straight wire, and evaluate
!" " µ0 I
µ0 I
#! B • dl = 2" r circle
#! dl = 2" r 2" r
!" "
#! B • dl = µ0 I
!" "
#! B • dl
Note: Independent of r.
Non-circular path
!" "
B • dl = Bdl cos !
dl cos ! = rd"
!" "
µ0 I
#! B • dl = #! 2" r rd#
µ0 I
=
d#
#
!
2"
!" "
#! B • dl = µ0 I
µ0 I
B=
2 !r
!# d! = 2"
Ampere’s law
#!
!" "
B • dl = µ0 I encl
closed
path
• In general, can be more than one current-carrying
conductor, need to make algebraic sum of currents.
• Any wires outside the integration path can affect B at
a particular point, but the line integral around the
whole surface will be zero.
• Ampere’s law works for any shape.
Wire outside the surface
!" " µ0 I
B • dl =
d"
2!
!" "
µ0 I
B • dl = !
d#
2"
Sign from right-hand rule.
(Thumb in direction of current,
+ve if fingers in same direction
as integration.)
Another right hand rule
!" "
#! B • dl > 0
!" "
#! B • dl < 0
Step 1: Curl the fingers of your RH in the direction of Integration
Step 1: If Integral > 0, then current flows in the direction of thumb
Step 1: If Integral < 0, then current flows in the direction
opposite that of the thumb
Toroidal solenoid (non-magnetic core)
!" "
#! B • dl = µ0 I encl
closed
path
path 1 - No net current through loop
path 3 - No net current through loop
path 2 -
2! rB = µ NI
µ0 NI
B=
2! r
Field is entirely within toroid.
Chapter 28 Summary
Chapter 28 Summary cont.
Chapter 28 summary cont.
End of Chapter 28
You are responsible for the material covered in Y&F Sections 28.1-28.7.
You are expected to:
•
Understand the following terms:
•
Law of Biot and Savart, Ampere’s Law, Solenoid
•
Be able to calculate the magnetic field for simple geometries by
integrating the law of Biot and Savart
•
Be able to calculate the magnetic field for more complex geometries
using Ampere’s law
Recommended Y&F Exercises chapter 28:
11,15,16,18,20,25,32,33