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Creating magnetic fields
Gravitational fields acted on masses, and
masses set up gravitational fields:
Fgravity = mg where g = GM/r2
Electric fields acted on charges, and charges
set up electric fields:
Felectric = qE where E = kQ/r2
Magnetic fields acted on moving charges; do
moving charges set up magnetic fields?
Creating Magnetic Fields
Magnetic fields acted on moving charges; do
moving charges set up magnetic fields?
YES! From the gravitational and electric
cases, we can guess that we will need:
a constant that describes the strength
similar to G and k;
an inverse square relationship with
distance (1/r2); and
a dependence on what is acted upon
(like m and q) - in this case, qv.
Creating Magnetic Fields
But we also have the right hand rule in the
magnetic force equation, and we’ll need a
right hand rule in the field generating
equation also:
Magnetic Fields
B = (mo/4p) q v sin(qvr) / r2
direction: right hand rule
thumb = hand x fingers
Point your hand in the direction of v, curl you
fingers in the direction of r, and the field, B
will be in the direction of your thumb; if
the charge is negative, the field direction
is opposite that of your thumb.
Magnetic Constant
The constant (mo/4p) is a seemingly strange
way of writing a constant that serves the
same purpose as G and k, but that is exactly
what it does.
The value: (mo/4p) = 1 x 10-7 T*m*sec/Coul
(or 1 x 10-7 T*m/Amp).
In fact, the constant k is sometimes written as:
k = 1/(4peo). [In both cases, the sub zero on
m and e indicates the field is in vacuum.]
4p ?
Why does the combination of 4p enter into the
constant? Consider the field idea where the
mass, charge, or in this case the moving
charge, sets up the field by throwing out
field particles. The density of these field
particles, and hence the strength of the field,
depends on the number of field particles (a
constant) and the area they are going
through. This area is that of a sphere: 4pr2.
Thus the 4p really goes with the r2 in the
denominator: B = mo q v sin(qvr) / (4p r2)
Creating Magnetic Fields
A single moving charge does create a
magnetic field in the space around it, but
since both the charge and the constant are
very small, we usually don’t have to worry
about these effects.
But if we have a series of charges moving
(which means a current), then we can
generate an appreciable magnetic field.
Currents and Magnetic Fields
A moving charge (qv) actually can be thought
of as a current over a small length:
qv = q(DL/Dt) = (DQ/Dt)DL = I DL ,
so that we have for each small length:
B = (mo/4p) I DL sin(qIr) / r2
with direction: thumb(field) =
hand(current) x fingers(radius).
General Case
For the current over a complete circuit, the
field produced at any particular point in
space will depend on where the point is and
the shape of the circuit.
B = (mo I /4p)  dL sin(qIr) / r2
where both qIr and r depend on L (that is, they
depend on which dL we are at in the sum).
Keep in mind that the magnetic field is a
vector, so we need to use the RHR and
consider components!
Field at the Center of a Loop
B = (mo I /4p)  dL sin(qIr) / r2
At the center of a circle of current, r = R =
constant, and qIL = 90o = constant. Note:
all elements will give a field out of the loop!
Thus the integral is trivial and we have:
Bat center of circle = (mo I /4p)* (2pR)(1) / R2 =
mo I / 2R . If we have
several turns in the wire, each
R I
turn contributes: B = moNI / 2R .
Field at the Center of a Loop
B = moNI / 2R
Note that this field depends on the current
(both N and I) and the shape (R). [Note that
the only parameter associated with a circle is R.]
It has the magnetic constant (mo) in it as
well. Only the 2 in the formula can’t be
immediately guessed at.
Note also that it has the mo and I in the
numerator, and it has a distance in the
denominator. This should happen for all
the different geometries.
Example
What is the field strength at the center of a
loop that has 5 Amps running through 300
turns, where the radius of the loop is 6 cm?
Bat center of loop = mo N I / 2R
B = (4p x 10-7 T*m/A) * 300 * 5A / (2*.06m)
= .0157 T = 157 Gauss.
Field on axis of circular loop
What is the magnetic field on axis but not at
the center of a circular loop of current?
Using the RHR, the magnetic field due to the
element on the right (current going
into the page) is up and right
as shown on the diagram.
B
The horizontal field will cancel
a r
in pairs, but the vertical
R
components will add!
I
Field on axis of circular loop
By = B cos(q) where the red dot indicates the
angle q on the diagram. Again from the
diagram we can see that cos (q) = R/r.
From
B = (mo I /4p)  dL sin(qIr) / r2
with r = constant and qIr= 90o
B
we have:
By = cos(q)*(mo I /4p)*(2pR)(1)/r2
a r
= mo I R2 / 2r3
R
where r = [a2 + R2]1/2 .
I
Field on axis of circular loop
Bon axis = mo I R2 / 2r3
Note here also that there is a mo and I in the
numerator, and an inverse distance. The
geometry depends on both R and r, and both
parameters are included in the formula.
Note that as a becomes small, r approaches
R, and B approaches the result we had for
the field at the center of the coil:
Bloop = mo I / 2R .
Field off-axis
B = (mo I /4p)  dL sin(qIr) / r2
Note that if we try to find the field off-axis,
both qIr and r no longer remain constant,
and hence the integral becomes very hard to
integrate!
Magnetic Field due to a current
in a straight wire
Let’s now consider the magnetic field at some
point (x) a distance a above a current, I, in a
straight wire. We start with the Biot-Savart
Law: B = (mo I /4p)  dL sin(qIr) / r2.
By the RHR, the direction of B is out of the
screen.
x
r
a
I
Magnetic Field due to a current
in a straight wire
B = (mo I /4p)  dL sin(qIr) / r2.
From the diagram, we see that sin(qIr) = a/r .
We see that both qIr and r depend on which dL
we are at. Thus the integral is not trivial,
but it does look like one we have done
before!
x
r
a
-x
I
Magnetic Field due to a current
in a straight wire
B = (mo I /4p)  dL sin(qIr) / r2.
B = (mo I /4p) xLxR dx (a/r) / r2 where
r = [x2 + a2]1/2 . Using the same trick we did
before for this integral (method of
substitution), we get:
x
r
a
-x
I
Magnetic Field due to a current
in a straight wire
Bshort wire = (mo I /4pa)*[cos(qL) - cos(qR)] .
For a long wire (a « -xL, a « xR) this reduces
to: Blong wire = (mo I /2pa).
From the cylindrical symmetry of the wire, we
see that the magnetic field is directed
around the wire .
Note that we again have mo and I in the
numerator, and an inverse distance (1/a).
Magnetic Field due to a current
in a straight wire
Bshort wire = (mo I /4pa)*[cos(qL) - cos(qR)] .
Blong wire = (mo I /2pa).
For a long wire, the only geometric parameter
is a, the distance the point is from the wire.
For a short wire, both the length and distance
to the wire are contained in the angles qL
and qR.
Shortcut RHR rule #1
For the case of a circular loop of wire, the
magnetic field went straight through the
center. If you curl your right hand fingers in
the direction of the current around the loop,
your thumb points in the direction of the
field through the center of the loop.
Shortcut RHR rule #2
For the case of the straight wire, if you point
your right hand thumb in the direction of the
straight current, your fingers will curl
around in the direction of the circular
magnetic field around the wire.
Warning: make sure you use your right hand
when doing these RHR shortcuts!
Computer Homework
There is a computer homework assignment,
Vol. 4, #2, on Magnetic Fields that will
give you practice with calculating the
magnetic fields due to different current
geometries.
The Solenoid
The solenoid is an important special case for
creating magnetic fields. A solenoid is a
coil of wire that is wrapped around a
cylinder rather than around a circle. It
has elements of both the circular loop and
the straight wire. It can be solved the
straight-forward way, but there is an easier
(but trickier) way - we’ll talk about this
easier way next.
Gauss’ Law for Magnetism
Recall from Part I, Gauss’ Law for Electricity:
closed area E  dA = 4pkQenclosed .
This looked like a hard way to solve for E
(since E is inside the integral), but using
symmetry it sometimes turns out easy.
For magnetism, we cannot separate poles,
therefore we cannot enclose any poles. This
means that we have: closed area B  dA = 0.
Another Law
Recall from Part I that Electric Fields started
on positive charges and went towards
negative charges.
For magnetic fields, however, the magnetic
fields went around currents. This leads to:
closed loop B  dL = (constant) * Iencircled .
Again, this looks like it will be hard to solve
for B (since it is inside the integral), but
we’ll see that for some symmetric results, it
turns out to be simple.
Ampere’s Law
closed loop B  dL = (constant) * Iencircled .
To determine the constant, we use something
we already know: the field due to a long
straight wire: B = moI/2pa . Due to the
symmetry of the wire, the integral is easy to
do and gives:
(moI/2pa) * 2pa = constant * I . Thus the
constant is simply mo . Thus Ampere’s
Law is: closed loop B  dL = mo Iencircled .
Back to the solenoid
closed loop B  dL = mo Iencircled .
In using Ampere’s Law, we need to specify a
particular closed loop (just as we needed to
specify a particular closed area in using Gauss’
Law for Electric Fields). I coming out of page
Let’s first consider
generally how the
field should look:


I going into page
The Solenoid
Inside the solenoid, the field should be in the
same direction as for a circular coil: bend
your fingers around in the direction of the
circular current, and your thumb will point
in the direction of the field:

Binside

Inside the Solenoid
 If we choose a
Binside rectangular loop

inside the solenoid and
go around this loop in a CCW sense, then
Ampere’s Law closed loop B  dL = mo Iencircled
indicates that, since no current is encircled
by our rectangle (Iencircled = 0), and since the
up and down lengths are  to the field, the
field near the top and the field in the
middle must be the same!
Outside the solenoid
This means that the field inside the solenoid
is constant (just like the Electric Field was
constant between parallel plates).
The same argument applies for outside the
solenoid, and since we know the field very
far away from the long solenoid should be
zero, the field near the solenoid (but
outside of it) should also be near zero
(again like the E field outside of the parallel
plates).
Value of Field Inside
 To find the field inside
Binside we choose a

rectangular loop half
outside and half inside. Since the field is
 to the field on the up and down sides, and
since the field is zero on the top, we have:
Binside L = mo N I where N is the number of
loops in the rectangle of length L.
B inside solenoid
Binside L = mo N I becomes when we move
the L over and define n = N/L:
Binside-long = mo n I . This holds for a very
long solenoid. If we do this exactly for a
shorter solenoid, we get:
Binside-short = ½mo n I [cos(qR) - cos(qL)] .
As the diameter of the solenoid becomes
small compared to its length, qR 0o and
qL 180o giving the long result (see next
slide for diagram showing angles).
Solenoid
 a = R = Diameter/2
x
xR where xR-xL=L

= qR
 a = R = Diameter/2
x
-xL where xR-xL=L

= qL
(Note: For this position, xL < 0.)
Solenoid
Binside-long = mo n I
Binside-short = ½mo n I [cos(qR) - cos(qL)]
Note that in both cases, we have mo and I in
the numerator. But where is the inverse
distance?
Solenoid
Binside-long = mo n I
Binside-short = ½mo n I [cos(qR) - cos(qL)]
Note that in both cases, we have mo and I in
the numerator. But where is the inverse
distance? It is inside the n (n = N/L)!
Note that the radius for a long solenoid
doesn’t matter. For a short solenoid, the
radius and length are both involved in
the angles qR and qL.
Force between Currents
Masses create gravitational fields that act on other
masses, so masses act on other masses.
Charges create electric fields that act on other
charges, so charges act on other charges.
How about currents?
Does the mass in one wire affect another wire?
Does the charge in one wire affect another wire?
Does the current in one wire affect another wire?
Forces between wires
Does the mass in one wire affect another wire?
Yes, but the amount is insignificant.
Does the charge in one wire affect another wire?
Normally wires are not charged, so they normally do
not affect another wire via this way. Even when
they carry current, the current leaving the wire
equals the current entering the wire. This
normally leaves the wire uncharged even when
current is flowing through it!
Forces between currents
The current in one wire does set up a
magnetic field in the area around the wire,
and this magnetic field will act on another
current in a neighboring wire and possibly
exert a force.
Generally, though, this is a small force and a
very hard calculation, since the magnetic
field will vary across the other wire.
Example
I1 = 0.5 Amps
What is the force on the
I2 = 15 Amps rectangular circuit due
to the presence of the
8 cm current in the long wire
direction & magnitude?
4 cm
7 cm
Example
I1 = 0.5 Amps
By symmetry, the force
I2 = 15 Amps on the top and bottom
parts cancel. Since the
8 cm left side is closer than
right, the force on the
4 cm
left side will “win”.
7 cm
Example
I1 = 0.5 Amps
I2 = 15 Amps
B1
Fleft
8 cm
4 cm
7 cm
By the RHR, the B field
due to the long wire at
the left side wire is into
the screen. The force
on the left side by the
RHR is then to the right
(repulsive).
Example
I1 = 0.5 Amps
Bleft = moI1 / 2p (.04 m)
I2 = 15 Amps = (2 x 10-7)(0.5)/(.04) T
= 2.5 x 10-6 T
Fleft
8 cm Fleft = I2 L Bleft
Fright
= (15)(.08)(2.5 x 10-6) N
4 cm
= 3.0 x 10-6 Nt left
7 cm
Similarly, Fright
= 1.71 x 10-6 Nt. right.
Example
I1 = 0.5 Amps
FL = 3.0 x 10-6 Nt left
I2 = 15 Amps FR = 1.7 x 10-6 Nt. rt
Fleft
8 cm FT = 1.3 x 10-6 Nt. left
Fright
4 cm
7 cm