Download PES 1120 Spring 2014, Spendier Lecture 31/Page 1 Today (finish

PES 1120 Spring 2014, Spendier
Lecture 31/Page 1
Today (finish chapter 29):
- Magnetic field strength inside an ideal solenoid and toroid
- A Current-Carrying Coil as a Magnetic Dipole
- HW7 due HW8 given out
Last time: Ampere’s Law
 
B
  ds  0 I encl
How to chose Amperian loop:


- angle θ between dl and B should be 0, 90, or 180 degrees!
Solenoid
A solenoid is a long coil of wire tightly wound in the helical form. The figure below shows
the magnetic field lines of a solenoid carrying a steady current I. We see that if the turns are
closely spaced, the resulting magnetic field inside the solenoid becomes fairly uniform,
provided that the length of the solenoid is much greater than its diameter. For an “ideal”
solenoid, which is infinitely long with turns tightly packed, the magnetic field inside the
solenoid is uniform and parallel to the axis, and vanishes outside the solenoid.
The right figure indicates that the magnetic field of a solenoid looks like the magnetic field of
a permanent bar magnet. However, a permanent magnet can never be turned off. The
magnetic field of a solenoid can be turned off simply by turning the current off. Solenoids
form the basis of many "electromagnets" that can be controlled by currents.
DEMO:
OBSERVATIONS:
1) B-field almost uniform inside
2) B-field almost zero outside
PES 1120 Spring 2014, Spendier
Lecture 31/Page 2
Now we will apply Ampere's law to calculate the magnetic field strength inside a solenoid
Magnetic field strength inside an ideal solenoid
We can use Ampere’s law to calculate the magnetic field strength inside an ideal solenoid.
The cross-sectional view of an ideal solenoid is shown.

To compute B , we consider a rectangular path of length
l and width w and traverse the path

in a counterclockwise manner. The line integral of B along this loop is










 B  dl   B  dl   B  dl   B  dl   B  dl
1
2
3
4
1) B=0 outside a solenoid


2) angle between B and dl is 90 ==> cos(90º)=0


3) angle between B and dl is 0 ==> cos(0º)=1


4) angle between B and dl is 90 ==> cos(90º)=0
 
B
  dl  0  0  Bl  0  0 I enc
On the other hand, the total current enclosed by the Amperian loop is
I enc  IN
where N is the total number of turns. Applying Ampere’s law yields
 
B
  dl  Bl  0 IN
B
0 IN
l
using N/l = n to represent the number of turns per unit length, the magnetic field inside an
ideal solenoid is
B  0 nI
So for an ideal solenoid the B field is constant over the cross-section of a solenoid.
PES 1120 Spring 2014, Spendier
Lecture 31/Page 3
We can increase B by:
- increasing he current in the wire
- increasing the density of the windings n (for example could have layers of windings)
Example: from our Demo:
l = 0.6m, I = 3 A, N=28
 IN  4 10   3 28 
B 0 
 1.8  104 T  0.18 mT
l
0.6
7
We measured B = 0.2 mT - so not bad!
USE Demo to show that when we decrease l the B field inside increases!
Toroid
Another Current configuration that has a symmetry that we can take advantage of is a
toroid. It has a rotational symmetry around its axis. To apply Ampere’s Law to calculate
the magnetic field, we find a path where we expect the B-field to be constant.
Magnetic Field inside a Toroid
We will use an Amperian loop in the interior of the toroid.
 
B
  dl  B  2 r   0 IN
B
0 IN
2 r
In contrast to the situation for a solenoid, B is not constant over the cross section of a
toroid. It is easy to show, with Ampere’s law, that B = 0 for points outside an ideal
toroid (as if the toroid were made from an ideal solenoid). The direction of the magnetic
field within a toroid follows from our curled–straight right-hand rule: Grasp the toroid
with the fingers of your right hand curled in the direction of the current in the windings;
your extended right thumb points in the direction of the magnetic field.
PES 1120 Spring 2014, Spendier
Lecture 31/Page 4
Example (similar to HW 8): The figure below shows a cross section of an infinite
conducting sheet carrying a current per unit x-length of λ; the current flows
perpendicularly into the page.
Use Ampere’s law to calculate the magnitude of the magnetic field at all points P and P’.
PES 1120 Spring 2014, Spendier
Lecture 31/Page 5
A Current-Carrying Coil as a Magnetic Dipole

A coil will behave like a magnetic dipole. If we place it in an external magnetic field B ,

a torque t given by
  
t  m B

acts on it. Here m is the magnetic dipole moment of the coil and has the magnitude
μ= NIA,
where N is the number of turns, I is the current in each turn, and A is the area enclosed by
each turn. (Caution: Don’t confuse the magnetic dipole moment with the permeability
constant μ0.)
A current loop produces a magnetic field like that of a bar magnet and thus has associated

north and south poles. The magnetic dipole moment m of the loop, its direction given by
a curled–straight right-hand rule, points from the south pole to the north pole, in the
direction of the field B within the loop.
We turn now to the other aspect of a current-carrying coil as a magnetic dipole.
What magnetic field does it produce at a point in the surrounding space? The problem
does not have enough symmetry to make Ampere’s law useful; so we must turn to the
law of Biot and Savart. For simplicity, we first consider only a coil with a single circular
loop and only points on its perpendicular central axis, which we take to be the x axis.
PES 1120 Spring 2014, Spendier
Lecture 31/Page 6
in which a is the radius of the circular loop and x is the distance of the point in question
from the center of the loop. Furthermore, the direction of the magnetic field B is the

same as the direction of the magnetic dipole moment m of the loop.
For axial points far from the loop, we have x >> a. With that approximation, the equation
reduces to
0 Ia 2
B  x 
2 x3
Recalling that pa 2 is the area A of the loop and extending our result to include a coil of N
turns, we can write this equation as
PES 1120 Spring 2014, Spendier
B  x 
Lecture 31/Page 7
0 NIA
2 x 3


Further, because B and m have the same direction, we can write the equation in vector
form, substituting from the identity
μ= NiA,


0 
B  x 
2 x 3
(current-carrying coil)
Thus, we have two ways in which we can regard a current-carrying coil as a magnetic
dipole:
(1) it experiences a torque when we place it in an external magnetic field;
(2) it generates its own intrinsic magnetic field for distant points along its axis given by
the equation above.