1. Chapter 30: Sources of the Magnetic Field
Chapter 28 opener. A long coil of wire with many closely spaced loops is called a solenoid. When a long solenoid carries an electric current, a nearly uniform magnetic field is produced within the loops as suggested by the alignment of the iron filings in this photo. The magnitude of the field inside a solenoid is readily found using Ampère’s law, one of the great general laws of electromagnetism, relating magnetic fields and electric currents. We examine these connections in detail in this Chapter, as well as other means for producing magnetic fields.

2. Chapter Outline Ampère’s Law Biot-Savart Law
Magnetic Field Due to a Straight Wire
Force between Two Parallel Wires
Definitions: The Ampere & the Coulomb
Ampère’s Law
Magnetic Field of a Solenoid & of a Toroid.
Biot-Savart Law
Magnetic Materials – Ferromagnetism
Electromagnets and Solenoids – Applications
Magnetic Fields in Magnetic Materials; Hysteresis
Paramagnetism and Diamagnetism

3. Biot-Savart Law & Ampère’s Law.
Note! We’ll cover topics in Ch. 30 in a different order than your textbook!
Specifically, there are 2 main (equivalent)
methods to calculate magnetic fields.
These are the
Biot-Savart Law & Ampère’s Law.
We’ll cover Ampère’s Law followed by the Biot-Savart Law. This is the opposite order than is in the textbook.

4. Use the right-hand rule to determine the
Magnetic Field Due to a Long, Straight Current Carrying Wire: Direction
Ch. 29: The magnetic field lines
for a long, straight, current
carrying wire are circles
concentric with the wire.
The field lines are in planes
perpendicular to the wire.
The magnitude of the field is
constant on a circle of radius a.
Use the right-hand rule to determine the
direction of the field, as shown.

5. Magnetic Field Due to a Current Carrying Wire
A compass can be used to
detect the magnetic field.
When there is no current
in the wire, there is no
field due to the current.
In this case, the compass
needles all point toward
the Earth’s north pole. This is due to the
Earth’s magnetic field

6. Magnetic Field Due to a Current Carrying Wire
When the wire carries a
current, the compass
needles deflect in a direction
tangent to the circle.
This shows the direction of
the magnetic field produced
by the wire.
If the current is reversed, the direction of the
needles also reverses.

7. The circular magnetic field around the wire is shown by the iron filings.

8. Magnetic Field Due to a Straight Wire
Experimental Results show that the
Magnetic field B due to a straight,
current carying wire is proportional
to the Current I & inversely
proportional to the distance r from
the wire:
μ0 is a constant, called the permeability
of free space. It’s value is
μ0 = 4π  10-7 T·m/A.
It plays a similar role for magnetic
fields that ε0 plays for electric fields!
Figure Same as Fig. 27–8b. Magnetic field lines around a long straight wire carrying an electric current I.

9. Calculation of B Near a Wire
An electric wire in the wall of a
building carries a dc current
I = 25 A vertically upward.
Calculate the magnetic field B
due to this current at a point
P = 10 cm in the radial
direction from the wire.
Solution: B = μ0I/2πr = 5.0 x 10-5 T

10. Calculation of B Near a Wire
An electric wire in the wall of a
building carries a dc current
I = 25 A vertically upward.
Calculate the magnetic field B
due to this current at a point
P = 10 cm in the radial
direction from the wire.
Solution
Use
This gives B = 5.0  10-5 T
Solution: B = μ0I/2πr = 5.0 x 10-5 T

11. Magnetic Field Midway Between Two Currents.
Example
Magnetic Field Midway Between Two Currents.
Two parallel straight wires a distance 10.0 cm apart carry
currents in opposite directions. Current I1 = 5.0 A is out
of the page, & I2 = 7.0 A is into the page.
Calculate the magnitude & direction of the magnetic
field halfway between the 2 wires.
Figure Example 28–2. Wire 1 carrying current I1 out towards us, and wire 2 carrying current I2 into the page, produce magnetic fields whose lines are circles around their respective wires.
Solution: As the figure shows, the two fields are in the same direction midway between the wires. Therefore, the total field is the sum of the two, and points upward: B1 = 2.0 x 10-5 T; B2 = 2.8 x 10-5 T; so B = 4.8 x 10-5 T.

12. Magnetic Field Midway Between Two Currents.
Example
Magnetic Field Midway Between Two Currents.
Two parallel straight wires a distance 10.0 cm apart
carry currents in opposite directions. Current I1 = 5.0 A
is out of the page, & I2 = 7.0 A is into the page.
Calculate the magnitude & direction of the magnetic
field halfway between the 2 wires.
Solution:
Use
for each wire. Then
add them as vectors
B = B1 + B2
B = 4.8  10-5 T
Figure Example 28–2. Wire 1 carrying current I1 out towards us, and wire 2 carrying current I2 into the page, produce magnetic fields whose lines are circles around their respective wires.
Solution: As the figure shows, the two fields are in the same direction midway between the wires. Therefore, the total field is the sum of the two, and points upward: B1 = 2.0 x 10-5 T; B2 = 2.8 x 10-5 T; so B = 4.8 x 10-5 T.

13. Magnetic Field Due to 4 Wires
Conceptual Example
Magnetic Field Due to 4 Wires
The figure shows 4 long parallel wires carrying equal currents
into or out of the page. In which configuration, (a) or (b), is
the magnetic field greater at the center of the square? a b a a a
Solution: The fields cancel in (b) but not in (a); therefore the field is greater in (a).

14. Magnetic Field Due to 4 Wires
Conceptual Example
Magnetic Field Due to 4 Wires
The figure shows 4 long parallel wires carrying equal currents
into or out of the page. In which configuration, (a) or (b), is
the magnetic field greater at the center of the square? a b
Solution: The fields cancel in (b) but not in (a); therefore the field is greater in (a).

15. Magnetic Force Between Two Parallel Wires
Recall from Ch. 29 that the force F
on a wire carrying current I in a
magnetic field B is:
We’ve just seen that the magnetic
field B produced at the position of
wire 2 due to the current in wire 1 is:
Figure (a) Two parallel conductors carrying currents I1 and I2. (b) Magnetic field B1 produced by I1 (Field produced by I2 is not shown.) B1 points into page at position of I2.
The magnetic force between
2 currents is analogous to
the Coulomb electric force
between 2 charges!!
The force F this field exerts
on a length ℓ2 of wire 2 is

16. Parallel Currents ATTRACT Anti-Parallel Currents REPEL.
Using the cross product
form of the force:
& applying the right
hand rule shows that
Parallel Currents ATTRACT
and
Anti-Parallel Currents REPEL.
Figure (a) Parallel currents in the same direction exert an attractive force on each other. (b) Antiparallel currents (in opposite directions) exert a repulsive force on each other.

17. Example Force Between Two Current-carrying Wires.
The two wires of a 2.0-m
long appliance cord are
d = 3.0 mm apart & carry
a current I = 8.0 A dc.
Calculate the force one
wire exerts on the other.
Solution: F = 8.5 x 10-3 N, and is repulsive. Appliances powered by 12V dc current are marketed to truckers, campers, people running on solar power (especially off-grid), among others.

18. Example Force Between Two Current-carrying Wires
The two wires of a 2.0-m
long appliance cord are
d = 3.0 mm apart & carry
a current I = 8.0 A dc.
Calculate the force one
wire exerts on the other.
Use:
Resulting in
F = 8.5  10-3 N
Solution: F = 8.5 x 10-3 N, and is repulsive. Appliances powered by 12V dc current are marketed to truckers, campers, people running on solar power (especially off-grid), among others.

19. Suspending a wire with a current.
Example
Suspending a wire with a current.
A horizontal wire carries a current I1 = 80 A dc.
Calculate the current I2 that a second parallel wire d = 20
cm below it must carry so that it doesn’t fall due to gravity.
The lower wire has a mass per meter of (m/ℓ) = 0.12 g/m.
Solution: The magnetic force due to the current in the wire must be equal and opposite to the gravitational force on the second wire. We need a definite length of wire, as the forces on an infinitely long wire will be infinite; choose 1 meter for simplicity. Substitution gives I2 = 15 A.

20. Definition of the Ampere
The SI unit of current, the Ampere is officially
defined in terms of the force between two current
carrying wires:
One Ampere IS DEFINED as that current flowing in each of two long parallel wires 1 m apart, which results in a force of exactly F =2  10-7 N per meter of length of each wire.

21. Definition of the Coulomb
Given that definition of the Ampere
One Coulomb is IS DEFINED as exactly
1 Ampere-second.