Chapter 30. Sources of the Magnetic Field 30.1 The Biot-Savart Law

Magnetic field is generated by a conductor carrying a steady current.

!

dB =µ04"

Ids# ˆ r r2

The vector

!

dB is perpendicular both to

!

ds (direction of current) and

!

ˆ r . µ0 = 4! " 10-7 T•m/A , is the permeability of free space.

Total magnetic field B due to a current-carrying conductor:

B =µ0 I4!

ds" ˆ r r 2#

Example: A thin, straight wire carrying a constant current I. Calculate the total magnetic field at point P. Solution: Using the Right-Hand Rule, the direction of the field at P is into the paper.

r = s2 + R2 sin! = sin(" #!) = R

s2 + R2

B = 2 ! µ0 I4"

Rds(s2 + R2)3/ 20

#

$

=µ0 I2"R

s(s2 + R2 )1/ 2%

& '

(

) * 0

#

B = µ0 I

2!R"1R

Important Result: (1) The magnetic field lines generated by a long straight

current-carrying wire are circles concentric with the wire and lie in planes perpendicular to the wire.

(2) The direction of the field can be determined by a Right-hand rule.

(3) The magnitude of the magnetic field decreases with the increase of the distance from the wire.

Example: Calculate the magnitude of B 4 cm from a long, straight wire carrying a current of 5.0 A.

Solution: B = 2 (10!7T "m / A)(5A)4#10!2m

= 2.5#10!5T

Example: A circular loop of wire of radius R carrying a current I. What is the magnetic field at the center of the loop? Solution: The direction of B at the center is out of the paper.

B = µ0 I4!

dsR2" =

µ0 I4!

(2!R)R2

=µ0 I2R

B = µ0 I

2R

I

R

ds

30.2 The Magnetic Force Between Two Parallel Conductors Two long, straight, parallel wires separated by a distance a and carrying currents I1 and I2 in the same direction. The magnetic field created by wire 2 at wire 1 is

!

B2 =µ0I22"d

The force on wire 1 (attractive):

!

F12 = I1lB2 =µ0I1I2l2"a

(a) Parallel conductors carrying currents in the same direction

attract each other. (b) Parallel conductors carrying currents in opposite directions

repel each other.

Definition of the ampere:

!

F12l

=µ0I1I22"a

= 2#10$7 I1I2a

If two long, parallel wire 1 m apart carrying the same

current and the force per unit length on each wire is 2 " 10-7 N/m, then the current is defined to be 1 A.

30.3 Ampere's Law A thin, straight conductor carrying current I,

B = µ0 I2!R

Line integral B !" ds = µ0 I2#R

ds" = µ0 I

Ampere's Law:

The line integral of B !ds around any closed path equals µ0 I . I is the total steady current passing through any surface bounded by the closed path.

B !" ds = µ0 I • Ampere's Law is useful for calculating

the magnetic field in a symmetry system

Example: Find values of B !" ds. a. B !" ds = b. B !" ds = c. B !" ds = d. B !" ds =

Example: Find values of B !" ds.

(a) B !" ds = (b) B !" ds = (c) B !" ds = (d) B !" ds =

Example: A long, straight wire carries a current I0 that is uniformly distributed through the cross-section. Calculate B. (a) Outside the wire: Use loop 1.

B !" ds = B ds" = B(2#r) = µ0 I0

B = µ0 I0

2!r (r # R).

(b) Inside the wire: Use loop 2.

Let Iin be the current inside the loop 2:

!

IinI0

= ratio of cross-sectional area =

!

"r2

"R2 $

!

Iin =r2

R2I0

!

B "# ds = B(2$r) = µ0Iin = µ0r2

R2I0

%

& '

(

) *

B =µ0 I02!R2" # $

% & ' r (r < R).

r

B%1/r

R

B

B%r

I0

2

1

r

R

Example: The magnetic field created by a Toroid. N: the total number of turns of wire. Apply Ampere's Law,

!

B "# ds = B ds# = B(2$r)= µ0NI

B = µ0NI

2!r"1r

• B % 1/r . B is not uniform within the coil. • If r >> cross sectional radius, B is approximately uniform. Example: Calculate the force exerted on the loop carrying current I2. For the top and bottom half circles, the force is 0. Because B || ds, ds! B = 0 . Forces on the two straight portions: Ftotal = 2(I2 L ! B)

Since B = µ0 I12!R

,

Ftotal =µ0 I1I2L!R

I1

L

R

I2 F F

30.4 The Magnetic Field of a Solenoid Solenoid: A long wire wound in the form of a helix to generate uniform magnetic field inside. An ideal solenoid: The turns are closely spaced and the length is much greater than the radius of the turns. For an ideal solenoid, line integral around the rectangular loop 2:

!

B "ds =# B "dsab# = B ds = Bla

b# Let n be the number of turns per unit

length of the solenoid. The current passing through the loop is nIl.

Apply Ampere's Law:

!

Bl = µ0nIl B = µ0nI (Ideal solenoid)

Example: A superconducting solenoid. B = 10 T, n = 2000 /m. What is the required current?

I = Bµ0n

=10 T

(2000 /m)(4! "10#7 T $m/A)= 3979 A

30.5 Gauss’s Law in Magnetism Define: Magnetic field flux through a surface: !B = B "dA# Unit: weber (Wb), 1 Wb = 1 T•m2 For Electric field:

Field lines start from positive charges (or !), and end at negative charges (or !).

!E = E "dA# =Qtotal

$ 0

For Magnetic field: All lines are continuous and form closed loops. The number of lines entering a surface = the number of lines leaving.

Gauss’s law in magnetism: The net magnetic flux through any closed surface is always zero. B !dA" = 0 30.6 Magnetism in Matter The magnetic Moments of Atoms (a) Classical Model: An electron moving in a circular orbit about the nucleus. The effective current of moving electron:

I = eT=e!2"

=ev2"r

The magnetic moment: µ = IA = 1

2 evr The orbital angular momentum: L = mvr

µ =e2m! " #

$ % & L (Classical)

L

r v

µ

(b) Quantum Physics: The magnetic moment is quantized.

µ =

e2m!i (Quantum Mechanics)

where, ! = h / 2! = 1.06 "10#34 J $ s, h is Planck's constant. i = 0, 1, 2, 3, ….. (a) Paramagnetism (most materials) Such materials contain atomic magnetic moments that are randomly oriented, and the net magnetic moment is zero. An external magnetic field, can partially align the atomic magnetic moments, and result a weak net magnetic moment. The net magnetic moment disappears when external field is removed (b) Ferromagnetism (iron, nickel, …) Such materials contain atomic magnetic moments that tend to align parallel to each other. They form micro-domains within which all magnetic moments are aligned. In un-magnetized samples, the domains are randomly oriented, and the net magnetic moment is zero. An external magnetic field, can align some domains and result a strong net magnetic moment. The magnetic moment partially remains when external field is removed (magnetized).

(c) Diamagnetism (silver, superconductors…) Such materials has no permanent atomic magnetic moments. An external magnetic field can induce a weak magnetic moment in the opposite direction of the applied field. The induced magnetic moment disappears when external field is removed. The Curie temperature is the critical temperature above which a ferromagnetic material loses its residual magnetism, and the material will become paramagnetic.

Above the Curie temperature, the thermal agitation is great enough to cause a random orientation of the moments. The Magnetic Field of the Earth

Paramagnetic M

0 Tc T

Ferromagnetic

Meissner Effect

Certain types of superconductors also exhibit perfect diamagnetism in the superconducting state. This is called the Meissner effect.

If a permanent magnet is brought near a superconductor, the two objects repel each other.