General Physics 2

Lecture 11 Magnetic Fields

➢Sources of the Magnetic Field
✓ The Biot–Savart Law
✓ The Magnetic Force Between Two Parallel Conductors
✓ Ampère’s Law
✓ The Magnetic Field of a Solenoid
✓ Gauss’s Law in Magnetism

1
 30.1. The Biot–Savart Law
Biot and Savart arrived at a mathematical expression
that gives the magnetic field at some point in space
in terms of the current that produces the field:

0 is a constant called the permeability of free space

The Biot–Savart Law: The magnetic field described by the

Biot–Savart law is the field due to a given current-carrying
conductor. Do not confuse this field with any external field
that may be applied to the conductor from some other source 2
Magnetic Field due to Currents

The passage of a steady current in a wire produces a

magnetic field around the wire.
◼ Field form concentric lines around the wire
◼ Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in the
direction of the current, the fingers will curl in the direction of
the field (second right-hand rule).
◼ Magnitude of the field
I
o I
B=
2 r

3
Magnitude of the field I

o I r
B=
2 r B

o called the permeability of free space

0 = 4  10 T  m A
−7

4
Example 30.1: Magnetic Field Surrounding a Thin, Straight Conductor
Consider a thin, straight wire of finite length carrying a constant current I and
placed along the x axis. Determine the magnitude and direction of the magnetic
field at point P due to this current

Figure 30.3. (a) A thin, straight wire carrying a current I.

(b) The angles 1 and 2 used for determining the net field
5
Example 30.1: Magnetic Field Surrounding a Thin, Straight Conductor
Consider a thin, straight wire of finite length carrying a constant current I and
placed along the x axis. Determine the magnitude and direction of the magnetic
field at point P due to this current

6
Example 30.1: Magnetic Field Surrounding a Thin, Straight Conductor
Consider a thin, straight wire of finite length carrying a constant current I and
placed along the x axis. Determine the magnitude and direction of the magnetic
field at point P due to this current

We can use this result to find the magnitude of

the magnetic field of any straight current-carrying
wire if we know the geometry and hence the
angles 1 and 2 (an infinitely long, straight wire)
→ 1 = /2 and 2 = - /2 for length elements
ranging between positions x = -  and x = +  →
(sin 1 - sin 2 ) = 2

This mathematical form as the expression for the magnitude

of the electric field due to a long charged wire 7
Example 30.2. Magnetic Field Due to a Curved Wire
Segment
Calculate the magnetic field at point O for the current-carrying wire segment.
The wire consists of two straight portions and a circular arc of radius a, which
subtends an angle 

Figure 30.4 The length of the curved segment AC is s

8
Example 30.2. Magnetic Field Due to a Curved Wire Segment
Calculate the magnetic field at point O for the current-carrying wire segment. The
wire consists of two straight portions and a circular arc of radius a, which subtends an
angle 

9
Example 30.2. Magnetic Field Due to a Curved Wire Segment
Calculate the magnetic field at point O for the current-carrying wire segment. The
wire consists of two straight portions and a circular arc of radius a, which subtends an
angle 

To find the magnetic field at the center of a circular

wire loop of radius R that carries a current I.

➢ The straight wires in Figure do not contribute to

the magnetic field. The only contribution is
from the curved segment.
➢ As the angle  increases, the curved segment
becomes a full circle when  = 2
➢ Therefore, find the magnetic field at the center
of a wire loop by letting  = 2

10
Example 30.3. Magnetic Field on the Axis of a Circular Current Loop
Consider a circular wire loop of radius a located in the yz plane and carrying a steady
current I. Calculate the magnetic field at an axial point P a distance x from the center
of the loop

11
Example 30.3. Magnetic Field on the Axis of a Circular Current Loop
Consider a circular wire loop of radius a located in the yz plane and carrying a steady
current I. Calculate the magnetic field at an axial point P a distance x from the center
of the loop

The x component of the field element

12
 Example 30.3. Magnetic Field on the Axis of a Circular Current Loop
Consider a circular wire loop of radius a located in the yz plane and carrying a steady
current I. Calculate the magnetic field at an axial point P a distance x from the center
of the loop

To find the magnetic field at the center of the

loop, set x = 0

If we consider points on the x axis very far from the

loop? How does the magnetic field behave at these
distant points

The magnitude of the magnetic moment  of the loop:

13
 30.2. The Magnetic Force Between Two Parallel Conductors

Figure 30.7 Two parallel wires that ➢ The forces are the same on both wires,
each carry a steady current exert a we denote the magnitude of the
magnetic force between the wires as
magnetic force on each other. The
simply FB
force is attractive if the currents are
parallel and repulsive if the currents
are antiparallel 14
30.2. The Magnetic Force Between Two Parallel Conductors

➢ The forces are the same on both wires, we denote the magnitude of the
magnetic force between the wires as simply FB

The force between two parallel wires is used to define the ampere

I1 = I2 = 1 A and a = 1 m → FB/l = 2 x 10-7 N./m

Definition of the ampere

When the magnitude of the force per unit length between two long, parallel
wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the
current in each wire is defined to be 1 A

15
 Example 1: Levitating a wire

Two wires, each having a weight per units length of 1.0x10-4 N/m, are
strung parallel to one another above the surface of the Earth, one directly
above the other. The wires are aligned north-south. When their distance
of separation is 0.10 mm what must be the current in each in order for the
lower wire to levitate the upper wire. (Assume the two wires carry the
same current).

l
1
I1
2 d
I2

16
 Two wires, each having a weight per
F1 units length of 1.0x10-4 N/m, are strung
1
parallel to one another above the surface
B2 I1 of the Earth, one directly above the
other. The wires are aligned north-south.
mg/l
2 d When their distance of separation is 0.10
mm what must be the current in each in
I2 order for the lower wire to levitate the
upper wire. (Assume the two wires carry
l the same current).

F1 mg o I 2
Weight of wire per unit = =
length: l l 2 d
mg/l = 1.0x10-4 N/m
1.0  10 −4 N / m =
( 4  10 −7
Tm A )( )
I 2

Wire separation: ( 2 )( 0.10m )

d=0.1 m
I1 = I 2
I = 7.1 A
17
Example 2: magnetic field between the wires

The two wires in the figure below carry currents of 3.00A and 5.00A in
the direction indicated. Find the direction and magnitude of the magnetic
field at a point midway between the wires.

3.00 A 5.00 A

20.0 cm

18
Example 30.4 Suspending a Wire
Two infinitely long, parallel wires are lying on the ground a distance a = 1.00 cm apart. A third
wire, of length L = 10.0 m and mass 400 g, carries a current of I1 = 100 A and is levitated
above the first two wires, at a horizontal position midway between them. The infinitely long
wires carry equal currents I2 in the same direction, but in the direction opposite that in the
levitated wire. What current must the infinitely long wires carry so that the three wires form an
equilateral triangle?

19
Example 30.4 Suspending a Wire
Two infinitely long, parallel wires are lying on the ground a distance a = 1.00 cm apart. A third
wire, of length L = 10.0 m and mass 400 g, carries a current of I1 = 100 A and is levitated above
the first two wires, at a horizontal position midway between them. The infinitely long wires carry
equal currents I2 in the same direction, but in the direction opposite that in the levitated wire.What
current must the infinitely long wires carry so that the three wires form an equilateral triangle?

The particle in equilibrium model by adding the

forces and setting the net force equal to zero

The current in the wires on the ground

20
30.3. Ampère’s Law

Andre-Marie Ampère
French Physicist (1775–1836)
➢ Ampère is credited with the
discovery of electromagnetism,
which is the relationship between
electric currents and magnetic fields.
➢ Ampère’s genius, particularly in
mathematics, became evident by
the time he was 12 years old

21
30.3. Ampère’s Law

Figure 30.9 The right- Figure 30.10 (a) and (b) Compasses show the effects of
hand rule for determining
the current in a nearby wire. (c) Circular magnetic field
the direction of the
lines surrounding a current-carrying conductor,
magnetic field
displayed with iron filings
surrounding a long,
straight wire carrying a
current. Notice that the
magnetic field lines form
circles around the wire 22
30.3. Ampère’s Law

is the circumference of the circular path of radius r

23
30.3. Ampère’s Law

➢ Ampère’s law describes the creation of magnetic fields by all

continuous current configurations, but at our mathematical level
it is useful only for calculating the magnetic field of current
configurations having a high degree of symmetry.
➢ Its use is similar to that of Gauss’s law in calculating electric
fields for highly symmetric charge distributions
24
Example 30.5 The Magnetic Field Created by a Long Current-Carrying Wire
A long, straight wire of radius R carries a steady current I that is uniformly distributed
through the cross section of the wire. Calculate the magnetic field a distance r from
the center of the wire in the regions r  R and r < R

Figure 30.13 A long, straight wire of radius R carrying a steady current I uniformly
distributed across the cross section of the wire. The magnetic field at any point
can be calculated from Ampère’s law using a circular path of radius r, concentric
with the wire

25
Example 30.5 The Magnetic Field Created by a Long Current-Carrying Wire
A long, straight wire of radius R carries a steady current I that is uniformly distributed
through the cross section of the wire. Calculate the magnetic field a distance r from
the center of the wire in the regions r  R and r < R

➢ The interior of the wire, where r < R

→ the current I’ passing through the plane
of circle 2 is less than the total current I
➢ The ratio of the current I’ enclosed by circle 2 to the entire current I equal
to the ratio of the area r2 enclosed by circle 2 to the cross-sectional area
R2 of the wire

26
Example 30.5 The Magnetic Field Created by a Long Current-Carrying Wire
A long, straight wire of radius R carries a steady current I that is uniformly distributed
through the cross section of the wire. Calculate the magnetic field a distance r from
the center of the wire in the regions r  R and r < R

Figure 30.14 Magnitude of the

magnetic field versus r for the wire.
The field is proportional to r inside
the wire and varies as 1/r outside
the wire
27
30.4. The Magnetic Field of a Solenoid

➢ A solenoid is a long wire wound in the

form of a helix.
➢ Areasonably uniform magnetic field can
be produced in the space surrounded by
the turns of wire—which we shall call
the interior of the solenoid—when the
solenoid carries a current.
➢ When the turns are closely spaced, each
can be approximated as
a circular loop
➢ The net magnetic field is the vector sum
of the fields resulting from all the turns
Figure 30.16 The magnetic
field lines for a loosely wound
solenoid 28
30.4. The Magnetic Field of a Solenoid

Solenoid magnet consists of a wire coil with multiple

loops.
It is often called an electromagnet.

29
30.4. The Magnetic Field of a Solenoid
➢ The solenoid is ideal, B in the interior space is uniform and
parallel to the axis and the magnetic field lines in the exterior
space form circles around the solenoid. The planes of these
circles are perpendicular to the page.
➢ The rectangular path (loop 2) of length ℓ, and width w.
➢ Let’s apply Ampère’s law to this path
➢ The contribution along side 3 is zero (the external magnetic
field lines are perpendicular to the path in this region).
➢ The contributions from sides 2 and 4 are both zero (B ⊥ dԦs both
inside and outside the solenoid)
➢ Side 1 gives a contribution to the integral because along this
path B is uniform and parallel to dԦs. The integral over the
closed rectangular path:

Figure 30.18 Cross-sectional view of an ideal

solenoid, where the interior magnetic field is
uniform and the exterior field is close to zero 30
30.4. The Magnetic Field of a Solenoid

The wire passes through

the loop N times, so the
total current through the
loop is NI

is the number of turns per unit length

31
Solenoid Magnet

Field lines inside a solenoid magnet are parallel, uniformly spaced

and close together.
The field inside is uniform and strong.
The field outside is non uniform and much weaker.
One end of the solenoid acts as a north pole, the other as a south
pole.
For a long and tightly looped solenoid, the field inside has a value:

B = o nI
32
Solenoid Magnet

B = o nI
n = N/l : number of (loop) turns per unit length.

I : current in the solenoid.

o = 4  10 Tm / A
−7

33
Example: Magnetic Field inside a Solenoid.

Consider a solenoid consisting of 100 turns of wire and

length of 10.0 cm. Find the magnetic field inside when it
carries a current of 0.500 A.

34
Example: Magnetic Field inside a Solenoid.

Consider a solenoid consisting of 100 turns of wire and

length of 10.0 cm. Find the magnetic field inside when it
carries a current of 0.500 A.

N = 100 N 100turns
l = 0.100 m n= = = 1000turns / m
I = 0.500 A l 0.10m
o = 4  10 −7 Tm / A
( )
B = o nI = 4  10 Tm / A (1000turns / m )( 0.500 A )
−7

B = 6.28 10−4 T

35
 30.5. Gauss’s Law in Magnetism
If the magnetic field at this element is B, the magnetic flux through the element is
B.dA , where dA is a vector that is perpendicular to the surface and has a
magnitude equal to the area dA. Therefore, the total magnetic flux B through the
surface is

The value of magnetic flux is

proportional to the total number
of magnetic field lines passing Figure 30.19 The magnetic flux through an area element
through the loop. dA is B.dA = B dA cos, where dA is a vector
perpendicular to the surface

➢ The unit of magnetic flux is T.m2, which is defined as a weber (Wb)

1 Wb = 1 T.m2 36
 Problem: determining a flux

A square loop 2.00m on a side is placed in a magnetic field of

strength 0.300T. If the field makes an angle of 50.0° with the
normal to the plane of the loop, determine the magnetic flux
through the loop.

37
 Problem: determining a flux
A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the
field makes an angle of 50.0° with the normal to the plane of the loop, determine the
magnetic flux through the loop.

Given: From what we are given, we use

L = 2.00 m
B = 0.300 T
F = BA cos  = ( 0.300T )( 2.00 m ) cos 50.0
2
 = 50.0˚

= 0.386 Tm 2
Find:

F=?

38
30.5. Gauss’s Law in Magnetism

Figure 30.20 Magnetic flux through a plane lying in a magnetic fiel

39
Example 30.7 Magnetic Flux Through a Rectangular Loop
A rectangular loop of width a and length b is located near a long wire carrying a
current I. The distance between the wire and the closest side of the loop is c. The
wire is parallel to the long side of the loop. Find the total magnetic flux through
the loop due to the current in the wire

Figure 30.21 The magnetic field due to the wire carrying a

current I is not uniform over the rectangular loop 40
Example 30.7 Magnetic Flux Through a Rectangular Loop
A rectangular loop of width a and length b is located near a long wire carrying a current I. The
distance between the wire and the closest side of the loop is c. The wire is parallel to the long
side of the loop. Find the total magnetic flux through the loop due to the current in the wire
B is parallel to dA at any point within the loop

→ The flux depends on the size of the loop.

If c becomes large such that the loop is very far from the wire, the flux approaches zero
If c goes to zero, the flux becomes infinite 41
 30.5. Gauss’s Law in Magnetism
➢ Remind: The electric flux through a closed surface surrounding a net charge is
proportional to that charge (Gauss’s law) → the number of electric field lines
leaving the surface depends only on the net charge within it.
➢ For any closed, the number of lines entering the surface equals the number
leaving the surface → the net magnetic flux is zero. In contrast, for a closed
surface surrounding one charge of an electric dipole, the net electric flux is not
zero
➢ Gauss’s law in magnetism
The net magnetic flux through any closed surface is always zero

Figure 30.22 The magnetic field lines of a

bar magnet form closed loops. (The
dashed line represents the intersection Figure 30.23 The electric field lines surrounding an electric dipole
42
of a closed surface with the page.) begin on the positive charge and terminate on the negative charge
Electric Field vs Magnetic Field
Electric Magnetic
Source Charges Moving Charges
Acts on Charges Moving Charges
Force F = Eq F = q v B sin()
Direction Parallel E Perpendicular to v,B

Field Lines

Opposites Charges Attract Currents Repel

43