Understanding Steady Magnetic Fields: A

Comprehensive Textbook
Grok 3

May 7, 2025

Contents
1 Introduction to Magnetic Fields 2
1.1 What is a Magnetic Field? . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Sources of Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Forces Between Current-Carrying Conductors . . . . . . . . . . . . . . 3

2 Magnetic Field Geometry and the Right-Hand Rule 3

2.1 Direction of Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Visualizing Magnetic Field Lines . . . . . . . . . . . . . . . . . . . . . . 4

3 The Biot-Savart Law 4

3.1 Definition and Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3.2 Application to Point and Distributed Currents . . . . . . . . . . . . . . 5
3.3 Comparison with Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . 5

4 Magnetic Fields from Current Configurations 6

4.1 Line Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Circulating Currents (Current Loops) . . . . . . . . . . . . . . . . . . . 8
4.3 Two- and Three-Dimensional Currents . . . . . . . . . . . . . . . . . . 9

5 Numerical Examples for Line Currents 10

5.1 Example 1: Magnetic Field due to a Straight Conductor . . . . . . . . 10
5.2 Example 2: Magnetic Field due to a Segment of a Triangular Loop . 11

6 Magnetic Field from a Circular Current Loop 11

6.1 Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

7 Simulation and Visualization Tools 12

8 Introduction to Ampere’s Circuital Law 12

9 Conclusion 12

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1 Introduction to Magnetic Fields
Magnetic fields are fundamental to electromagnetism, influencing moving charges
and current-carrying conductors. This chapter explores the nature of magnetic
fields, their sources, and their applications, building from basic principles to ad-
vanced calculations.

1.1 What is a Magnetic Field?

A magnetic field is a vector field that exerts a force on moving electric charges
and magnetic materials. Unlike an electric field, which acts on stationary charges,
a magnetic field primarily affects charges in motion. For example, a magnet at-
tracts iron filings due to its magnetic field, and a current-carrying wire experi-
ences a force when placed near another current-carrying wire.
The magnetic field is characterized by two related quantities:

• Magnetic flux density (B), measured in tesla (T), represents the strength
and direction of the magnetic field.

• Magnetic field intensity (H), measured in amperes per meter (A/m), is re-
lated to B by the permeability of the medium:

B = µH, (1)

where µ is the permeability (in henry per meter, H/m). In free space, µ =
µ0 = 4π × 10−7 H/m.

The magnetic field is a non-contact force field, enabling phenomena like elec-
tric motors, transformers, and magnetic resonance imaging (MRI). Understand-
ing magnetic fields is crucial for electrical engineering and physics.

1.2 Sources of Magnetic Fields

Magnetic fields are produced by:

• Moving charges: A single moving charge (e.g., an electron) generates a

magnetic field.

• Electric currents: A steady current in a conductor, such as a wire, is a

common source of magnetic fields.

• Permanent magnets: Materials like iron have aligned magnetic moments,

creating a magnetic field.

• Time-varying electric fields: According to Maxwell’s equations, a chang-

ing electric field induces a magnetic field (not covered in this steady-state
chapter).

This chapter focuses on magnetic fields produced by steady (DC) currents in

conductors, as they are foundational to understanding electromagnetic devices.

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1.3 Forces Between Current-Carrying Conductors
Consider two parallel wires carrying currents I1 and I2 . Experimental observa-
tions show:

• If the currents are in the same direction, the wires attract each other.

• If the currents are in opposite directions, the wires repel each other.

• If the currents are perpendicular, there is no force between the wires.

These interactions are due to the magnetic field produced by one wire acting on
the current in the other wire.
The force on a current-carrying conductor in a magnetic field is given by:

F = I(

L ×B), (2) where:

• I is the current (in amperes, A).

• L is the length vector of the conductor (in meters, m), aligned with the cur-
rent direction.

• B is the magnetic flux density (in tesla, T) at the conductor’s location.

The cross product indicates that the force is perpendicular to both the current
direction and the magnetic field. Since B = µ0 H, the force can also be written as:

F = Iµ0 (L × H). (3)

Imagine two parallel wires separated by distance d, each carrying current I

in the same direction. The magnetic field produced by wire 1 at wire 2’s position
causes a force on wire 2, and vice versa. This interaction is the basis for defining
the ampere in the SI system.

2 Magnetic Field Geometry and the Right-Hand Rule

The magnetic field produced by a current has a specific geometry. Unlike electric
field lines, which radiate outward from a charge, magnetic field lines form closed
loops around a current-carrying conductor.

2.1 Direction of Magnetic Fields

The direction of the magnetic field H around a current is determined by the right-
hand rule:

• Point the thumb of your right hand in the direction of the current I.

• Your fingers curl in the direction of the magnetic field H.

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For a straight wire carrying current along the positive z-axis, the magnetic field
at a point in the x-y plane circles the wire in the azimuthal direction (along aϕ in
cylindrical coordinates).
For a wire along the z-axis with current in the +z direction, at point (x, y, 0),
the magnetic field points in the −ay direction if x > 0, and in the +ay direction if
x < 0, consistent with the right-hand rule.

2.2 Visualizing Magnetic Field Lines

Magnetic field lines:

• Form closed loops around the current.

• Are denser near the conductor, indicating stronger field strength.

• Never intersect, as the magnetic field has a unique direction at each point.

Wire
z

I H
x

Figure 1: Magnetic field lines around a straight wire carrying current I along the
z-axis.

Figure 1 shows the magnetic field lines circling a wire, with arrows indicating
the direction determined by the right-hand rule.

3 The Biot-Savart Law

The Biot-Savart law is the cornerstone for calculating the magnetic field pro-
duced by a current. It is analogous to Coulomb’s law for electric fields but ac-
counts for the directional nature of magnetic fields.

3.1 Definition and Derivation

The differential magnetic field intensity dH at a point due to a differential current
element IdL is given by:
IdL × aR
dH = , (4)
4πR2
where:

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 • IdL is the differential current element (in A·m), with dL being the vector
length element along the current’s direction.

• R is the vector from the current element to the field point, with magnitude
R.

• aR = R/R is the unit vector pointing from the current element to the field
point.

The units of H are A/m, consistent with the right-hand side of the equation.
The cross product dL × aR ensures that the magnetic field is perpendicular to
both the current element and the line connecting the element to the field point.
The 1/R2 term indicates an inverse-square dependence, similar to gravitational
and electric fields.
The Biot-Savart law is derived from experimental observations of magnetic
fields around currents, combined with the requirement that the field obeys Maxwell’s
equations for steady currents. The 4π in the denominator ensures consistency
with the SI unit system.

3.2 Application to Point and Distributed Currents

For a single differential current element, Equation 4 gives the contribution to
the magnetic field. For a finite current distribution, the total field is obtained by
integration: Z
IdL × aR
H= . (5)
4πR2
Currents can be distributed over surfaces or volumes:

• Surface current: Described by surface current density K (A/m). The dif-

ferential current element is IdL = KdS, where dS is a differential surface
area. The magnetic field is:
Z
KdS × aR
H= . (6)
S 4πR2

• Volume current: Described by volume current density J (A/m²). The dif-

ferential current element is IdL = JdV , where dV is a differential volume.
The magnetic field is: Z
JdV × aR
H= . (7)
V 4πR2

3.3 Comparison with Coulomb’s Law

The Biot-Savart law resembles Coulomb’s law for the electric field due to a point
charge:
dQaR
dE = , (8)
4πϵ0 R2
where dQ is a differential charge, and ϵ0 is the permittivity of free space (8.854 ×
10−12 F/m). Both laws feature:

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 • An inverse-square distance dependence (1/R2 ).

• A 4π factor in the denominator for spherical symmetry.

• A dependence on the source (charge for E, current for H).

Key differences include:

• The electric field is along aR , while the magnetic field is perpendicular to

both dL and aR due to the cross product.

• The electric field acts on stationary charges, while the magnetic field affects
moving charges or currents.

4 Magnetic Fields from Current Configurations

This section applies the Biot-Savart law to calculate magnetic fields for various
current configurations, including straight wires, current loops, and distributed
currents.

4.1 Line Currents

Consider a straight conductor of finite length along the z-axis from z = z1 to
z = z2 , carrying current I in the +z direction. We want to find the magnetic field
H at a point P (ρ, 0, 0) in cylindrical coordinates (i.e., at (x, y, z) = (ρ, 0, 0)).
The differential current element is:

dL = dzaz . (9)

The position vector from the current element at (0, 0, z) to the field point P (ρ, 0, 0)
is:
R = (ρ, 0, 0) − (0, 0, z) = ρaρ − zaz . (10)
The distance R is: p
R= ρ2 + z 2 . (11)
The unit vector is:
ρaρ − zaz
aR = p . (12)
ρ2 + z 2
Calculate:

dL × R = (dzaz ) × (ρaρ − zaz ) = ρdz(az × aρ ) + 0 = ρdzaϕ , (13)

since az × aρ = aϕ and az × az = 0. Thus:

I(ρdzaϕ ) Iρdz
dH = 2 2 3/2
= aϕ . (14)
4π(ρ + z ) 4π(ρ2 + z 2 )3/2

The total field is: Z z2

Iρdz
H= aϕ . (15)
z1 4π(ρ2 + z 2 )3/2

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To evaluate the integral, use the substitution:

z = ρ cot α, dz = −ρ csc2 αdα. (16)

Then:
ρ2 + z 2 = ρ2 (1 + cot2 α) = ρ2 csc2 α, (ρ2 + z 2 )3/2 = ρ3 csc3 α. (17)
The integrand becomes:

ρdz ρ(−ρ csc2 αdα) csc2 α sin α

= = − dα = − 2 dα. (18)
2 2
(ρ + z ) 3/2 ρ csc α
3 3 ρ csc α
2 3 ρ

Thus: Z   Z α1
I α1
sin α I
H= − dα aϕ = − sin αdαaϕ , (19)
4π α2 ρ 4πρ α2
where α1 and α2 correspond to z1 and z2 . The integral is:
Z
sin αdα = − cos α, [− cos α]αα12 = − cos α1 + cos α2 . (20)

Therefore:
I
H= (cos α2 − cos α1 )aϕ . (21)
4πρ
The angles α1 and α2 are defined as the angles between the line from P to the
ends of the conductor and the perpendicular to the conductor.
The azimuthal unit vector aϕ can be found using:

aϕ = aℓ × aρ , (22)

where aℓ is the unit vector along the current (here, az ), and aρ is the unit vector
from the current to the field point (here, aρ ). Thus:

aϕ = az × aρ . (23)

• Semi-infinite conductor: If the conductor extends from z = 0 to z = ∞,

the angles are α1 = 90◦ (at z = 0) and α2 = 0◦ (as z → ∞). Thus:

I I I
H= (cos 0◦ − cos 90◦ )aϕ = (1 − 0)aϕ = aϕ . (24)
4πρ 4πρ 4πρ

• Infinite conductor: If the conductor extends from z = −∞ to z = ∞, the

angles are α1 = 180◦ and α2 = 0◦ . Thus:

I I I
H= (cos 0◦ − cos 180◦ )aϕ = (1 − (−1))aϕ = aϕ . (25)
4πρ 4πρ 2πρ

Figure 1 illustrates the field circling the wire, with field strength decreasing
as 1/ρ.

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4.2 Circulating Currents (Current Loops)
Consider a circular loop of radius a in the x-y plane, centered at the origin, car-
rying current I in the azimuthal direction (aϕ ). We calculate the magnetic field
H at a point on the z-axis, (0, 0, z0 ).
The magnetic field is: Z
IdL × aR
H= . (26)
4πR2
The differential current element in cylindrical coordinates is:

dL = adϕaϕ , (27)

where ϕ is the azimuthal angle. A point on the loop is at (a cos ϕ, a sin ϕ, 0). The
vector from this point to the field point (0, 0, z0 ) is:

R = (0, 0, z0 ) − (a cos ϕ, a sin ϕ, 0) = −a cos ϕax − a sin ϕay + z0 az . (28)

In cylindrical coordinates, −a cos ϕax − a sin ϕay = −aaρ . Thus:

R = −aaρ + z0 az . (29)

The distance R is: q

R= a2 + z02 . (30)
The unit vector is:
−aaρ + z0 az
aR = p . (31)
a2 + z02
Calculate:
−aaρ + z0 az
dL × aR = (adϕaϕ ) × p . (32)
a2 + z02
Using aϕ × aρ = az and aϕ × az = −aρ :

adϕ adϕ
dL × aR = p [(−a)(a ϕ × aρ ) + z0 (aϕ × az )] = p (−aaz − z0 aρ ). (33)
a2 + z02 a2 + z02

Thus:
I adϕ Iadϕ
dH = 2
·p (−aaz − z 0 aρ ) = (−aaz − z0 aρ ). (34)
4π(a2 + z0 ) a2 + z02 4π(a2 + z02 )3/2

The radial unit vector aρ = cos ϕax + sin ϕay varies with ϕ. The radial compo-
nent integrates to zero:
Z 2π Z 2π
z0 cos ϕdϕ = 0, z0 sin ϕdϕ = 0. (35)
0 0

The z-component is:

Z 2π Z 2π
Ia(−a)dϕ Ia2 Ia2 Ia2
Hz = = − dϕ = − ·2π = − .
0 4π(a2 + z02 )3/2 4π(a2 + z02 )3/2 0 4π(a2 + z02 )3/2 2(a2 + z02 )3/2
(36)

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However, checking the direction via the right-hand rule, for current in the +aϕ
direction, the field on the positive z-axis points in the +az direction, so:
Ia2
H= az . (37)
2(a2 + z02 )3/2
The numerator Ia2 can be written as I(πa2 )/π, where πa2 is the loop’s area.
The magnetic moment of the loop is:
m = I(πa2 )az . (38)
Thus:
m
H= . (39)
2π(a2 + z02 )3/2
Figure 2 (inspired by Figure 7.8 in the PDF) shows the loop and its magnetic
field lines, resembling a magnetic dipole.
z
H
(0, 0, z0 )

Loop, radius a

Figure 2: Magnetic field due to a circular current loop.

4.3 Two- and Three-Dimensional Currents

A surface current is a current flowing in a thin sheet, described by surface cur-
rent density K (A/m). For a sheet of width b with uniform K, the total current
is:
I = Kb. (40)
The differential current element is IdL = KdS, and the magnetic field is:
Z
KdS × aR
H= . (41)
S 4πR2
A volume current is described by current density J (A/m²). The differential
current element is IdL = JdV , and the magnetic field is:
Z
JdV × aR
H= . (42)
V 4πR2
Figure 3 (inspired by Figure 7.4 in the PDF) illustrates:
• (a) Line current: A thin wire.
• (b) Surface current: A thin sheet with current flow.
• (c) Volume current: A three-dimensional region with current density.

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 J
(c) Volume
(a) Line (b) Surface
K
I

Figure 3: Types of current distributions.

5 Numerical Examples for Line Currents

Numerical examples reinforce theoretical concepts by applying formulas to prac-
tical scenarios.

5.1 Example 1: Magnetic Field due to a Straight Conductor

Calculate the magnetic field H at point P (5, 0, 0) due to a straight conductor along
the z-axis from z = −2 m to z = 3 m, carrying a current I = 10 A in the +z direc-
tion.
Use Equation 21:
I
H= (cos α2 − cos α1 )aϕ . (43)
4πρ
Here, ρ = 5 m, I = 10 A. Determine the angles:

• √
At z1 = −2, √
the vector from P (5, 0, 0) to (0, 0, −2) is (−5, 0, 2). The distance is
52 + 22 = 29. The angle α1 is between the z-axis and the line from P to
the point. The cosine is:
−2
cos α1 = √ ≈ −0.3714. (44)
29
√ √
• At z2 = 3, the vector to (0, 0, 3) is (−5, 0, −3), distance 52 + 3 2 = 34. The
cosine is:
3
cos α2 = √ ≈ 0.5145. (45)
34
Calculate:
cos α2 − cos α1 = 0.5145 − (−0.3714) = 0.8859. (46)
The field magnitude is:

I 10 10
= = ≈ 0.1592 A/m. (47)
4πρ 4π · 5 62.8319

Thus:
H = 0.1592 · 0.8859aϕ ≈ 0.1410aϕ A/m = 141.0 mA/m. (48)
Determine aϕ :
aℓ = az , aρ = ax , aϕ = az × ax = −ay . (49)
So:
H = −141.0ay mA/m. (50)

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5.2 Example 2: Magnetic Field due to a Segment of a Triangu-
lar Loop
A triangular loop in the x-y plane has a side along the x-axis from (0, 0, 0) to
(2, 0, 0), carrying a current of 10 A in the +x direction. Find H at (0, 0, 5).
Treat the side as a straight conductor. Use Equation 21. The field point is at
ρ = 5 m. The angles are:

• At (0, 0, 0), the vector to (0, 0, 5) is (0, 0, −5). The angle α1 = 90◦ , so cos α1 = 0.
√ √
• At (2, 0, 0), the vector is (−2, 0, −5), distance 22 + 52 = 29. The cosine is:
2
cos α2 = √ ≈ 0.3714. (51)
29

Calculate:
cos α2 − cos α1 = 0.3714 − 0 = 0.3714. (52)
The field magnitude is:

I 10
= ≈ 0.1592 A/m. (53)
4πρ 4π · 5

Thus:
H = 0.1592 · 0.3714aϕ ≈ 0.0591aϕ A/m = 59.1 mA/m. (54)
Determine aϕ :
aℓ = ax , aρ = az , aϕ = ax × az = −ay . (55)
So:
H = −59.1ay mA/m. (56)
Figure 7.6 in the PDF likely shows the triangular loop and a close-up of the
side, with angles α1 , α2 , and ρ labeled.

6 Magnetic Field from a Circular Current Loop

The magnetic field of a circular loop is a key application, often used in solenoids
and electromagnets. The derivation was provided earlier, resulting in:

Ia2
H= az . (57)
2(a2 + z02 )3/2

6.1 Numerical Example

A circular loop lies in the plane x2 +y 2 = 9, z = 0, carrying 10 A in the aϕ direction.
Find H at (0, 0, 4) and√(0, 0, −4).
The radius is a = 9 = 3 m, I = 10 A. Use:

Ia2
H= az . (58)
2(a2 + z02 )3/2

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 • At (0, 0, 4), z0 = 4:

a2 + z02 = 9 + 16 = 25, (a2 + z02 )3/2 = 253/2 = 125. (59)

10 · 9 90
H= az = az = 0.36az A/m. (60)
2 · 125 250
• At (0, 0, −4), z0 = −4, so a2 + z02 = 25. The magnitude is the same, and the
field direction remains az (depends on current direction):

H = 0.36az A/m. (61)

7 Simulation and Visualization Tools

Simulations visualize magnetic fields, aiding understanding and design. Tools
like Finite Element Method Magnetics (FEMM) solve 2D magnetic problems, show-
ing field lines and magnitudes.
FEMM can model the field around a straight wire or a current loop, validating
analytical results like those derived above.

8 Introduction to Ampere’s Circuital Law

Ampere’s circuital law relates the magnetic field to the current it encloses:
I
H · dl = Ienc , (62)

where Ienc is the total current enclosed by a closed path. This law is useful for
symmetric current distributions (e.g., infinite wires, solenoids).
Ampere’s law simplifies magnetic field calculations in cases with high sym-
metry, complementing the Biot-Savart law.

9 Conclusion
This chapter has provided a comprehensive exploration of steady magnetic fields,
from the basic concept of magnetic forces to advanced calculations for various
current configurations. By mastering the Biot-Savart law, right-hand rule, and
numerical applications, you are equipped to analyze magnetic fields in engineer-
ing and physics contexts.

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