Static Magnetic Fields (Cheng Ch 6.1 to 6.4) — Complete Master Study Note

Master Note Overview & Exam Quick Reference

This standalone study guide covers Sections 6-1 to 6-4 of David K. Cheng’s Field and Wave Electromagnetics. Designed for rapid mastery and A+ performance, it integrates core concepts, physical intuition, step-by-step derivations, textbook examples, and 100% of all past-year questions (PYQs) from 2015 to 2025 mapped to their exact year(s) of appearance.

🗝️ Core Formula Quick Reference

Physical Quantity	Governing Equation (SI Units)	Cheng Reference / PYQ Tag
Lorentz Force Equation	$\vec{F}=q(\vec{E}+\vec{u}\times \vec{B})[\text{N}]$	Sec 6-1 [PYQ: 2018, 2021]
Magnetic Force on Current Element	$d{\vec{F}}_m=Id\vec{l}\times \vec{B}[\text{N}]$	Sec 6-1
Solenoidal Postulate (Divergence)	$\nabla \cdot \vec{B}=0⟺{\oint }_S\vec{B}\cdot d\vec{s}=0$	Sec 6-2 [PYQ: 2015, 2017, 2020, 2025]
Curl Postulate & Ampere’s Law	$\nabla \times \vec{B}={\mu }_0\vec{J}⟺{\oint }_C\vec{B}\cdot d\vec{l}={\mu }_{0}I$	Sec 6-2 [PYQ: 2015, 2016, 2017, 2018, 2021, 2023, 2025]
Vector Magnetic Potential ($\vec{A}$)	$\vec{B}=\nabla \times \vec{A}[\text{Wb/m}\text{ or }\text{T}\cdot \text{m}]$	Sec 6-3 [PYQ: 2019, 2022, 2024, 2025]
Vector Poisson’s Equation	${\nabla }^2\vec{A}=-{\mu }_0\vec{J}(\text{Coulomb Gauge: }\nabla \cdot \vec{A}=0)$	Sec 6-3 [PYQ: 2020]
Integral Formula for $\vec{A}$	$\vec{A}=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }}{R}[\text{Wb/m}]$	Sec 6-3 [PYQ: 2020, 2024]
Magnetic Flux & Vector Potential	$\Phi ={\oint }_C\vec{A}\cdot d\vec{l}[\text{Wb}]$	Sec 6-3 [PYQ: 2019]
Biot-Savart Law	$d\vec{B}=\frac{{\mu }_{0}I}{4\pi R^2}(d{\vec{l}}^{\prime }\times {\hat{a}}_R)=\frac{{\mu }_{0}I}{4\pi R^3}(d{\vec{l}}^{\prime }\times \vec{R})$	Sec 6-4 [PYQ: 2016, 2017, 2021, 2024]
Finite Straight Wire ($2L$)	$\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}IL}{2\pi r\sqrt{L^2+r^2}}\stackrel{L\gg r}{\to }{\hat{a}}_{\varphi }\frac{{\mu }_{0}I}{2\pi r}$	Sec 6-4 [PYQ: 2015, 2016, 2021, 2023, 2025]
Center of Square Loop ($w$)	$B=\frac{2\sqrt{2}{\mu }_{0}I}{\pi w}$	Sec 6-4 [PYQ: 2017]
Axis of Circular Loop ($b$)	$\vec{B}={\hat{a}}_z\frac{{\mu }_{0}I{b}^2}{2(b^2+z^2{)}^{3/2}}\stackrel{z=0}{\to }{\hat{a}}_z\frac{{\mu }_{0}I}{2b}$	Sec 6-4 [PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025]

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0. Physical Foundations & Prerequisites

Conceptual Bridge: Electrostatics vs. Magnetostatics

• Electrostatics (Chapter 3): Concerned with electric charges at rest ($\vec{u}=0$). The fundamental field vector is Electric Field Intensity $\vec{E}$ ($\text{V/m}$), generated by charge density ${\rho }_v$.
• Magnetostatics (Chapter 6): Concerned with electric charges moving at a steady velocity $\vec{u}$ (DC currents, $\partial {\rho }_v/\partial t=0$, $\nabla \cdot \vec{J}=0$). The fundamental field vector in free space is Magnetic Flux Density $\vec{B}$ ($\text{Tesla}$ or ${\text{Wb/m}}^2$).
• Electromagnetic Waves (Chapter 7+): Generated when charges accelerate or decelerate ($\partial \vec{J}/\partial t\ne 0$).

                      [ ELECTRIC CHARGES ]
                               |
       +-----------------------+-----------------------+
       |                                               |
[ Charges at Rest ]                           [ Steady DC Motion ]
  (u = 0, ∂ρ/∂t = 0)                          (u = const, ∇·J = 0)
       |                                               |
  Electrostatic Field                             Magnetostatic Field
   Electric Field E                                Magnetic Flux Density B
  (∇·E = ρ/ε₀, ∇×E = 0)                         (∇·B = 0, ∇×B = μ₀J)

A. The Magnetic Flux Density Vector ($\vec{B}$)

• SI Unit: Tesla ($\text{T}$) or Weber per square meter (${\text{Wb/m}}^2$). Note that $1\text{ Weber}=1\text{ Volt-second}⟹1\text{ T}=1{\text{ Wb/m}}^2=1\text{ V}\cdot {\text{s/m}}^2$.
• CGS Unit: Gauss ($\text{G}$). Conversion: $1\text{ Tesla}=10^4\text{ Gauss}$. (Earth’s magnetic field $\approx 0.5\text{ G}=0.5\times 10^{-4}\text{ T}$).
• Permeability of Free Space (${\mu }_0$): Universal constant defined as: ${\mu }_0=4\pi \times 10^{-7}[\text{H/m}\text{ or }{\text{N/A}}^2]$

B. Lorentz Force Equation [PYQ: 2018, 2021]

When a test charge $q$ moves with velocity $\vec{u}$ in a region containing both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, it experiences a total electromagnetic force $\vec{F}$ given by:

$\vec{F}={\vec{F}}_e+{\vec{F}}_m=q\vec{E}+q(\vec{u}\times \vec{B})=q(\vec{E}+\vec{u}\times \vec{B})[\text{N}]\text{[PYQ: 2018, 2021]}$

Key Physical Properties of Magnetic Force (${\vec{F}}_m=q(\vec{u}\times \vec{B})$):

1. Magnitude: $|{\vec{F}}_m|=quB\sin \theta$, where $\theta$ is the angle between velocity $\vec{u}$ and magnetic flux density $\vec{B}$.
2. Direction: Perpendicular to both velocity $\vec{u}$ and magnetic field $\vec{B}$ according to the right-hand rule.
3. Zero Work Theorem: The magnetic force does zero work on a free moving charge: $dW={\vec{F}}_m\cdot d\vec{l}={\vec{F}}_m\cdot (\vec{u}dt)=q(\vec{u}\times \vec{B})\cdot \vec{u}dt=0$ Physical Consequence: A magnetic field can alter the direction of motion of a charged particle, but it cannot change its speed or kinetic energy.

Magnetic Force on a Differential Current Element:

Consider $N$ charge carriers per unit volume moving with velocity $\vec{u}$ along a thin wire of cross-sectional area $S$. The differential force on volume $d{v}^{\prime }=Sd{l}^{\prime }$ containing $NSd{l}^{\prime }$ carriers is: $d{\vec{F}}_m=(NSd{l}^{\prime })q(\vec{u}\times \vec{B})=(NqSu)d\vec{l}\times \vec{B}$ Since steady current $I=NqSu$, we obtain the fundamental current-element force equation:

$d{\vec{F}}_m=Id\vec{l}\times \vec{B}[\text{N}]$

What to Memorize

1. Lorentz Force Equation: $\vec{F}=q(\vec{E}+\vec{u}\times \vec{B})$.
2. Current Element Force: $d{\vec{F}}_m=Id\vec{l}\times \vec{B}$.
3. Permeability: ${\mu }_0=4\pi \times 10^{-7}\text{ H/m}$.
4. Unit Conversion: $1\text{ T}=1{\text{ Wb/m}}^2=10^4\text{ Gauss}$.

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1. Section 6-2: Fundamental Postulates of Magnetostatics in Free Space

Magnetostatics in free space is completely defined by two fundamental differential postulates specifying the divergence and the curl of $\vec{B}$.

A. The Two Fundamental Postulates

Postulate	Differential (Point) Form	Integral Form	Physical Name / Principle
Divergence Postulate	$\nabla \cdot \vec{B}=0$	${\oint }_S\vec{B}\cdot d\vec{s}=0$	Law of Conservation of Magnetic Flux / No Isolated Monopoles [PYQ: 2020, 2025]
Curl Postulate	$\nabla \times \vec{B}={\mu }_0\vec{J}$	${\oint }_C\vec{B}\cdot d\vec{l}={\mu }_{0}I$	Ampere’s Circuital Law [PYQ: 2015, 2016, 2017, 2018, 2021, 2023, 2025]

Where:

• $\vec{B}$ = Magnetic flux density ($\text{T}$)
• $\vec{J}$ = Volume current density (${\text{A/m}}^2$)
• ${\mu }_0=4\pi \times 10^{-7}\text{ H/m}$ = Permeability of free space
• $I={\int }_S\vec{J}\cdot d\vec{s}$ = Total free current passing through open surface $S$ enclosed by contour $C$.

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B. Physical Significance & Proofs

1. “Magnetic Poles Cannot Be Isolated” / Solenoidal Nature [PYQ: 2020, 2025]

Verbatim PYQ [PYQ: 2025 (08 Marks)]

“Magnetic poles cannot be isolated” — justify the statement.

Proof & Derivation:

1. Start with the differential divergence postulate: $\nabla \cdot \vec{B}=0$
2. Take the volume integral over an arbitrary volume $V$ bounded by closed surface $S$: ${\int }_V(\nabla \cdot \vec{B})dv=0$
3. Apply the Divergence Theorem (${\int }_V(\nabla \cdot \vec{A})dv={\oint }_S\vec{A}\cdot d\vec{s}$): ${\oint }_S\vec{B}\cdot d\vec{s}=0\text{[PYQ: 2020, 2025]}$
4. Physical Conclusion:
   • The total outward magnetic flux through any closed surface is identically zero.
   • There are no magnetic flow sources or sinks (no isolated magnetic charges/monopoles exist in nature).
   • Magnetic flux lines always form continuous closed loops upon themselves. If a permanent bar magnet is cut in half, each piece immediately forms a complete new pair of North and South poles.

                  GAUSS'S LAW FOR MAGNETISM: ∮ B · ds = 0

          +-------------------------------------------------+
          |                                                 |
          |       /-------\                 /-------\       |
          |      /    N    \--------------->/    S    \      |
          +----->|  North  |   Magnetic    |  South  |<-----+
                 |  Pole   |  Flux Lines   |  Pole   |
                 \    |    /--------------->\    |    /
                  \---+---/                 \---+---/
                      |                         ^
                      |   Continuous Closed     |
                      +-------------------------+

2. Consistency with Continuity Equation for Steady Currents

Taking the divergence of the curl postulate ($\nabla \times \vec{B}={\mu }_0\vec{J}$): $\nabla \cdot (\nabla \times \vec{B})={\mu }_0(\nabla \cdot \vec{J})$ Since the divergence of the curl of any vector field is identically zero ($\nabla \cdot (\nabla \times \vec{A})\equiv 0$): ${\mu }_0(\nabla \cdot \vec{J})=0⟹\nabla \cdot \vec{J}=0$ This is completely consistent with the continuity equation for steady DC currents ($\nabla \cdot \vec{J}=-\partial {\rho }_v/\partial t=0$).

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C. Ampere’s Circuital Law [PYQ: 2015, 2016, 2017, 2018, 2021, 2023, 2025]

Applying Stokes’s Theorem (${\int }_S(\nabla \times \vec{A})\cdot d\vec{s}={\oint }_C\vec{A}\cdot d\vec{l}$) to the curl postulate: ${\int }_S(\nabla \times \vec{B})\cdot d\vec{s}={\mu }_0{\int }_S\vec{J}\cdot d\vec{s}⟹{\oint }_C\vec{B}\cdot d\vec{l}={\mu }_{0}I[\text{PYQ: 2015, 2016, 2017, 2018, 2021, 2023, 2025}]$

Operational Rule

The circulation of magnetic flux density $\vec{B}$ around any closed contour $C$ equals ${\mu }_0$ times the total enclosed current $I$. The direction of contour $C$ and positive current $I$ follow the right-hand rule (thumb points along current, fingers curl along contour $C$).

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D. Applications & Textbook Examples from Section 6-2

1. Cheng Example 6-1: Infinitely Long Circular Conductor [PYQ: 2017, 2018, 2022]

Verbatim PYQ [PYQ: 2017 (11m), 2018 (11m), 2022 (10m)]

An infinitely long, straight conductor with a circular cross section of radius $b$ carries a steady current $I$. Determine the magnetic flux density $B$ both inside and outside the conductor.

                CIRCULAR CONDUCTOR CROSS-SECTION (Radius b)

                          Y ^
                            |   . - ~ - .
                            |  /    |                                | /   r |                                 ||   .--+      |  <-- Conductor Surface (r = b)
                            ||  /   |      |
                    --------+|--+---o------+---------> X
                            ||  \ Amperian |
                            ||   '--+ Loop |
                            | \     | r<b /
                            |  \    |    /
                            |   ' - ~ - '
                            |

Step 1: Symmetry Analysis

By cylindrical symmetry around the $z$-axis (aligned with the conductor), $\vec{B}$ is purely $\varphi$-directed ($\vec{B}={\hat{a}}_{\varphi }{B}_{\varphi }$) and depends only on radial distance $r$.

Step 2: Inside the Conductor ($r\le b$)

Construct a circular Amperian contour $C_1$ of radius $r<b$ in the $xy$-plane. Assuming uniform current density $J=\frac{I}{\pi b^2}$, the enclosed current is: $I_{\text{enc}}=J\cdot (\pi r^2)=I\frac{\pi r^2}{\pi b^2}=I\frac{r^2}{b^2}$

Applying Ampere’s Circuital Law: ${\oint }_{C_1}\vec{B}\cdot d\vec{l}=B_{\varphi }(2\pi r)={\mu }_0{I}_{\text{enc}}={\mu }_{0}I\frac{r^2}{b^2}$ $B_{\varphi }=\frac{{\mu }_{0}Ir}{2\pi b^2}⟹{\vec{B}}_{\text{in}}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{2\pi b^2}(r\le b)\text{[Linear growth with }r\text{]}$

Step 3: Outside the Conductor ($r\ge b$)

Construct a circular Amperian contour $C_2$ of radius $r>b$. The contour encloses the entire conductor current $I_{\text{enc}}=I$.

Applying Ampere’s Circuital Law: ${\oint }_{C_2}\vec{B}\cdot d\vec{l}=B_{\varphi }(2\pi r)={\mu }_{0}I$ $B_{\varphi }=\frac{{\mu }_{0}I}{2\pi r}⟹{\vec{B}}_{\text{out}}={\hat{a}}_{\varphi }\frac{{\mu }_{0}I}{2\pi r}(r\ge b)\text{[Inverse decay }1/r\text{]}$

   Magnetic Flux Density B_φ
         ^
         |
  B_max  |-----------o  <-- Peak Field at Surface r = b : B_max = μ₀I / (2πb)
         |          /          |         /            |        /     ` .
         |       /          ` . <-- Inverse Decay (~ 1/r)
         |      / (Linear)      ` .
       --+-----o--------------------> Radial Distance r
               0         b

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2. Cheng Example 6-2: Closely Wound Air-Core Toroidal Coil

Textbook Problem

Determine the magnetic flux density inside a closely wound toroidal coil with an air core having $N$ turns and carrying current $I$. The toroid has mean radius $b$ and cross-sectional radius $a$.

                            TOROIDAL COIL GEOMETRY
                                  
                                 .-'""'-.
                               .'   ||   '.
                              /  .--||--.                               |  /   ||   \  |   <-- Mean Radius b
                             | |    o-----+--|-----> Center Axis
                             |  \   ||   /  |   <-- Core Radius a
                              \  '--||--'  /
                               '.   ||   .'
                                 '-....-'

Derivation:

1. Cylindrical symmetry ensures $\vec{B}={\hat{a}}_{\varphi }{B}_{\varphi }$.
2. Construct a circular contour $C$ of radius $r$ centered at the origin.
3. Inside the Core ($(b-a)<r<(b+a)$): The contour links with $N$ turns, each carrying current $I⟹I_{\text{enc}}=NI$. ${\oint }_C\vec{B}\cdot d\vec{l}=B_{\varphi }(2\pi r)={\mu }_{0}NI⟹\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}NI}{2\pi r}(b-a<r<b+a)$
4. Outside the Core ($r<b-a$ or $r>b+a$): The net enclosed current is zero ($I_{\text{enc}}=0$) because equal currents flow in opposite directions across the contour boundary $⟹\vec{B}=0$.

---

3. Cheng Example 6-3: Infinitely Long Solenoid

Textbook Problem

Determine the magnetic flux density inside an infinitely long solenoid with an air core having $n$ turns per unit length carrying current $I$.

                         INFINITE SOLENOID CONTOUR

                     +-----------------------------------+
       Outside (B=0) | 1                               4 |
                     + - - - - - - - - - - - - - - - - - +  <-- Contour C (Length L)
                     | 2 (In)                        3   |
   ==================|===================================|=====================> z-axis
                     |   Inside Solenoid: B = a_z μ₀nI   |
   ===========================================================================

Derivation:

1. By symmetry, for an infinite solenoid, the internal magnetic field is strictly axial ($\vec{B}={\hat{a}}_z{B}_z$) and uniform, while the external field is zero (${\vec{B}}_{\text{out}}=0$).
2. Construct a rectangular contour $C$ of length $L$ with side 2-3 inside the solenoid and side 4-1 outside.
3. Evaluating ${\oint }_C\vec{B}\cdot d\vec{l}$:
   • Side 2-3 (inside): ${\int }_2^3\vec{B}\cdot d\vec{l}=B_{z}L$.
   • Sides 3-4 and 1-2 (perpendicular to field): $\int \vec{B}\cdot d\vec{l}=0$.
   • Side 4-1 (outside, $B=0$): ${\int }_4^1\vec{B}\cdot d\vec{l}=0$.
4. Total enclosed current in length $L$: $I_{\text{enc}}=(nL)I$.
5. Applying Ampere’s Law: $B_{z}L={\mu }_{0}nLI⟹\vec{B}={\hat{a}}_z{\mu }_{0}nI[\text{T}]$

---

2. Section 6-3: Vector Magnetic Potential ($\vec{A}$)

Because $\vec{B}$ is divergence-free ($\nabla \cdot \vec{B}=0$), it is solenoidal. By vector identity II ($\nabla \cdot (\nabla \times \vec{A})\equiv 0$), $\vec{B}$ can always be expressed as the curl of an intermediary vector field $\vec{A}$:

$\vec{B}=\nabla \times \vec{A}[\text{SI Unit: Weber/meter }(\text{Wb/m})\text{ or }\text{Tesla-meter }(\text{T}\cdot \text{m})][\text{PYQ: 2019, 2022, 2024, 2025}]$

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A. Coulomb Gauge & Vector Poisson’s Equation [PYQ: 2020]

To uniquely specify a vector field, both its curl and divergence must be defined (Helmholtz’s Theorem). Substitute $\vec{B}=\nabla \times \vec{A}$ into Ampere’s point postulate ($\nabla \times \vec{B}={\mu }_0\vec{J}$):

$\nabla \times (\nabla \times \vec{A})={\mu }_0\vec{J}$

Using the vector expansion identity $\nabla \times (\nabla \times \vec{A})=\nabla (\nabla \cdot \vec{A})-{\nabla }^2\vec{A}$:

$\nabla (\nabla \cdot \vec{A})-{\nabla }^2\vec{A}={\mu }_0\vec{J}$

The Coulomb Gauge Choice:

To simplify the equation for static fields, we choose the divergence of $\vec{A}$ to be zero: $\nabla \cdot \vec{A}=0\text{(Coulomb Gauge Condition for Static Fields)}$

This reduces the relationship directly to Vector Poisson’s Equation [PYQ: 2020]:

${\nabla }^2\vec{A}=-{\mu }_0\vec{J}[\text{PYQ: 2020}]$

In Cartesian coordinates, this vector equation breaks down into three independent scalar Poisson’s equations: ${\nabla }^2{A}_x=-{\mu }_0{J}_x,{\nabla }^2{A}_y=-{\mu }_0{J}_y,{\nabla }^2{A}_z=-{\mu }_0{J}_z$

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B. General Integral Formula for $\vec{A}$ [PYQ: 2020, 2024]

Since each scalar Cartesian component equation (${\nabla }^2{A}_x=-{\mu }_0{J}_x$) is mathematically identical to electrostatics Poisson’s equation (${\nabla }^{2}V=-{\rho }_v/{ϵ}_0$), which has particular solution $V=\frac{1}{4\pi {ϵ}_0}{\int }_{V^{\prime }}\frac{{\rho }_v}{R}d{v}^{\prime }$, the general volume integral solution for $\vec{A}$ is:

$\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{V^{\prime }}\frac{\vec{J}}{R}d{v}^{\prime }[\text{Wb/m}][\text{PYQ: 2020, 2024}]$

For a thin filamentary current wire carrying current $I$ ($\vec{J}d{v}^{\prime }=Id{\vec{l}}^{\prime }$):

$\vec{A}=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }}{R}[\text{Wb/m}]$

---

C. Physical Significance & Magnetic Flux Proof [PYQ: 2019]

Verbatim PYQ [PYQ: 2019 (06 Marks)]

Prove mathematically that the line integral of any vector magnetic potential around any closed path equals the total magnetic flux passing through the area enclosed by the path.

Proof & Derivation:

1. The total magnetic flux $\Phi$ passing through an open surface $S$ bounded by contour $C$ is: $\Phi ={\int }_S\vec{B}\cdot d\vec{s}$
2. Substitute $\vec{B}=\nabla \times \vec{A}$: $\Phi ={\int }_S(\nabla \times \vec{A})\cdot d\vec{s}$
3. Apply Stokes’s Theorem (${\int }_S(\nabla \times \vec{A})\cdot d\vec{s}={\oint }_C\vec{A}\cdot d\vec{l}$):

$\Phi ={\oint }_C\vec{A}\cdot d\vec{l}[\text{Weber},\text{Wb}][\text{PYQ: 2019}]$

Physical Meaning: Vector magnetic potential $\vec{A}$ is the circulation density of magnetic flux. Its closed line integral around any path directly gives the linked magnetic flux $\Phi$.

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D. Vector Magnetic Potential ($\vec{A}$) vs. Scalar Magnetic Potential ($V_m$) [PYQ: 2015, 2019]

Verbatim PYQ [PYQ: 2015 (05m), 2019 (05m)]

Distinguish between vector magnetic potential ($\vec{A}$) and scalar magnetic potential ($V_m$).

Property / Feature	Vector Magnetic Potential ($\vec{A}$)	Scalar Magnetic Potential ($V_m$)
Fundamental Definition	Defined via solenoidal property: $\vec{B}=\nabla \times \vec{A}$.	Defined via irrotational property in current-free space: $\vec{H}=-\nabla V_m$.
Governing Equation	Vector Poisson’s Eq: ${\nabla }^2\vec{A}=-{\mu }_0\vec{J}$.	Laplace’s Eq: ${\nabla }^2{V}_m=0$ (in region with $\vec{J}=0$).
Region of Validity	Everywhere in space (both current-carrying $\vec{J}\ne 0$ and current-free $\vec{J}=0$).	Restricted strictly to current-free regions ($\vec{J}=0$).
Single-Valuedness	Always single-valued and continuous across space.	Multi-valued if integration path encircles a current.
Physical Link	Line integral gives magnetic flux: $\Phi ={\oint }_C\vec{A}\cdot d\vec{l}$.	Potential difference gives MMF: $V_{m2}-V_{m1}=-{\int }_{P1}^{P2}\vec{H}\cdot d\vec{l}$.
SI Unit	Weber per meter ($\text{Wb/m}$) or Tesla-meter ($\text{T}\cdot \text{m}$).	Ampere ($\text{A}$).

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E. Vector Potential $\vec{A}$ Inside and Outside a Circular Conductor [PYQ: 2020]

Verbatim PYQ [PYQ: 2020 (13 Marks)]

An infinitely long, straight conductor with a circular cross section of radius $b$ carries a steady current $I$. Determine the vector magnetic potential $\vec{A}$ both inside and outside the conductor.

Derivation:

1. Align the conductor with the $z$-axis. Current flows along $+{\hat{a}}_z⟹\vec{A}={\hat{a}}_z{A}_z(r)$.
2. We know $\vec{B}=\nabla \times \vec{A}=-{\hat{a}}_{\varphi }\frac{\partial A_z}{\partial r}={\hat{a}}_{\varphi }{B}_{\varphi }$.

Outside the Conductor ($r\ge b$):

$-\frac{\partial A_z}{\partial r}=B_{\varphi }=\frac{{\mu }_{0}I}{2\pi r}⟹A_z(r)=-\frac{{\mu }_{0}I}{2\pi }\ln r+C_1(r\ge b)$

Inside the Conductor ($r\le b$):

$-\frac{\partial A_z}{\partial r}=B_{\varphi }=\frac{{\mu }_{0}Ir}{2\pi b^2}⟹A_z(r)=-\frac{{\mu }_{0}I{r}^2}{4\pi b^2}+C_2(r\le b)$

Boundary Continuity at $r=b$:

$-\frac{{\mu }_{0}I{b}^2}{4\pi b^2}+C_2=-\frac{{\mu }_{0}I}{2\pi }\ln b+C_1⟹C_2=C_1-\frac{{\mu }_{0}I}{2\pi }\ln b+\frac{{\mu }_{0}I}{4\pi }$

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3. Section 6-4: Biot-Savart Law and Applications

A. Derivation of Biot-Savart Law from Vector Potential [PYQ: 2021, 2024]

Verbatim PYQ [PYQ: 2021 (09m), 2024 (12m)]

Derive the equation for Biot-Savart Law starting from vector magnetic potential. Point out its applications in magnetostatics.

                   SOURCE ELEMENT AND OBSERVATION POINT GEOMETRY

                           Field Point P(x,y,z)
                                 .
                                /|
                               / |
                   Vector R   /  |
                   (m)       /   |
                            /    |
                           /     |
                          v      |
     Source Element I dl' o------+---------------------> Line Contour C'
     at P'(x',y',z')

Step-by-Step Derivation:

1. Start with the vector potential integral formula for a filamentary circuit: $\vec{A}=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }}{R}$
2. Take the curl of both sides with respect to field coordinates $(x,y,z)$: $\vec{B}=\nabla \times \vec{A}=\nabla \times \left[\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }}{R}\right]=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\nabla \times \left(\frac{d{\vec{l}}^{\prime }}{R}\right)$
3. Apply the vector identity $\nabla \times (f\vec{G})=f(\nabla \times \vec{G})+(\nabla f)\times \vec{G}$ (where $f=\frac{1}{R}$ and $\vec{G}=d{\vec{l}}^{\prime }$): $\nabla \times \left(\frac{d{\vec{l}}^{\prime }}{R}\right)=\frac{1}{R}(\nabla \times d{\vec{l}}^{\prime })+\left[\nabla \left(\frac{1}{R}\right)\right]\times d{\vec{l}}^{\prime }$
4. Since $d{\vec{l}}^{\prime }$ depends strictly on primed source coordinates $(x^{\prime },y^{\prime },z^{\prime })$, the unprimed spatial derivatives operate as zero: $\nabla \times d{\vec{l}}^{\prime }=0$.
5. Evaluate $\nabla \left(\frac{1}{R}\right)$: $\nabla \left(\frac{1}{R}\right)=-\frac{{\hat{a}}_R}{R^2}=-\frac{\vec{R}}{R^3}(\text{where }{\hat{a}}_R\text{ points from source to field point})$
6. Substitute back: $\nabla \times \left(\frac{d{\vec{l}}^{\prime }}{R}\right)=\left(-\frac{{\hat{a}}_R}{R^2}\right)\times d{\vec{l}}^{\prime }=\frac{d{\vec{l}}^{\prime }\times {\hat{a}}_R}{R^2}$
7. Substitute this integrand into the flux density equation to obtain Biot-Savart Law:

$\vec{B}=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }\times {\hat{a}}_R}{R^2}=\frac{{\mu }_{0}I}{4\pi }{\oint }_{C^{\prime }}\frac{d{\vec{l}}^{\prime }\times \vec{R}}{R^3}[\text{T}][\text{PYQ: 2016, 2017, 2021, 2024}]$

Differential form due to current element $Id{\vec{l}}^{\prime }$:

$d\vec{B}=\frac{{\mu }_{0}I}{4\pi R^2}(d{\vec{l}}^{\prime }\times {\hat{a}}_R)=\frac{{\mu }_{0}I}{4\pi R^3}(d{\vec{l}}^{\prime }\times \vec{R})[\text{T}]$

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B. Comparison: Biot-Savart Law vs. Ampere’s Circuital Law [PYQ: 2015, 2018, 2023, 2025]

Verbatim PYQ [PYQ: 2015, 2018, 2023, 2025 (10 Marks)]

Compare the usefulness and importance of Biot-Savart law against Ampere’s circuital law in determining magnetic flux density.

Parameter	Ampere’s Circuital Law ($\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$)	Biot-Savart Law ($d\vec{B}=\frac{{\mu }_{0}I}{4\pi R^2}d{\vec{l}}^{\prime }\times {\hat{a}}_R$)
Mathematical Nature	Integral relation linking circulation along a closed path to enclosed current.	Differential line-element summation integrated over source path.
Symmetry Requirement	Requires high geometric symmetry (cylindrical, planar, toroidal) to factor $B$ out of integral.	No symmetry required; universally applicable to any arbitrary wire geometry.
Computational Effort	Extremely fast, simple 1-line solution when symmetry exists.	Requires complex vector integration over source coordinates.
Electrostatic Analog	Analogous to Gauss’s Law ($\oint \vec{D}\cdot d\vec{s}=Q$).	Analogous to Coulomb’s Law ($d\vec{E}=\frac{dq}{4\pi {ϵ}_0{R}^2}{\hat{a}}_R$).
Primary Use Case	Infinite conductors, coaxial lines, solenoids, toroids.	Finite wires, square/polygonal loops, circular loops on axis.

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C. Applications & Textbook Examples from Section 6-4

1. Cheng Example 6-4: Straight Wire of Length $2L$ [PYQ: 2015, 2016, 2021, 2023, 2025]

Verbatim PYQ [PYQ: 2015 (15m), 2016 (18m), 2021 (15m), 2023 (10m), 2025 (12m)]

A direct current $I$ flows in a straight wire of length $2L$. Find the magnetic flux density $\vec{B}$ at a point located at a distance $r$ from the wire in the bisecting plane: (a) By determining the vector magnetic potential $\vec{A}$ first [PYQ: 2016, 2021, 2025], and (b) By applying Biot-Savart law [PYQ: 2015, 2016, 2021, 2023].

                    STRAIGHT WIRE OF LENGTH 2L IN BISECTING PLANE

                                  Z ^
                                    | +L
                                    |
                        Source dz' o|
                                    | \  Vector R = a_r r - a_z z'
                                    |  \ R = √(r² + z'²)
                                    |                            -----------+----+------------------> Field Point P(r, 0, 0)
                                    |   /  Distance r
                                    |  /
                                    | /
                                    | -L

Method (a): Via Vector Magnetic Potential $\vec{A}$ First [PYQ: 2016, 2021, 2025]

1. Align wire along $z$-axis from $z^{\prime }=-L$ to $z^{\prime }=+L$. Field point $P(r,0,0)$.
2. Source element $d{\vec{l}}^{\prime }={\hat{a}}_{z}d{z}^{\prime }$, distance $R=\sqrt{r^2+z^{\prime 2}}$.
3. Vector potential $\vec{A}={\hat{a}}_z{A}_z$: $A_z=\frac{{\mu }_{0}I}{4\pi }{\int }_{-L}^L\frac{d{z}^{\prime }}{\sqrt{r^2+z^{\prime 2}}}=\frac{{\mu }_{0}I}{4\pi }{\left[\ln \left(z^{\prime }+\sqrt{r^2+z^{\prime 2}}\right)\right]}_{-L}^L$ $A_z=\frac{{\mu }_{0}I}{4\pi }\ln \left(\frac{\sqrt{L^2+r^2}+L}{\sqrt{L^2+r^2}-L}\right)=\frac{{\mu }_{0}I}{2\pi }\ln \left(\frac{\sqrt{L^2+r^2}+L}{r}\right)$
4. Compute $\vec{B}=\nabla \times \vec{A}=-{\hat{a}}_{\varphi }\frac{\partial A_z}{\partial r}$: $\frac{\partial A_z}{\partial r}=\frac{{\mu }_{0}I}{2\pi }\cdot \frac{r}{\sqrt{L^2+r^2}+L}\cdot \left[\frac{\frac{r}{\sqrt{L^2+r^2}}\cdot r-(\sqrt{L^2+r^2}+L)}{r^2}\right]=-\frac{{\mu }_{0}IL}{2\pi r\sqrt{L^2+r^2}}$

$\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}IL}{2\pi r\sqrt{L^2+r^2}}[\text{T}][\text{PYQ: 2016, 2021, 2025}]$

Method (b): Via Biot-Savart Law Directly [PYQ: 2015, 2016, 2021, 2023]

1. Source element $d{\vec{l}}^{\prime }={\hat{a}}_{z}d{z}^{\prime }$, distance vector $\vec{R}={\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime }$.
2. Cross product: $d{\vec{l}}^{\prime }\times \vec{R}=({\hat{a}}_{z}d{z}^{\prime })\times ({\hat{a}}_{r}r-{\hat{a}}_z{z}^{\prime })={\hat{a}}_{\varphi }rd{z}^{\prime }$.
3. Apply Biot-Savart Law: $\vec{B}=\frac{{\mu }_{0}I}{4\pi }{\int }_{-L}^L\frac{{\hat{a}}_{\varphi }rd{z}^{\prime }}{(r^2+z^{\prime 2}{)}^{3/2}}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{4\pi }{\left[\frac{z^{\prime }}{r^2\sqrt{r^2+z^{\prime 2}}}\right]}_{-L}^L$ $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{4\pi }\cdot \frac{2L}{r^2\sqrt{L^2+r^2}}={\hat{a}}_{\varphi }\frac{{\mu }_{0}IL}{2\pi r\sqrt{L^2+r^2}}[\text{T}]$

Infinite Wire Limit ($L\gg r$):

${\lim }_{L\to \infty }\frac{L}{\sqrt{L^2+r^2}}=1⟹\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}I}{2\pi r}[\text{T}]$

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2. Cheng Example 6-5: Center of a Square Loop [PYQ: 2017]

Verbatim PYQ [PYQ: 2017 (12 Marks)]

State and explain Biot-Savart law. With the help of this law, find the magnetic flux density at the center of a square loop with side $w$ carrying a direct current $I$.

                        SQUARE LOOP CARRYING CURRENT I
                                 Y ^
                                   |  Side 3 (Length w)
                               +---+---+
                               |   |   |
              Side 4 (Length w)|   o---+-----> X (Center O)
                               |       | Side 2 (Length w)
                               +---+---+
                                  Side 1 (Length w)

Derivation:

1. Place loop in $xy$-plane centered at origin. Side length $w⟹L=w/2$, distance from side to center $r=w/2$.
2. The field at the center due to one side of length $w$ (using straight wire formula ${\vec{B}}_1={\hat{a}}_n\frac{{\mu }_{0}IL}{2\pi r\sqrt{L^2+r^2}}$): $B_1=\frac{{\mu }_{0}I(w/2)}{2\pi (w/2)\sqrt{(w/2{)}^2+(w/2{)}^2}}=\frac{{\mu }_{0}I}{2\pi (w/2)\sqrt{2}}=\frac{{\mu }_{0}I}{\sqrt{2}\pi w}$
3. By right-hand rule, all 4 sides produce fields pointing in the same normal direction $+{\hat{a}}_z$ out of the page:

$B_{\text{total}}=4B_1=4\times \frac{{\mu }_{0}I}{\sqrt{2}\pi w}⟹\vec{B}={\hat{a}}_z\frac{2\sqrt{2}{\mu }_{0}I}{\pi w}[\text{T}][\text{PYQ: 2017}]$

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3. Cheng Example 6-6: Axis of a Circular Loop [PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025]

Verbatim PYQ [PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025 (10-15 Marks)]

Find the magnetic flux density at a point $P(0,0,z)$ on the axis of a circular loop of radius $b$ carrying direct current $I$.

                      CIRCULAR LOOP ON Z-AXIS

                                 Z ^ P(0,0,z) Field Point
                                   |  .
                                   | /|
                       Vector R    |/ | Vector R = a_z z - a_r b
                       (m)        /|  | R = √(b² + z²)
                                 / |  |
  ------------------------------+--+--+------------------------> Y
                               /   | b
                              o----+  Source Element dl' = a_ϕ b dφ'
                             X (Circular Loop Radius b)

Derivation:

1. Loop of radius $b$ lies in $xy$-plane centered at origin.
2. Source element $d{\vec{l}}^{\prime }={\hat{a}}_{\varphi }bd{\varphi }^{\prime }$ at $(b,{\varphi }^{\prime },0)$.
3. Field point $P(0,0,z)⟹\vec{R}={\hat{a}}_{z}z-{\hat{a}}_{r}b$, magnitude $R=\sqrt{b^2+z^2}$.
4. Cross product: $d{\vec{l}}^{\prime }\times \vec{R}=({\hat{a}}_{\varphi }bd{\varphi }^{\prime })\times ({\hat{a}}_{z}z-{\hat{a}}_{r}b)={\hat{a}}_{r}bzd{\varphi }^{\prime }+{\hat{a}}_z{b}^{2}d{\varphi }^{\prime }$
5. Symmetry Cancellation: Diametrically opposite source elements produce equal radial components ${\hat{a}}_r$ pointing in opposite directions $⟹$ radial component integrates to zero (${\int }_0^{2\pi }{\hat{a}}_{r}d{\varphi }^{\prime }=0$).
6. Integrate axial component ${\hat{a}}_z$: $B_z=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{2\pi }\frac{b^{2}d{\varphi }^{\prime }}{(b^2+z^2{)}^{3/2}}=\frac{{\mu }_{0}I{b}^2}{4\pi (b^2+z^2{)}^{3/2}}(2\pi )$

$\vec{B}={\hat{a}}_z\frac{{\mu }_{0}I{b}^2}{2(b^2+z^2{)}^{3/2}}[\text{T}][\text{PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025}]$

Center of Loop Limit ($z=0$):

$\vec{B}(0,0,0)={\hat{a}}_z\frac{{\mu }_{0}I}{2b}[\text{T}]$

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4. Exam Strategy, Traps & PYQ Checklist

Exam Traps to Avoid

1. Zero Work Error: Remember that ${\vec{F}}_m=q(\vec{u}\times \vec{B})$ does no work. Never write that magnetic force speeds up a particle.
2. Unprimed vs. Primed Derivative Error: In $\vec{B}=\nabla \times \vec{A}$, the curl $\nabla$ operates strictly on unprimed field coordinates $(x,y,z)$. $d{\vec{l}}^{\prime }$ depends on primed source coordinates $(x^{\prime },y^{\prime },z^{\prime })$, so $\nabla \times d{\vec{l}}^{\prime }=0$.
3. Cylindrical Conductor Limits: Do not confuse inside ($r\le b⟹B\propto r$) and outside ($r\ge b⟹B\propto 1/r$).
4. Vector Units: Always include units ($\text{T}$, ${\text{Wb/m}}^2$, $\text{Wb/m}$, $\text{N}$, $\text{A/m}$).

📋 Complete PYQ Checklist (Sections 6-1 to 6-4)

• State Lorentz force equation [PYQ: 2018, 2021]
• Write differential & integral postulates of magnetostatics [PYQ: 2015, 2016, 2017, 2020, 2025]
• Prove “Magnetic poles cannot be isolated” / $\oint \vec{B}\cdot d\vec{s}=0$ [PYQ: 2020, 2025]
• State & explain Ampere’s Circuital Law [PYQ: 2015, 2016, 2017, 2018, 2021, 2023, 2025]
• Derive $B$ inside and outside circular conductor of radius $b$ [PYQ: 2017, 2018, 2022]
• Define vector magnetic potential $\vec{A}$ [PYQ: 2019, 2022, 2024, 2025]
• Deduce vector Poisson’s equation ${\nabla }^2\vec{A}=-{\mu }_0\vec{J}$ [PYQ: 2020]
• Derive integral expression for $\vec{A}$ [PYQ: 2020, 2024]
• Prove magnetic flux $\Phi ={\oint }_C\vec{A}\cdot d\vec{l}$ [PYQ: 2019]
• Distinguish vector magnetic potential $\vec{A}$ vs scalar magnetic potential $V_m$ [PYQ: 2015, 2019]
• Determine $\vec{A}$ inside and outside circular conductor [PYQ: 2020]
• State & derive Biot-Savart Law from $\vec{A}$ [PYQ: 2016, 2017, 2021, 2024]
• Compare Biot-Savart Law vs Ampere’s Circuital Law [PYQ: 2015, 2018, 2023, 2025]
• Derive $B$ for finite wire ($2L$) via vector potential $\vec{A}$ and Biot-Savart Law [PYQ: 2015, 2016, 2021, 2023, 2025]
• Derive $B$ at center of square loop [PYQ: 2017]
• Derive $B$ on axis of circular loop [PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025]