03 magnetostatics_master_notes

ECE 2105: Topic 3 — Magnetostatics Master Study Notes

Targeting A+ Performance — Standalone, Highly Comprehensive, and Rigorous

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Section 1: Fundamental Postulates & Laws of Magnetostatics

1.1 The Postulates of Magnetostatics in Free Space

Magnetostatics deals with magnetic fields that are constant in time, typically produced by a steady flow of direct currents (DC). Just as electrostatics is governed by two postulates (Gauss’s and Faraday’s laws), magnetostatics is governed by two fundamental postulates in both differential and integral forms.

A. Postulate I: No Isolated Magnetic Monopoles (Divergence of $\vec{B}$)

• Differential Form: $\nabla \cdot \vec{B}=0$
• Integral Form: (Applying the Divergence Theorem) ${\oint }_S\vec{B}\cdot d\vec{s}=0$
• Physical Significance:
  1. The total outward magnetic flux passing through any closed surface is always zero.
  2. This indicates that there are no magnetic “charges” (monopoles) or sources/sinks of magnetic flux. Magnetic flux lines always close upon themselves, forming continuous loops [Ref: L 10.pdf Fig 6-1, successive division of a bar magnet].
  3. Newton’s Third Law Connection: Unlike electric charges, cutting a magnet in half does not isolate a North or South pole; instead, it creates a new magnet with its own North and South poles.

B. Postulate II: Ampere’s Law (Curl of $\vec{B}$ / $\vec{H}$)

• Differential Form: $\nabla \times \vec{B}={\mu }_0\vec{J}⟹\nabla \times \vec{H}=\vec{J}$
• Integral Form: (Applying Stokes’s Theorem) ${\oint }_C\vec{H}\cdot d\vec{l}=I_{\text{enclosed}}$
• Physical Significance: The circulation of the magnetic field intensity ($\vec{H}$) around any closed path $C$ is exactly equal to the total free current $I$ passing through the surface bounded by the path. The direction of circulation and current follow the Right-Hand Rule.

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1.2 The Lorentz Force Equation

A static charge experiences only an electric force ${\vec{F}}_e=q\vec{E}$. However, when a test charge $q$ is in motion with velocity $\vec{u}$ in a magnetic field $\vec{B}$, it experiences an additional magnetic force ${\vec{F}}_m=q(\vec{u}\times \vec{B})$.

When both electric and magnetic fields coexist, the total electromagnetic force on the charge is the vector sum: $\vec{F}={\vec{F}}_e+{\vec{F}}_m=q(\vec{E}+\vec{u}\times \vec{B})$ This is known as the Lorentz Force Equation [PYQ: 2018, 2021] [Ref: L 10.pdf Slide 3].

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1.3 Biot-Savart Law vs. Ampere’s Circuital Law

Students often confuse when to use which law. This comparison table is crucial for scoring full marks on comparative questions [PYQ: 2015, 2018, 2023, 2025].

Feature	Biot-Savart Law ($\vec{B}=\frac{{\mu }_{0}I}{4\pi }\oint \frac{d\vec{l}\times {\hat{a}}_R}{R^2}$)	Ampere’s Circuital Law ($\oint \vec{H}\cdot d\vec{l}=I$)
Basic Concept	Sums the magnetic field contributions of individual infinitesimal current elements ($Id\vec{l}$).	Relates net magnetic circulation along a closed path directly to the enclosed current.
Applicability	General. Applies to any current geometry (arbitrary lines, loops, finite wires).	General, but mathematically practical only for highly symmetric current distributions.
Symmetry Dependency	None. Can solve asymmetric cases via explicit integration.	Requires high coordinate symmetry (infinite lines, cylinders, solenoids, toroids).
Electrostatic Analogy	Analogous to Coulomb’s Law (integrating point charges).	Analogous to Gauss’s Law (pillbox / symmetrical surfaces).

A+ Exam Hack: If asked to compare their usefulness, write: “Ampere’s law is a powerful tool to find $\vec{B}$ quickly in highly symmetric systems without integration, whereas Biot-Savart is the only recourse for systems lacking high spatial symmetry.”

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Section 2: High-Yield Field Derivations (Biot-Savart Applications)

Derivation 2.1: Magnetic Field of a Finite Straight Wire of Length $2L$

Problem: A direct current $I$ flows in a straight wire of length $2L$. Find the magnetic flux density $\vec{B}$ at point $P$ located at a distance $r$ in the bisecting plane [PYQ: 2015, 2016, 2021, 2023, 2025].

[Figure: Finite straight wire of length 2L lying along the z-axis, with observation point P(r, 0, 0) in the xy-plane. Distance vector R and angle theta shown. Cheng Fig 6-5 / L 10 Slide 15]

Step-by-Step Derivation:

1. Set up the Geometry: Let the wire lie along the z-axis from $z=-L$ to $z=+L$. Let the observation point $P$ be in the xy-plane at coordinates $(r,0,0)$.
2. Define the Source Element:
   • Differential current element: $Id{\vec{l}}^{\prime }=I{\hat{a}}_{z}d{z}^{\prime }$
   • Source coordinate: $(0,0,z^{\prime })$
   • Field coordinate: $(r,0,0)$
   • Distance vector: $\vec{R}=r{\hat{a}}_r-z^{\prime }{\hat{a}}_z$
   • Distance magnitude: $R=\sqrt{r^2+(z^{\prime }{)}^2}$
   • Unit vector: ${\hat{a}}_R=\frac{r{\hat{a}}_r-z^{\prime }{\hat{a}}_z}{\sqrt{r^2+(z^{\prime }{)}^2}}$
3. Evaluate the Cross Product: $d{\vec{l}}^{\prime }\times {\hat{a}}_R=({\hat{a}}_{z}d{z}^{\prime })\times \left(\frac{r{\hat{a}}_r-z^{\prime }{\hat{a}}_z}{R}\right)={\hat{a}}_{\varphi }\frac{rd{z}^{\prime }}{R}$ (Since ${\hat{a}}_z\times {\hat{a}}_r={\hat{a}}_{\varphi }$ and ${\hat{a}}_z\times {\hat{a}}_z=0$)
4. Set up the Biot-Savart Integral: $\vec{B}=\frac{{\mu }_{0}I}{4\pi }{\int }_{-L}^L\frac{d{\vec{l}}^{\prime }\times {\hat{a}}_R}{R^2}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{4\pi }{\int }_{-L}^L\frac{d{z}^{\prime }}{(r^2+(z^{\prime }{)}^2{)}^{3/2}}$
5. Evaluate the Integration: Using the standard integration formula: $\int \frac{dz}{(r^2+z^2{)}^{3/2}}=\frac{z}{r^2\sqrt{r^2+z^2}}$ $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{4\pi }{\left[\frac{z^{\prime }}{r^2\sqrt{r^2+(z^{\prime }{)}^2}}\right]}_{-L}^L$ $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}Ir}{4\pi r^2}\left(\frac{L}{\sqrt{r^2+L^2}}-\frac{-L}{\sqrt{r^2+L^2}}\right)$ $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}IL}{2\pi r\sqrt{r^2+L^2}}$

Transition to an Infinitely Long Wire ($L\gg r$):

As $L\to \infty$, the term $\frac{L}{\sqrt{r^2+L^2}}\approx 1$. The expression simplifies to: $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}I}{2\pi r}$ This matches the exact expression obtained much more simply via Ampere’s Circuital Law!

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Derivation 2.2: Magnetic Field at the Center of a Square Loop of Side $w$

Problem: Use the Biot-Savart Law to find the magnetic flux density at the center of a square loop of side $w$ carrying direct current $I$ [PYQ: 2017].

[Figure: Square loop in the xy-plane centered at the origin, with sides of length w carrying counter-clockwise current I. Observation point is the origin. Cheng Fig 6-6]

Step-by-Step Derivation:

1. Exploit Symmetry: A square loop consists of 4 straight wire segments, each of length $w$. By symmetry, the magnetic field contribution ${\vec{B}}_i$ from each side points in the same direction (${\hat{a}}_z$, out of the page using the Right-Hand Rule). ${\vec{B}}_{\text{total}}=4{\vec{B}}_1$
2. Map to the Finite Wire Formula: The magnetic field of a single side of length $2L$ at a perpendicular distance $r$ is: $B=\frac{{\mu }_{0}IL}{2\pi r\sqrt{r^2+L^2}}$
3. Identify Parameters for the Square Side:
   • Total length of one side is $w⟹2L=w⟹L=w/2$
   • Perpendicular distance from the center of the side to the center of the square is $r=w/2$.
4. Substitute and Solve: $B_1=\frac{{\mu }_{0}I(w/2)}{2\pi (w/2)\sqrt{(w/2{)}^2+(w/2{)}^2}}=\frac{{\mu }_{0}I}{2\pi \sqrt{2(w/2{)}^2}}=\frac{{\mu }_{0}I}{2\pi \sqrt{2}(w/2)}=\frac{{\mu }_{0}I}{\sqrt{2}\pi w}$
5. Multiply by 4: $B_{\text{total}}=4\left(\frac{{\mu }_{0}I}{\sqrt{2}\pi w}\right)=\frac{2\sqrt{2}{\mu }_{0}I}{\pi w}$ Vector form: $\vec{B}={\hat{a}}_z\frac{2\sqrt{2}{\mu }_{0}I}{\pi w}$

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Derivation 2.3: Magnetic Field along the Axis of a Circular Loop of Radius $b$

Problem: Find the magnetic flux density at a point on the axis of a circular loop of radius $b$ that carries a direct current $I$ [PYQ: 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2025].

[Figure: Circular current loop of radius b in the xy-plane carrying current I. Observation point P(0, 0, z) on the z-axis. Distance R, angle theta, and differential element dl shown. Cheng Fig 6-7]

Step-by-Step Derivation:

1. Set up Coordinates: Let the loop of radius $b$ lie in the xy-plane, centered at the origin. Let the observation point be $P(0,0,z)$ on the z-axis.
2. Define elements:
   • Differential length: $d{\vec{l}}^{\prime }={\hat{a}}_{\varphi }bd{\varphi }^{\prime }$
   • Vector from source to field point: $\vec{R}=z{\hat{a}}_z-b{\hat{a}}_r$
   • Distance magnitude: $R=\sqrt{z^2+b^2}$
3. Evaluate the Cross Product: $d{\vec{l}}^{\prime }\times \vec{R}=({\hat{a}}_{\varphi }bd{\varphi }^{\prime })\times (z{\hat{a}}_z-b{\hat{a}}_r)={\hat{a}}_{r}bzd{\varphi }^{\prime }+{\hat{a}}_z{b}^{2}d{\varphi }^{\prime }$ (Since ${\hat{a}}_{\varphi }\times {\hat{a}}_z={\hat{a}}_r$ and ${\hat{a}}_{\varphi }\times {\hat{a}}_r=-{\hat{a}}_z$)
4. Exploit Symmetry: As we integrate around the circle from ${\varphi }^{\prime }=0$ to $2\pi$, the radial component ${\hat{a}}_r$ systematically cancels out because diametrically opposite elements produce opposing radial fields. Only the axial component (${\hat{a}}_z$) survives.
5. Biot-Savart Integral for $B_z$: $B_z=\frac{{\mu }_{0}I}{4\pi }\oint \frac{[d{\vec{l}}^{\prime }\times \vec{R}{]}_z}{R^3}=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{2\pi }\frac{b^{2}d{\varphi }^{\prime }}{(z^2+b^2{)}^{3/2}}$ $B_z=\frac{{\mu }_{0}I{b}^2}{4\pi (z^2+b^2{)}^{3/2}}{\int }_0^{2\pi }d{\varphi }^{\prime }=\frac{{\mu }_{0}I{b}^2(2\pi )}{4\pi (z^2+b^2{)}^{3/2}}$ $\vec{B}={\hat{a}}_z\frac{{\mu }_{0}I{b}^2}{2(z^2+b^2{)}^{3/2}}$

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Section 3: Vector Magnetic Potential ($\vec{A}$)

3.1 Definition and Physical Concept

In electrostatics, we defined a scalar potential $V$ such that $\vec{E}=-\nabla V$ because $\nabla \times \vec{E}=0$. In magnetostatics, since $\nabla \cdot \vec{B}=0$ (solenoidal field), we can define a vector field $\vec{A}$ such that its curl equals the magnetic flux density: $\vec{B}=\nabla \times \vec{A}\text{[Weber/m] or [Tesla}\cdot \text{m]}$ This vector field $\vec{A}$ is called the Vector Magnetic Potential [PYQ: 2016, 2018, 2019, 2022, 2024, 2025].

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3.2 Derivation of Vector Poisson’s Equation

How is $\vec{A}$ related to the source current density $\vec{J}$? Let’s derive it!

1. Start with Ampere’s postulate in differential form: $\nabla \times \vec{B}={\mu }_0\vec{J}$
2. Substitute the definition $\vec{B}=\nabla \times \vec{A}$: $\nabla \times (\nabla \times \vec{A})={\mu }_0\vec{J}$
3. Apply the vector triple product identity (“BAC-CAB” rule): $\nabla (\nabla \cdot \vec{A})-{\nabla }^2\vec{A}={\mu }_0\vec{J}$
4. Gauge Selection (Lorentz/Coulomb Gauge): We have the freedom to specify the divergence of our vector potential. To decouple the equation, we choose: $\nabla \cdot \vec{A}=0$
5. This yields Vector Poisson’s Equation [PYQ: 2020]: ${\nabla }^2\vec{A}=-{\mu }_0\vec{J}$

In Cartesian coordinates, this is equivalent to three independent scalar Poisson’s equations: ${\nabla }^2{A}_x=-{\mu }_0{J}_x,{\nabla }^2{A}_y=-{\mu }_0{J}_y,{\nabla }^2{A}_z=-{\mu }_0{J}_z$ Its integral solution is: $\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}\frac{\vec{J}}{R}d{v}^{\prime }$

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3.3 Physical Significance of $\vec{A}$

Proof: Show that the line integral of $\vec{A}$ around any closed path equals the total magnetic flux ($\Phi$) passing through the area enclosed [PYQ: 2019]:

1. Consider a closed path $C$ bounding an open surface $S$.
2. The magnetic flux $\Phi$ is defined as: $\Phi ={\int }_S\vec{B}\cdot d\vec{s}$
3. Substitute $\vec{B}=\nabla \times \vec{A}$: $\Phi ={\int }_S(\nabla \times \vec{A})\cdot d\vec{s}$
4. Apply Stokes’s Theorem to convert the surface integral to a line integral: $\Phi ={\oint }_C\vec{A}\cdot d\vec{l}$ This proves that the line integral of $\vec{A}$ around $C$ is exactly equal to the enclosed magnetic flux.

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3.4 Vector vs. Scalar Magnetic Potential

Property	Vector Magnetic Potential ($\vec{A}$)	Scalar Magnetic Potential ($V_m$) [PYQ: 2018]
Governing Equation	$\vec{B}=\nabla \times \vec{A}$	$\vec{B}=-{\mu }_0\nabla V_m$
Condition of Validity	Always valid everywhere.	Valid only in current-free regions ($\vec{J}=0$).
Field Nature	Solenoidal ($\nabla \cdot \vec{B}=0$).	Irrotational ($\nabla \times \vec{B}=0$ when $\vec{J}=0$).
Mathematical Type	Vector field.	Scalar field.

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Section 4: Magnetization & Material Behavior

4.1 The Magnetization Vector ($\vec{M}$)

When an external magnetic field is applied to a material, it aligns the microscopic atomic magnetic dipoles (due to orbiting and spinning electrons) and induces macroscopic magnetic effects.

We define the Magnetization Vector ($\vec{M}$) as the volume density of magnetic dipole moments: $\vec{M}={\lim }_{\Delta v\to 0}\frac{{\sum }_{k=1}^{n\Delta v}{\vec{m}}_k}{\Delta v}\text{[A/m]}$

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4.2 Derivation of Bound Magnetization Currents

A magnetized material can be mathematically replaced by equivalent “bound” currents: a volume current density ${\vec{J}}_m$ and a surface current density ${\vec{J}}_{ms}$ [PYQ: 2016, 2018, 2022, 2025].

Derivation:

1. Consider a volume $v^{\prime }$ containing magnetized material. The differential vector potential $d\vec{A}$ due to a magnetization dipole element $d\vec{m}=\vec{M}d{v}^{\prime }$ is: $d\vec{A}=\frac{{\mu }_0\vec{M}\times {\hat{a}}_R}{4\pi R^2}d{v}^{\prime }=\frac{{\mu }_0}{4\pi }\vec{M}\times {\nabla }^{\prime }\left(\frac{1}{R}\right)d{v}^{\prime }$
2. Integrating over the entire volume: $\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}\vec{M}\times {\nabla }^{\prime }\left(\frac{1}{R}\right)d{v}^{\prime }$
3. Using the vector identity $\vec{M}\times {\nabla }^{\prime }\left(\frac{1}{R}\right)=\frac{{\nabla }^{\prime }\times \vec{M}}{R}-{\nabla }^{\prime }\times \left(\frac{\vec{M}}{R}\right)$: $\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}\frac{{\nabla }^{\prime }\times \vec{M}}{R}d{v}^{\prime }-\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}{\nabla }^{\prime }\times \left(\frac{\vec{M}}{R}\right)d{v}^{\prime }$
4. Applying a vector corollary of the Divergence Theorem to the second term: ${\int }_{v^{\prime }}{\nabla }^{\prime }\times \vec{F}d{v}^{\prime }=-{\oint }_{S^{\prime }}\vec{F}\times d{\vec{s}}^{\prime }={\oint }_{S^{\prime }}\vec{F}\times {\hat{a}}_{n}d{s}^{\prime }$ $\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}\frac{{\nabla }^{\prime }\times \vec{M}}{R}d{v}^{\prime }+\frac{{\mu }_0}{4\pi }{\oint }_{S^{\prime }}\frac{\vec{M}\times {\hat{a}}_n}{R}d{s}^{\prime }$
5. Comparing this with the general formula for vector potentials from current densities: $\vec{A}=\frac{{\mu }_0}{4\pi }{\int }_{v^{\prime }}\frac{{\vec{J}}_m}{R}d{v}^{\prime }+\frac{{\mu }_0}{4\pi }{\oint }_{S^{\prime }}\frac{{\vec{J}}_{ms}}{R}d{s}^{\prime }$ We identify: ${\vec{J}}_m=\nabla \times \vec{M}\text{[Volume current density, A/m²]}$ ${\vec{J}}_{ms}=\vec{M}\times {\hat{a}}_n\text{[Surface current density, A/m]}$

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4.3 Classification of Magnetic Materials

The alignment of atomic dipoles depends on material atomic structures. The magnetic susceptibility ${\chi }_m$ (where $\vec{M}={\chi }_m\vec{H}$) acts as our differentiator [Ref: L 8.pdf Slide 20].

Material Type	Susceptibility (${\chi }_m$)	Relative Permeability (${\mu }_r$)	Atomic/Physical Origin	Example Materials
Diamagnetism	Negative, very small ($\sim -10^{-5}$)	${\mu }_r\lesssim 1$	Orbital electron motion; vanishes when external field is removed.	Copper, Gold, Bismuth, Diamond
Paramagnetism	Positive, very small ($\sim 10^{-5}$)	${\mu }_r\gtrsim 1$	Unpaired electron spins weakly aligning with applied field.	Aluminum, Magnesium, Tungsten
Ferromagnetism	Positive, extremely large (up to $10^6$)	${\mu }_r\gg 1$	Spontaneous alignment of localized “magnetic domains”.	Iron, Cobalt, Nickel, Steel

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Section 5: Magnetostatic Boundary Conditions

Boundary conditions define how magnetic field vectors ($\vec{B}$ and $\vec{H}$) behave at the interface between two distinct magnetic media.

[Figure: Interface between Medium 1 (mu1) and Medium 2 (mu2). A cylindrical pillbox of height dh and surface area ds across the boundary. A rectangular closed loop abcd of height dh and width dw across the boundary. Cheng Fig 6-11 / L 8 Slide 11]

Derivation 5.1: Normal Boundary Condition (Continuity of $\vec{B}$)

1. Apply the first postulate of magnetostatics (${\oint }_S\vec{B}\cdot d\vec{s}=0$) to a small cylindrical pillbox placed symmetrically across the interface.
2. Let the top face of area $\Delta s$ be in Medium 1, and the bottom face in Medium 2. The height of the pillbox $\Delta h\to 0$.
3. The contribution of the curved side walls vanishes as $\Delta h\to 0$. ${\oint }_S\vec{B}\cdot d\vec{s}=B_{1n}\Delta s-B_{2n}\Delta s=0⟹B_{1n}=B_{2n}$ Vector form: ${\hat{a}}_{n2}\cdot ({\vec{B}}_1-{\vec{B}}_2)=0$

• Result: The normal component of magnetic flux density ($\vec{B}$) is always continuous across any interface.

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Derivation 5.2: Tangential Boundary Condition (Discontinuity of $\vec{H}$)

1. Apply Ampere’s circuital law (${\oint }_C\vec{H}\cdot d\vec{l}=I$) to a small rectangular loop $abcd$ of width $\Delta w$ and height $\Delta h\to 0$.
2. The line integral evaluates to: $\oint \vec{H}\cdot d\vec{l}=H_{1t}\Delta w-H_{2t}\Delta w=I_{\text{enclosed}}=J_s\Delta w$ $⟹H_{1t}-H_{2t}=J_s$ Vector form: ${\hat{a}}_{n2}\times ({\vec{H}}_1-{\vec{H}}_2)={\vec{J}}_s$

• Result: The tangential component of magnetic field intensity ($\vec{H}$) is discontinuous across an interface by an amount equal to the free surface current density ${\vec{J}}_s$.
• Note: Since almost all physical media have finite conductivities, ${\vec{J}}_s=0$ at the boundary. Hence, $H_{1t}=H_{2t}$ for all practical boundaries (except perfect conductors).

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5.3 The Law of Refraction for Magnetic Fields

From $B_{1n}=B_{2n}⟹{\mu }_1{H}_{1n}={\mu }_2{H}_{2n}$ and $H_{1t}=H_{2t}$ (assuming ${\vec{J}}_s=0$): $\tan {\alpha }_1=\frac{H_{1t}}{H_{1n}},\tan {\alpha }_2=\frac{H_{2t}}{H_{2n}}$ Dividing these two equations: $\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{H_{1t}/H_{1n}}{H_{2t}/H_{2n}}=\frac{H_{2n}}{H_{1n}}=\frac{{\mu }_1}{{\mu }_2}$ $\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{\mu }_{r1}}{{\mu }_{r2}}$

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Section 6: Magnetic Forces, Torques & Energy

Derivation 6.1: Force between Two Parallel Wires & Newton’s Third Law

Problem: Calculate the force per unit length between two infinitely long parallel wires separated by distance $d$ carrying currents $I_1$ and $I_2$, and prove it satisfies Newton’s third law [PYQ: 2017, 2018, 2020, 2022, 2023, 2024, 2025].

[Figure: Two parallel straight wires along the z-axis, wire 1 carrying current I1, wire 2 carrying current I2, separated by distance d in the xy plane. Magnetic field lines B1 around wire 1 shown. L 10 Slide 18]

Step-by-Step Derivation:

1. Field of Wire 1 at Wire 2: The magnetic field intensity produced by current $I_1$ at a distance $d$ is: ${\vec{B}}_{21}=-{\hat{a}}_x\frac{{\mu }_0{I}_1}{2\pi d}$ (Assuming wire 1 lies along the z-axis at $x=0$, and wire 2 lies along the z-axis at $x=d$)
2. Force Exerted on Wire 2: Using the current force formula $d\vec{F}=Id\vec{l}\times \vec{B}$: ${\vec{F}}_{21}=I_2{\int }_0^L({\hat{a}}_{z}dz)\times \left(-{\hat{a}}_x\frac{{\mu }_0{I}_1}{2\pi d}\right)$ ${\vec{F}}_{21}=-{\hat{a}}_y\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}$
3. Force Per Unit Length: $\frac{{\vec{F}}_{21}}{L}=-{\hat{a}}_y\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$ (Negative sign indicates attraction when currents are in the same direction!)
4. Proof of Newton’s Third Law: Applying the same steps for the force on wire 1 due to the magnetic field ${\vec{B}}_{12}$ from wire 2: ${\vec{B}}_{12}={\hat{a}}_x\frac{{\mu }_0{I}_2}{2\pi d}$ ${\vec{F}}_{12}=I_1{\int }_0^L({\hat{a}}_{z}dz)\times \left({\hat{a}}_x\frac{{\mu }_0{I}_2}{2\pi d}\right)={\hat{a}}_y\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}$ Clearly: ${\vec{F}}_{12}=-{\vec{F}}_{21}$ This proves that the magnetic forces acting on the current-carrying conductors are equal in magnitude and opposite in direction, rigorously satisfying Newton’s Third Law.

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Derivation 6.2: Torque on a Rectangular Loop in a Uniform Field

Problem: Derive the torque experienced by a rectangular current loop of dimensions $w\times l$ carrying current $I$ placed in a uniform magnetic field $\vec{B}$ [PYQ: 2022].

[Figure: Labeled rectangular loop of length l and width w carrying current I. Loop plane makes an angle theta with magnetic field B. Force vectors F1 and F2 forming a torque couple are shown. L 10 Slide 19]

Step-by-Step Derivation:

1. Define the magnetic dipole moment of the rectangular loop: $\vec{m}={\hat{a}}_{n}IS={\hat{a}}_{n}I(w\cdot l)$
2. When placed in a magnetic field $\vec{B}$, the net translational force on any closed loop is zero (${\vec{F}}_{\text{net}}=0$).
3. However, the forces on opposite sides of the loop form a torque couple. The torque vector $\vec{\tau }$ is defined as: $\vec{\tau }=\vec{r}\times \vec{F}$
4. For a loop rotated at angle $\theta$ relative to the magnetic field, the torque magnitude is: $\tau =(F\sin \theta )\cdot w=(IlB\sin \theta )w=I(wl)B\sin \theta =mB\sin \theta$
5. In vector notation, this is represented by: $\vec{\tau }=\vec{m}\times \vec{B}$

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Derivation 6.3: Magnetic Energy of Coupled Inductors

Problem: Derive the total magnetic energy stored in two mutually coupled inductor circuits [PYQ: 2019, 2020].

[Figure: Two loops, C1 and C2, connected to current generators raising currents i1 and i2. Flux linkages Phi11, Phi12, Phi21, Phi22 shown. Cheng Fig 6-22 / L 9 Slide 14]

Step-by-Step Derivation:

1. Define Energy Stored in a Single Inductor: The work done to raise a current from $0$ to $I_1$ in an inductor with self-inductance $L_1$: $W_1=\int v_1{i}_{1}dt=L_1{\int }_0^{I_1}{i}_{1}d{i}_1=\frac{1}{2}{L}_1{I}_1^2$
2. Introduce the Second Inductor: With loop 1 current held constant at $I_1$, we increase current $i_2$ in loop 2 from $0$ to $I_2$.
3. As $i_2$ increases, it induces a voltage in loop 1 due to mutual coupling: $v_{12}=L_{12}\frac{d{i}_2}{dt}$
4. The generator on loop 1 must do additional work to keep the current constant at $I_1$: $W_{12}=\int v_{12}{I}_{1}dt=L_{12}{I}_1{\int }_0^{I_2}d{i}_2=L_{12}{I}_1{I}_2$
5. Simultaneously, self-energy is stored in loop 2: $W_2=\frac{1}{2}{L}_2{I}_2^2$
6. Calculate Total Energy: The total energy is the sum of these three work components: $W_{\text{total}}=W_1+W_{12}+W_2=\frac{1}{2}{L}_1{I}_1^2+L_{12}{I}_1{I}_2+\frac{1}{2}{L}_2{I}_2^2$

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Section 7: The Hall Effect

7.1 Physical Concept and Origin

The Hall Effect is a classic electromagnetic transport phenomenon. When a current-carrying conductor is placed in an external magnetic field perpendicular to the direction of current flow, the moving charge carriers experience a transverse magnetic force.

[Figure: Block of conductor of width d and thickness b. Current density J along the y-axis, magnetic field B along the z-axis. Depicted deflection of electrons toward the side, creating a charge separation and transverse Hall voltage VH across the sides. Cheng Fig 6-28 / L 9 Slide 16]

This force deflects the charge carriers (electrons or holes) toward one side of the conductor. As charges pile up on one face, they leave behind opposite charges on the opposite face, creating a transverse electric field known as the Hall Field (${\vec{E}}_H$).

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7.2 Mathematical Derivation of Hall Voltage

1. At steady state, equilibrium is reached when the electric force from the accumulated charge separation balances the magnetic deflection force exactly: ${\vec{F}}_e+{\vec{F}}_m=0⟹q{\vec{E}}_H+q(\vec{u}\times \vec{B})=0$ ${\vec{E}}_H=-(\vec{u}\times \vec{B})$
2. Let the current flow along the y-axis ($\vec{J}={\hat{a}}_y{J}_y$) and the magnetic field be along the z-axis ($\vec{B}={\hat{a}}_z{B}_0$).
3. For electrons (charge $q=-e$), their drift velocity is $\vec{u}=-{\hat{a}}_y{u}_0$. ${\vec{E}}_H=-(-{\hat{a}}_y{u}_0\times {\hat{a}}_z{B}_0)={\hat{a}}_x{u}_0{B}_0$
4. The potential difference (Hall Voltage $V_H$) generated across the width $d$ of the block is: $V_H={\int }_0^d{E}_{H}dx=u_0{B}_{0}d$
5. Since current density is related to drift velocity by $J_y=Nq{u}_0⟹u_0=\frac{J_y}{Nq}$, we can express the Hall Voltage as: $V_H=\left(\frac{J_y}{Nq}\right)B_{0}d$

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7.3 Applications of the Hall Effect

• Determining the Charge Carrier Type: The polarity of $V_H$ reveals whether the predominant charge carriers are holes (p-type) or electrons (n-type).
• Measuring Carrier Concentration ($N$): By measuring $V_H$, current, and magnetic field, the density of carriers can be directly calculated.
• Magnetic Field Sensor (Magnetometer): Used to measure unknown magnetic fields based on a calibrated $V_H$ output.

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Section 8: Standalone Solutions to Classic PYQ Problems

Problem 8.1: Air-Iron Interface Boundary Refraction [PYQ: 2017, 2023]

Question: A plane boundary ($y=0$) separates Air (Medium 1, ${\mu }_{r1}=1$) and Iron (Medium 2, ${\mu }_{r2}=5000$).

1. Assuming ${\vec{B}}_1={\hat{a}}_{x}0.5-{\hat{a}}_{y}10$ (mT), find ${\vec{B}}_2$ and the angle ${\vec{B}}_2$ makes with the interface.
2. Assuming ${\vec{B}}_2={\hat{a}}_{x}10+{\hat{a}}_{y}0.5$ (mT), find ${\vec{B}}_1$ and the angle ${\vec{B}}_1$ makes with the normal.

Detailed Solution:

Part 1:

1. The interface is $y=0$ (the xz-plane). Therefore, the normal direction is along the y-axis (${\hat{a}}_y$) and the tangential direction is along the x-axis (${\hat{a}}_x$).
2. Identify normal and tangential components of ${\vec{B}}_1$:
   • $B_{1n}=B_{1y}=-10$ mT
   • $B_{1t}=B_{1x}=0.5$ mY
3. Apply boundary conditions:
   • Normal component is continuous: $B_{2n}=B_{1n}=-10\text{ mT}⟹B_{2y}=-10\text{ mT}$
   • Tangential component of H is continuous: $H_{2t}=H_{1t}⟹\frac{B_{2t}}{{\mu }_2}=\frac{B_{1t}}{{\mu }_1}⟹B_{2t}=\left(\frac{{\mu }_{r2}}{{\mu }_{r1}}\right)B_{1t}$ $B_{2x}=\left(\frac{5000}{1}\right)(0.5\text{ mT})=2500\text{ mT}$
4. Construct the ${\vec{B}}_2$ vector: ${\vec{B}}_2={\hat{a}}_{x}2500-{\hat{a}}_{y}10\text{ (mT)}$
5. Calculate the angle ${\alpha }_2^{\prime }$ with the interface: $\tan {\alpha }_2^{\prime }=\frac{|B_{2y}|}{|B_{2x}|}=\frac{10}{2500}=0.004⟹{\alpha }_2^{\prime }=0.23^{\circ }$

Part 2:

1. Identify normal and tangential components of ${\vec{B}}_2$:
   • $B_{2n}=B_{2y}=0.5$ mT
   • $B_{2t}=B_{2x}=10$ mT
2. Apply boundary conditions:
   • Normal component is continuous: $B_{1n}=B_{2n}=0.5\text{ mY}⟹B_{1y}=0.5\text{ mT}$
   • Tangential component of H is continuous: $B_{1t}=\left(\frac{{\mu }_{r1}}{{\mu }_{r2}}\right)B_{2t}⟹B_{1x}=\left(\frac{1}{5000}\right)(10\text{ mT})=0.002\text{ mT}$
3. Construct the ${\vec{B}}_1$ vector: ${\vec{B}}_1={\hat{a}}_{x}0.002+{\hat{a}}_{y}0.5\text{ (mT)}$
4. Calculate the angle ${\alpha }_1$ with the normal: $\tan {\alpha }_1=\frac{|B_{1x}|}{|B_{1y}|}=\frac{0.002}{0.5}=0.004⟹{\alpha }_1=0.23^{\circ }$

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Problem 8.2: Vector Potential of a Finite Wire [PYQ: 2025]

Question: A direct current $I$ flows in a straight wire of length $2L$. Find the magnetic flux density $\vec{B}$ at a distance $r$ in the bisecting plane by determining the vector magnetic potential $\vec{A}$ first.

Detailed Solution:

1. Let the wire lie along the z-axis from $z=-L$ to $+L$. The current density is directed along ${\hat{a}}_z$.
2. The vector potential $\vec{A}$ has only a z-component: $A_z=\frac{{\mu }_{0}I}{4\pi }{\int }_{-L}^L\frac{d{z}^{\prime }}{\sqrt{r^2+(z^{\prime }{)}^2}}$
3. Using the standard integral $\int \frac{dz}{\sqrt{r^2+z^2}}=\ln (z+\sqrt{r^2+z^2})$: $A_z=\frac{{\mu }_{0}I}{4\pi }{\left[\ln \left(z^{\prime }+\sqrt{r^2+(z^{\prime }{)}^2}\right)\right]}_{-L}^L=\frac{{\mu }_{0}I}{4\pi }\ln \left(\frac{L+\sqrt{r^2+L^2}}{-L+\sqrt{r^2+L^2}}\right)$ $\vec{A}={\hat{a}}_z\frac{{\mu }_{0}I}{4\pi }\ln \left(\frac{\sqrt{r^2+L^2}+L}{\sqrt{r^2+L^2}-L}\right)$
4. Find $\vec{B}$ using the curl of $\vec{A}$ in cylindrical coordinates: $\vec{B}=\nabla \times \vec{A}=-{\hat{a}}_{\varphi }\frac{\partial A_z}{\partial r}$
5. Differentiating $A_z$ with respect to $r$ yields: $\vec{B}={\hat{a}}_{\varphi }\frac{{\mu }_{0}IL}{2\pi r\sqrt{r^2+L^2}}$ This demonstrates that finding $\vec{A}$ first is a mathematically elegant alternative to direct Biot-Savart integration.