Module IV: Electrodynamics - Question Solutions

Question 1: Divergence of Curl is Zero

Question

Show that the divergence of a curl of a vector is zero.

Statement to Prove

∇ · (∇ × A⃗) = 0

The divergence of curl of any vector field is always zero

Proof

Step-by-Step Derivation

Step 1: Let A⃗ = Aₓî + Aᵧĵ + Aᵤk̂ be a vector field.

Step 2: Calculate the curl of A⃗:

∇ × A⃗ = |î ĵ k̂|
|∂/∂x ∂/∂y ∂/∂z|
|Aₓ Aᵧ Aᵤ|

Expanding the determinant:

∇ × A⃗ = î(∂Aᵤ/∂y - ∂Aᵧ/∂z) - ĵ(∂Aᵤ/∂x - ∂Aₓ/∂z) + k̂(∂Aᵧ/∂x - ∂Aₓ/∂y)

Step 3: Let B⃗ = ∇ × A⃗, where:

Bₓ = ∂Aᵤ/∂y - ∂Aᵧ/∂z
Bᵧ = ∂Aₓ/∂z - ∂Aᵤ/∂x
Bᵤ = ∂Aᵧ/∂x - ∂Aₓ/∂y

Step 4: Calculate the divergence of B⃗:

∇ · B⃗ = ∂Bₓ/∂x + ∂Bᵧ/∂y + ∂Bᵤ/∂z

Step 5: Substitute the components:

∇ · (∇ × A⃗) = ∂/∂x(∂Aᵤ/∂y - ∂Aᵧ/∂z) + ∂/∂y(∂Aₓ/∂z - ∂Aᵤ/∂x) + ∂/∂z(∂Aᵧ/∂x - ∂Aₓ/∂y)

Step 6: Expand all terms:

= ∂²Aᵤ/∂x∂y - ∂²Aᵧ/∂x∂z + ∂²Aₓ/∂y∂z - ∂²Aᵤ/∂y∂x + ∂²Aᵧ/∂z∂x - ∂²Aₓ/∂z∂y

Step 7: Group the terms:

= (∂²Aᵤ/∂x∂y - ∂²Aᵤ/∂y∂x) + (∂²Aₓ/∂y∂z - ∂²Aₓ/∂z∂y) + (∂²Aᵧ/∂z∂x - ∂²Aᵧ/∂x∂z)

Step 8: By Clairaut's theorem (equality of mixed partial derivatives):

∂²f/∂x∂y = ∂²f/∂y∂x

Therefore, each bracketed term equals zero:

∇ · (∇ × A⃗) = 0 + 0 + 0 = 0

Result

∇ · (∇ × A⃗) = 0

The divergence of the curl of any vector field is identically zero.

Physical Significance

This identity has important physical implications:

The magnetic field B⃗ = ∇ × A⃗ (where A⃗ is the vector potential) is always divergence-free: ∇ · B⃗ = 0
This confirms there are no magnetic monopoles
Curl represents rotation/circulation, and divergence measures source/sink - rotation has no net source

Question 2: Gradient Calculation

Question

What is gradient? If φ = 3(x²y - y²x), find ∇φ at the point (1, -2, -1).

Definition of Gradient

Gradient (∇φ or grad φ):

The gradient of a scalar field φ is a vector field that points in the direction of the greatest rate of increase of the scalar field, and its magnitude equals the rate of change in that direction.

∇φ = (∂φ/∂x)î + (∂φ/∂y)ĵ + (∂φ/∂z)k̂

Gradient in Cartesian coordinates

Properties of Gradient:

Points in the direction of maximum increase of φ
Magnitude |∇φ| gives the maximum rate of change
Perpendicular to equipotential surfaces (surfaces of constant φ)
If ∇φ = 0 at a point, it's a critical point (maximum, minimum, or saddle)

Solution

Given:

φ = 3(x²y - y²x) = 3x²y - 3y²x

Point: (x, y, z) = (1, -2, -1)

Step 1: Calculate ∂φ/∂x

∂φ/∂x = ∂/∂x(3x²y - 3y²x)

∂φ/∂x = 6xy - 3y²

Step 2: Calculate ∂φ/∂y

∂φ/∂y = ∂/∂y(3x²y - 3y²x)

∂φ/∂y = 3x² - 6yx

Step 3: Calculate ∂φ/∂z

Since φ does not contain z:

∂φ/∂z = 0

Step 4: Write the gradient

∇φ = (6xy - 3y²)î + (3x² - 6yx)ĵ + 0k̂

Step 5: Evaluate at point (1, -2, -1)

Substituting x = 1, y = -2:

∂φ/∂x = 6(1)(-2) - 3(-2)² = -12 - 12 = -24

∂φ/∂y = 3(1)² - 6(-2)(1) = 3 + 12 = 15

∂φ/∂z = 0

∇φ at (1, -2, -1) = -24î + 15ĵ + 0k̂ = -24î + 15ĵ

Question 3: Curl of Vector Fields

Question

Determine the curl of these vector fields:
(1) A⃗ = âₓ(2x² + y²) + âᵧ(xy - y²)
(2) A⃗ = -yz âₓ + 4xy âᵧ + y âᵤ

Curl Formula

∇ × A⃗ = |î ĵ k̂|
|∂/∂x ∂/∂y ∂/∂z|
|Aₓ Aᵧ Aᵤ|

∇ × A⃗ = î(∂Aᵤ/∂y - ∂Aᵧ/∂z) - ĵ(∂Aᵤ/∂x - ∂Aₓ/∂z) + k̂(∂Aᵧ/∂x - ∂Aₓ/∂y)

Part (1): A⃗ = (2x² + y²)î + (xy - y²)ĵ

Identify Components:

Aₓ = 2x² + y²
Aᵧ = xy - y²
Aᵤ = 0 (no z-component given)

Calculate partial derivatives:

∂Aᵤ/∂y = 0, ∂Aᵧ/∂z = 0

∂Aᵤ/∂x = 0, ∂Aₓ/∂z = 0

∂Aᵧ/∂x = y, ∂Aₓ/∂y = 2y

Calculate curl:

∇ × A⃗ = î(0 - 0) - ĵ(0 - 0) + k̂(y - 2y)

∇ × A⃗ = k̂(-y) = -yk̂

∇ × A⃗ = -y k̂ (or -y âᵤ)

Part (2): A⃗ = -yz î + 4xy ĵ + y k̂

Identify Components:

Aₓ = -yz
Aᵧ = 4xy
Aᵤ = y

Calculate partial derivatives:

∂Aᵤ/∂y = 1, ∂Aᵧ/∂z = 0

∂Aᵤ/∂x = 0, ∂Aₓ/∂z = -y

∂Aᵧ/∂x = 4y, ∂Aₓ/∂y = -z

Calculate curl:

∇ × A⃗ = î(1 - 0) - ĵ(0 - (-y)) + k̂(4y - (-z))

∇ × A⃗ = î(1) - ĵ(y) + k̂(4y + z)

∇ × A⃗ = î - yĵ + (4y + z)k̂

Question 4: Maxwell's Fourth Equation

Question

State and explain Maxwell's fourth equation.

Statement

Maxwell's Fourth Equation (Ampere-Maxwell Law)

∇ × H⃗ = J⃗ + ∂D⃗/∂t

Differential (Point) Form

∮ H⃗ · dl⃗ = I + ∫∫ (∂D⃗/∂t) · dS⃗

Integral Form

Terms Explanation

Symbol	Name	Description
H⃗	Magnetic Field Intensity	Magnetic field strength (A/m)
J⃗	Conduction Current Density	Current per unit area due to moving charges (A/m²)
∂D⃗/∂t	Displacement Current Density	Time rate of change of electric displacement (A/m²)
D⃗	Electric Displacement	D⃗ = εE⃗ (C/m²)

Physical Significance

Key Interpretations:

Modified Ampere's Law: This equation is Maxwell's modification of Ampere's circuital law by adding the displacement current term.
Source of Magnetic Field: Magnetic fields are produced by both conduction currents (moving charges) AND time-varying electric fields.
Displacement Current: Even in the absence of actual charge movement, a changing electric field acts like a current and produces a magnetic field.
Electromagnetic Waves: This equation, combined with Faraday's law, predicts the existence of electromagnetic waves.

Displacement Current in a Capacitor

Displacement current maintains continuity of current through a capacitor gap

Alternative Forms

∇ × B⃗ = μ₀J⃗ + μ₀ε₀(∂E⃗/∂t)

In terms of B⃗ and E⃗ (in free space)

∇ × B⃗ = μ₀(J⃗ + J⃗ᵈ)

where J⃗ᵈ = ε₀(∂E⃗/∂t) is the displacement current density

Why Maxwell Added the Displacement Current

Original Ampere's law (∇ × H⃗ = J⃗) violated the continuity equation in time-varying fields. Consider a capacitor being charged:

Current flows in the wires but no conduction current exists between the plates
However, the electric field between plates is changing
Maxwell realized that ∂D⃗/∂t must act as an equivalent current to maintain continuity
This insight led to the prediction of electromagnetic waves

Question 5: Ampere's Circuital Law and Maxwell's Fourth Equation Derivation

Question

State Ampere's Circuital law. Derive Maxwell's fourth equation in differential form.

Ampere's Circuital Law

Statement:

The line integral of magnetic field intensity H⃗ around any closed path is equal to the total current enclosed by that path.

∮ H⃗ · dl⃗ = I_enclosed

Original Ampere's Circuital Law (Integral Form)

In terms of current density J⃗:

∮ H⃗ · dl⃗ = ∫∫ J⃗ · dS⃗

Derivation of Maxwell's Fourth Equation

Step-by-Step Derivation

Step 1: Start with Ampere's law in integral form:

∮ H⃗ · dl⃗ = ∫∫_S J⃗ · dS⃗

Step 2: Apply Stokes' theorem to the left side:

∮ H⃗ · dl⃗ = ∫∫_S (∇ × H⃗) · dS⃗

Step 3: Equating the two surface integrals:

∫∫_S (∇ × H⃗) · dS⃗ = ∫∫_S J⃗ · dS⃗

Step 4: Since this holds for any arbitrary surface S:

∇ × H⃗ = J⃗

Ampere's law in differential form (for steady currents)

Step 5: Check for consistency using continuity equation. Taking divergence of both sides:

∇ · (∇ × H⃗) = ∇ · J⃗

Since ∇ · (∇ × H⃗) = 0 (divergence of curl is always zero):

∇ · J⃗ = 0

Step 6: But the continuity equation states:

∇ · J⃗ = -∂ρ/∂t

This is non-zero for time-varying charge distributions! This inconsistency shows Ampere's law needs modification.

Step 7: From Gauss's law: ∇ · D⃗ = ρ

Therefore: ∂ρ/∂t = ∂(∇ · D⃗)/∂t = ∇ · (∂D⃗/∂t)

The continuity equation becomes:

∇ · J⃗ + ∇ · (∂D⃗/∂t) = 0

∇ · (J⃗ + ∂D⃗/∂t) = 0

Step 8: Maxwell modified Ampere's law by adding the displacement current term:

Maxwell's Fourth Equation (Ampere-Maxwell Law)

∇ × H⃗ = J⃗ + ∂D⃗/∂t

Differential Form

Now taking divergence:

∇ · (∇ × H⃗) = ∇ · J⃗ + ∇ · (∂D⃗/∂t) = 0

This is consistent with the continuity equation!

Summary of Maxwell's Equations

Equation	Differential Form	Physical Law
1st	∇ · D⃗ = ρ	Gauss's Law (Electric)
2nd	∇ · B⃗ = 0	Gauss's Law (Magnetic)
3rd	∇ × E⃗ = -∂B⃗/∂t	Faraday's Law
4th	∇ × H⃗ = J⃗ + ∂D⃗/∂t	Ampere-Maxwell Law