Q.26 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 26 Solution

Topic: Electromagnetic Waves · Official NTA Booklet Code 45

Question

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

✓

Verified Against NTA Final Answer Key

Option (3) : non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

📚 Electromagnetic Waves Medium

📋 Given Parameters & Problem Setup

Capacitor Geometry Circular parallel plates of radius R

Charging Rate Surface charge density σ(t) increases at a constant rate with time

Target Property Spatial variation of induced magnetic field B(r) arising from displacement current

⚡

The 30-Second Exam Hall Shortcut: ⚡ Ampère-Maxwell Cylindrical Peak Rule

⏱️ Save ~50 seconds

Inside plates (r ≤ R): B ∝ r (linear growth); Outside plates (r ≥ R): B ∝ 1/r (hyperbolic decay) ⟹ B is MAXIMUM at the perimeter (r = R)!

1 Displacement current density J_d = ε₀ (dE/dt) is uniform across the circular plates of radius R.

2 Inside the plates (r ≤ R): Ampèrian loop of radius r encloses flux ∝ r² ⟹ B(2πr) = μ₀ J_d (πr²) ⟹ B(r) ∝ r.

3 Outside the plates (r ≥ R): Ampèrian loop encloses the entire total displacement current I_d ⟹ B(2πr) = μ₀ I_d ⟹ B(r) ∝ 1/r.

4 Thus B(r) starts from 0 at the center (r = 0), increases linearly to a MAXIMUM at the rim (r = R), and then decays as 1/r outside.

5 Therefore, B is non-zero everywhere with its maximum at the cylindrical surface connecting the peripheries (r = R)!

6 Select Option (3) in under 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Ampère-Maxwell Law (NCERT Physics · Electromagnetic Waves: Displacement Current I_d = ε₀ dΦ_E/dt).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Displacement Current Density Between the Plates

For circular parallel plates of radius R, the electric field between the plates is E = σ / ε₀. Since dσ/dt is constant, the time derivative of the electric field is uniform:

\frac{dE}{dt} = \frac{1}{\varepsilon_0} \frac{d\sigma}{dt} = \text{constant}

• Displacement current density J_d = ε₀ (dE/dt) is uniform throughout the cylindrical region between the plates.

Step 2

Magnetic Field Inside the Plates (r ≤ R)

Construct a circular Ampèrian loop of radius r ≤ R coaxial with the plates. Applying the Ampère-Maxwell law:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enclosed} \implies B(r) \cdot 2\pi r = \mu_0 \left( \varepsilon_0 \pi r^2 \frac{dE}{dt} \right) \implies B(r) = \frac{\mu_0 \varepsilon_0 r}{2} \frac{dE}{dt} \propto r

• The magnetic field grows linearly from zero on the central axis (r = 0) up to the edge r = R.

Step 3

Magnetic Field Outside the Plates (r ≥ R) & Maximum

For an Ampèrian loop of radius r ≥ R, the enclosed displacement current is the full total displacement current I_d = ε₀ π R² (dE/dt):

B(r) \cdot 2\pi r = \mu_0 I_d \implies B(r) = \frac{\mu_0 \varepsilon_0 R^2}{2r} \frac{dE}{dt} \propto \frac{1}{r}

• Because B increases for r ≤ R and decreases for r ≥ R, it attains its global maximum at r = R (the peripheral boundary).
• Thus B is non-zero everywhere and reaches maximum at the imaginary cylindrical surface connecting the peripheries, matching Option (3).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Maxwell discovered displacement current precisely because a changing electric field generates a non-zero magnetic field.

⚠️

Student Fallacy (Why candidates pick this trap): Believing that because no physical conduction electrons cross the vacuum gap, B must be zero.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. B grows linearly with r inside (B ∝ r) and decays as 1/r outside; it is neither constant nor zero outside.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing electric field (which is uniform) with induced magnetic field.

3 C

✓ 100% Correct

Verdict & Scientific Reason:

100% exact. B(r) ∝ r for r ≤ R and B(r) ∝ 1/r for r ≥ R, peaking at r = R.

💡

Why this is airtight: None; canonical Maxwell-Ampère electrodynamics.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. B is non-zero throughout the interior except on the singular line r = 0.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting spatial domains.

⚠️

NTA Trap Alert: Confusing Electric Field with Induced Magnetic Field

The electric field E between the plates is uniform (constant in space). But the INDUCED magnetic field B is created by enclosed displacement flux, so it grows with radius (B ∝ r) and peaks at the perimeter.

🛡️ How to avoid the -1 mark: Remember: E is uniform; B forms concentric loops that grow linearly to the edge!

📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Electromagnetic Waves Standard Benchmark

Core Scientific Principle Tested:

"Magnetic field inside charging capacitor is B(r) ∝ r (r ≤ R) and B(r) ∝ 1/r (r ≥ R), peaking at r = R."

Syllabus Focus: Displacement Current & Ampère-Maxwell Law

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Inside: B grows with r; Outside: B dies as 1/r ⟹ PEAK at the rim (r = R)!

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NEET 2025 Physics Question 26 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Ampère-Maxwell Cylindrical Peak Rule

Step-by-Step Rigorous Workflow & Mathematical Proof

Displacement Current Density Between the Plates

Magnetic Field Inside the Plates (r ≤ R)

Magnetic Field Outside the Plates (r ≥ R) & Maximum

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing Electric Field with Induced Magnetic Field

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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EM Wave in Dielectric Medium: Dual-Key Question Audit (Both Options 2 and 3 Accepted by NTA)

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