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CodeVID-P12-04-DCM-01
Displacement Current & Maxwell's Equations — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The displacement current arises due to a changing:
- A.magnetic flux
- B.electric flux
- C.charge at rest
- D.resistance
2.
In a charging capacitor, the displacement current between the plates is:
- A.zero
- B.equal to the conduction current
- C.twice the conduction current
- D.infinite
3.
Gauss's law for magnetism states that $\oint \vec{B}\cdot d\vec{A}$ equals:
- A.$\frac{q}{\epsilon_0}$
- B.$\mu_0 I$
- C.0
- D.$\mu_0\epsilon_0$
4.
EM waves are radiated by:
- A.a static charge
- B.an accelerating charge
- C.a constant current
- D.a permanent magnet
5.
The Ampère–Maxwell law connects the magnetic field to current and to a changing:
- A.temperature
- B.electric field
- C.mass
- D.potential energy
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define displacement current and write its expression.
7.
A capacitor's electric flux changes at $4\times10^{8}\ \text{V·m/s}$. Find the displacement current ($\epsilon_0=8.85\times10^{-12}\ \text{F/m}$).
8.
Why was Ampère's original law found to be inconsistent for a charging capacitor?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
State the four Maxwell equations in integral form and name each.
10.
A capacitor of $50\ \mu\text{F}$ has its voltage rising at $1500\ \text{V/s}$. Find the displacement current.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain, using the charging-capacitor argument, why Maxwell introduced the displacement current, and write the modified Ampère–Maxwell law. State how a changing field of one kind produces the other.
Answer Key
Section A — Multiple Choice Questions
- (B) electric flux
- (B) equal to the conduction current
- (C) 0
- (B) an accelerating charge
- (B) electric field
Section B — Short Answer (2 marks)
- The current associated with a changing electric flux; $I_d=\epsilon_0\frac{d\Phi_E}{dt}$. It produces a magnetic field even where no charge moves.
- $I_d=\epsilon_0\frac{d\Phi_E}{dt}=8.85\times10^{-12}\times4\times10^{8}\approx3.54\times10^{-3}\ \text{A}$.
- Different surfaces bounded by the same loop gave different enclosed currents (one through the wire, one through the gap), so the law gave two answers until the displacement current was added.
Section C — Short Answer (3 marks)
- Gauss (E): $\oint \vec{E}\cdot d\vec{A}=\frac{q}{\epsilon_0}$; Gauss (B): $\oint \vec{B}\cdot d\vec{A}=0$; Faraday: $\oint \vec{E}\cdot d\vec{l}=-\frac{d\Phi_B}{dt}$; Ampère–Maxwell: $\oint \vec{B}\cdot d\vec{l}=\mu_0 I+\mu_0\epsilon_0\frac{d\Phi_E}{dt}$.
- $I_d=C\frac{dV}{dt}=50\times10^{-6}\times1500=0.075\ \text{A}$.
Section D — Long Answer (5 marks)
- For a loop around the wire, a flat surface encloses conduction current $I$, but a surface dipping between the plates encloses no charge — Ampère's law gives two answers. Maxwell added $I_d=\epsilon_0\frac{d\Phi_E}{dt}$, which equals $I$ in the gap, giving $\oint \vec{B}\cdot d\vec{l}=\mu_0 I+\mu_0\epsilon_0\frac{d\Phi_E}{dt}$. Thus a changing electric field produces a magnetic field (Ampère–Maxwell) and a changing magnetic field produces an electric field (Faraday) — the basis of EM waves.
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