PHYS7021
Burdwan · Electricity, Magnetism & Wave Optics · 2018–2024

X · GE-2 Borderline PYQs — Fully Solved (Confirm with Lecturer)

All 7 borderline GE-2 questions with full numeric solutions. Outside PHYS7021 text — confirm with lecturer.

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These 7 questions (B1–B7) appeared in old GE-2 papers (2018, 2019, 2022, 2023) but are not verbatim in the PHYS7021 syllabus text. Your lecturer may or may not ask them. Revise them only after the core magnetostatics and EMI pages are strong. Every solution below is fully worked in SI units.

Outside PHYS7021. These questions belong to old GE-2 papers. Confirm with your lecturer whether they will be set on the PHYS7021 exam before prioritising them.

Constants: $e = 1.6\times 10^{-19}$ C, $\mu_0 = 4\pi\times 10^{-7}$ H/m, $\varepsilon_0 = 8.85\times 10^{-12}$ F/m.

Solved borderline PYQs

B12018

Dipole torque numeric: length 3 cm, charges $5\,\mu$C, $E = 2\times 10^4$ N/C → max and min torque.

Recall

Dipole moment $p = q\,d$ (C·m). Torque $\vec{\tau} = \vec{p}\times\vec{E}$, magnitude $|\tau| = pE\sin\theta$.

Steps

  1. Dipole moment: $q = 5\times 10^{-6}$ C, $d = 0.03$ m, so $p = 5\times 10^{-6}\times 0.03 = 1.5\times 10^{-7}$ C·m.
  2. Maximum torque at $\theta = 90°$, $\sin\theta = 1$: $\tau_{max} = pE = 1.5\times 10^{-7}\times 2\times 10^4 = 3.0\times 10^{-3}$ N·m.
  3. Minimum torque at $\theta = 0°$, $\sin 0 = 0$: $\tau_{min} = 0$.

Answer

$\boxed{\,\tau_{max} = 3.0\times 10^{-3}\,\text{N·m}\,}$ (broadside-on), $\boxed{\,\tau_{min} = 0\,}$ (aligned).

Exam tip

State both angles: max at $90°$, min at $0°$. Half the marks are for the angle.
Dipole p at angle to E with torque p cross E
Fig B1. Dipole $\vec{p}$ in field $\vec{E}$; torque $\vec{\tau} = \vec{p}\times\vec{E}$ twists it toward alignment.

B22023

Explain what happens when an electric dipole is placed in a uniform electric field.

Recall

A uniform $\vec{E}$ exerts equal and opposite forces on $+q$ and $-q$. The two forces cancel as a translation; their lines of action differ, so a torque remains.

Steps

  1. Net force: $\vec{F} = q\vec{E} + (-q\vec{E}) = \vec{0}$. The centre of the dipole does not translate in a uniform field.
  2. Net torque: $\vec{\tau} = \vec{p}\times\vec{E}$, with $|\tau| = pE\sin\theta$. It rotates the dipole toward $\theta = 0$ (alignment with the field).
  3. At $\theta = 0$ the torque is zero (stable equilibrium); at $\theta = 180°$ it is zero but unstable.
  4. A small disturbance from $\theta = 0$ gives torsional SHM (like a pendulum).

Answer

In a uniform field, an electric dipole has zero translation but a torque $\boxed{\,\vec{\tau} = \vec{p}\times\vec{E}\,}$ that aligns it with the field; stable equilibrium at $\theta = 0$.

Exam tip

One-line summary: "uniform → no translation, only rotation."

B32023

Dipole ($+1\,\mu$C, $-1\,\mu$C, 2 cm) in field $2.5\times 10^4$ N/C → torque to rotate it by 30°.

Recall

$|\tau| = pE\sin\theta$ with $\theta$ the angle between $\vec{p}$ and $\vec{E}$.

Steps

  1. Dipole moment: $p = 1\times 10^{-6}\times 0.02 = 2.0\times 10^{-8}$ C·m.
  2. Torque at 30°: $\tau = 2.0\times 10^{-8}\times 2.5\times 10^4\times \sin 30°$.
  3. Compute: $\sin 30° = 0.5$, so $\tau = 2.0\times 10^{-8}\times 2.5\times 10^4\times 0.5 = 2.5\times 10^{-4}$ N·m.

Answer

$\boxed{\,\tau = 2.5\times 10^{-4}\,\text{N·m}\,}$.

Exam tip

Write $\sin 30° = 0.5$ before multiplying — keeps the calculation transparent.

B42019

Lorentz force numeric: $q = 1.6\times 10^{-19}$ C, $\vec{v} = 3\hat{i}+2\hat{j}$ m/s, $\vec{E} = 6\hat{i}+6\hat{j}+3\hat{k}$ V/m, $\vec{B} = \hat{j}+2\hat{k}$ T.

Recall

Lorentz force: $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$. Expand the cross product as a $3\times 3$ determinant; remember the minus sign on the $\hat{j}$ component.

Steps

  1. Cross product $\vec{v}\times\vec{B}$: write $\vec{v}=(3,2,0)$ and $\vec{B}=(0,1,2)$. Determinant gives $\hat{i}(2\cdot2-0\cdot1) - \hat{j}(3\cdot2-0\cdot0) + \hat{k}(3\cdot1-2\cdot0) = (4,-6,3)$.
  2. So $\vec{v}\times\vec{B} = 4\hat{i} - 6\hat{j} + 3\hat{k}$ (units V/m — an effective electric field).
  3. Add $\vec{E}$: $\vec{E} + \vec{v}\times\vec{B} = (6+4)\hat{i} + (6-6)\hat{j} + (3+3)\hat{k} = 10\hat{i} + 0\hat{j} + 6\hat{k}$ V/m.
  4. Multiply by $q$: $\vec{F} = 1.6\times 10^{-19}(10\hat{i} + 6\hat{k}) = 1.6\times 10^{-18}\hat{i} + 9.6\times 10^{-19}\hat{k}$ N.
  5. Magnitude: $|\vec{F}| = \sqrt{(1.6\times 10^{-18})^2 + (9.6\times 10^{-19})^2} \approx 1.87\times 10^{-18}$ N.

Answer

$\boxed{\,\vec{F} = (1.6\times 10^{-18})\hat{i} + (9.6\times 10^{-19})\hat{k}\,\text{N}\,}$, magnitude $\approx 1.9\times 10^{-18}$ N.

Exam tip

Expand the $3\times 3$ determinant line by line — the minus sign on the $\hat{j}$ component is where marks are lost.

B52022

What is the work done by a magnetic field on a moving charge?

Recall

The magnetic force $\vec{F}_B = q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$. A force perpendicular to velocity does no work.

Steps

  1. Power delivered by the magnetic force: $P = \vec{F}_B\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$ (scalar triple product with a repeated vector is zero).
  2. Integrate over time: $W = \int P\,dt = 0$.
  3. Conclusion: the magnetic field changes the direction of motion but never the speed or kinetic energy.

Answer

$\boxed{\,W_B = 0\,}$ — a magnetic field does no work on a moving charge. It only bends the path.

Exam tip

Contrast with the electric force $\vec{F} = q\vec{E}$, which can do work. One line: "B bends, E speeds up."

B62022

Force between two parallel wires (10 cm long, 2 cm apart, currents 20 A and 30 A).

Recall

Field of wire 1 at wire 2: $B = \mu_0 I_1/2\pi d$. Force on length $l$ of wire 2: $F = I_2\,l\,B$. Parallel currents attract; antiparallel currents repel.

Steps

  1. Combined formula: $F = \mu_0 I_1 I_2\,l/(2\pi d)$ with $\mu_0/2\pi = 2\times 10^{-7}$ (in SI).
  2. Convert: $l = 0.10$ m, $d = 0.02$ m.
  3. Substitute: $F = 2\times 10^{-7}\times 20\times 30\times 0.10/0.02$.
  4. Compute: $20\times 30 = 600$, $0.10/0.02 = 5$, so $F = 2\times 10^{-7}\times 600\times 5 = 6.0\times 10^{-4}$ N.

Answer

$\boxed{\,F = 6.0\times 10^{-4}\,\text{N, attractive (parallel currents)}\,}$.

Exam tip

State "attractive because currents are parallel; antiparallel would repel" — that line earns the direction mark.
Two parallel wires 2 cm apart carrying currents 20 A and 30 A in the same direction; arrows of attraction
Fig B6. Parallel currents $I_1$, $I_2$ separated by $d$ attract with $F/l = \mu_0 I_1 I_2/(2\pi d)$.

B72023

Parallel-plate capacitor ($A = 0.25$ m², $d = 1$ cm, 10 V battery) → force of attraction between plates.

Recall

Each plate sits in the field of the other plate only, $E = \sigma/2\varepsilon_0$. For fixed voltage $V$ held by the battery, $F = \varepsilon_0 A V^2/(2 d^2)$.

Steps

  1. Field of a single sheet: $E_{one} = \sigma/(2\varepsilon_0)$. A plate does not pull on its own field, so the factor is $1/2$.
  2. Fixed-voltage formula: with $V$ held constant, $F = \varepsilon_0 A V^2/(2 d^2)$.
  3. Convert: $V = 10$ V, $V^2 = 100$, $d = 0.01$ m, $d^2 = 10^{-4}$ m², $A = 0.25$ m².
  4. Numerator: $\varepsilon_0 A V^2 = 8.85\times 10^{-12}\times 0.25\times 100 = 2.2125\times 10^{-10}$. Denominator: $2\times 10^{-4}$.
  5. Ratio: $F = 2.2125\times 10^{-10}/2\times 10^{-4} \approx 1.11\times 10^{-6}$ N.

Answer

$\boxed{\,F \approx 1.1\times 10^{-6}\,\text{N (attractive)}\,}$.

Exam tip

Never use the full $E = \sigma/\varepsilon_0$ to find the force on a plate — always halve it, or use the boxed fixed-$V$ formula.

Reference. Brij Lal & Subrahmanyam Ch-1, 4, 7; H.C. Verma Vol-1 Ch-29, 31, 34, 35; S.L. Arora Vol-1 Ch-1, 2, 4.

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