Quizzes · Electromagnetism

Integrative questions across the course — every concept page carries its own quiz too. Follow the lecture timeline to study in order.

Conceptual

Q1. Why does a negative charge move opposite to electric field lines?

Answer

Field direction is defined by the force on a positive test charge (\(\vec F = q\vec E\)); flip the sign of \(q\) and the force flips.

Q2. Why can two equipotential surfaces never cross?

Answer

A crossing point would have two different potentials at once — contradiction. (Equivalently: \(\vec E = -\nabla V\) would be undefined there.)

Q3. Why does changing magnetic flux induce an EMF, even with no battery anywhere?

Answer

Faraday's law: \(\mathcal{E} = -d\Phi_B/dt\) — a changing \(\vec B\) creates a circulating \(\vec E\) field, which drives charges around the loop.

Q4. A capacitor is charged, disconnected, and a dielectric is slid in. What happens to \(Q\), \(V\), and the stored energy?

Answer

\(Q\) fixed (isolated); \(C \to \kappa C\) so \(V = Q/C\) drops by \(\kappa\); \(U = Q^2/2C\) drops by \(\kappa\) too — the missing energy went into pulling the dielectric in.

Computational

Q5. Flux through a \(0.2\ \text{m}^2\) loop tilted \(60°\) from a uniform 0.3 T field?

Answer

\(\Phi = BA\cos\theta = 0.3\times0.2\times\cos 60° = 0.03\) Wb.

Q6. RC time constant for \(R = 5\ \text{k}\Omega\), \(C = 2\ \mu\text{F}\) — and how long to reach 95% charge?

Answer

\(\tau = RC = 10\) ms. 95% needs \(1 - e^{-t/\tau} = 0.95 \Rightarrow t = 3\tau = 30\) ms.

Q7. Two protons are 1 nm apart. Compare the electric repulsion with their gravitational attraction.

Answer

\(F_E = k e^2/r^2 \approx 2.3\times10^{-10}\) N; \(F_G = G m_p^2/r^2 \approx 1.9\times10^{-46}\) N. Ratio \(\sim 10^{36}\) — why gravity is irrelevant inside atoms.

Multiple choice

Q8. Unit of electric flux: (a) N/C (b) V/m (c) N·m²/C

Answer

(c) — field × area. (Equivalently V·m.)

Q9. Inside a long solenoid with \(n = 500\) turns/m carrying \(i = 2\) A, \(B\) equals: (a) \(\mu_0 n i\) (b) \(\mu_0 i/(2\pi n)\) (c) \(\mu_0 n^2 i\)

Answer

(a) — from Ampère's law; numerically \(B = 4\pi\times10^{-7}\times500\times2 \approx 1.3\) mT.

Q10. EMF induced in \(N = 50\) turns by \(\Phi = 0.1\sin(\pi t)\) Wb: (a) \(-5\pi\cos(\pi t)\) V (b) \(-10\pi\cos(\pi t)\) V (c) \(-50\pi\cos(\pi t)\) V

Answer

(a) — \(\mathcal{E} = -N\,d\Phi/dt = -50 \times 0.1\pi\cos(\pi t) = -5\pi\cos(\pi t)\) V.