Electromagnetism Practice (Interactive)
Intuition
The force that powers civilisation: Electromagnetism is the force that powers modern civilisation — from the generators that produce electricity to the motors that drive machines to the antennas that transmit data. Understanding it is understanding how the modern world works.
Why it matters: Every electrical device, every wireless communication, and every computer chip relies on electromagnetic principles. Understanding circuits, fields, and waves is essential for electrical engineering and physics.
The key insight: Maxwell equations are a set of four equations that describe all electromagnetic phenomena — once you understand these four equations, you understand everything from static electricity to light propagation.
University Physics — Electromagnetism Practice
10 auto-graded practice problems at medium to hard difficulty. Select an answer, submit, and review the explanation.
Covers electrostatics (Gauss’s law, electric potential, capacitors), magnetostatics (Ampere’s law, Biot-Savart, magnetic materials), electrodynamics (Faraday’s law, Maxwell’s equations, electromagnetic waves), circuits (AC circuits, impedance, resonance), and EM waves (polarization, Poynting vector, reflection and refraction).
Electrostatics and Magnetostatics
Q1. A point charge of +3.0 micro-C is placed at the origin. What is the electric flux through a sphere of radius 0.20 m centered at the origin? (epsilon_0 = 8.85 x 10^-12 C^2/(N*m^2))
A. 3.39 x 10^5 Nm^2/C B. 1.69 x 10^5 Nm^2/C C. 6.78 x 10^5 Nm^2/C D. 8.49 x 10^4 Nm^2/C’,
Show answer — A
Answer: A — By Gauss’s law, the flux through any closed surface enclosing the charge is Phi = q / epsilon_0 = 3.0 x 10^-6 C / 8.85 x 10^-12 C^2/(Nm^2) = 3.39 x 10^5 Nm^2/C. The flux depends only on the enclosed charge, not on the radius of the Gaussian surface.
Q2. A parallel-plate capacitor has plates of area 0.020 m^2 separated by a 2.0 mm gap filled with a dielectric (kappa = 5.0). What is its capacitance?
A. 443 pF B. 88.5 pF C. 221 pF D. 886 pF
Show answer — A
Answer: A — Capacitance with dielectric: C = kappa epsilon_0 A / d = 5.0 8.85 x 10^-12 F/m 0.020 m^2 / 0.002 m = 5.0 8.85 x 10^-12 10 = 4.425 x 10^-10 F = 442.5 pF, rounding to 443 pF. The dielectric increases the capacitance by a factor of kappa = 5.0 compared to vacuum.
Q3. A long straight wire carries a current of 15 A. What is the magnetic field magnitude at a perpendicular distance of 5.0 cm from the wire? (mu_0 = 4pi x 10^-7 T*m/A)
A. 30 micro-T B. 120 micro-T C. 60 micro-T D. 15 micro-T
Show answer — C
Answer: C — By Ampere’s law for a long straight wire: B = mu_0 * I / (2pir) = (4pi x 10^-7 T*m/A 15 A) / (2pi 0.050 m) = (6pi x 10^-7) / (0.1pi) = 6.0 x 10^-5 T = 60 micro-T. The field circles the wire, falling as 1/r with distance.
Q4. A solenoid of length 0.30 m and radius 0.05 m has 500 turns carrying 4.0 A. A second coil of 10 turns is wound around the solenoid at its center. What is the mutual inductance between them?
A. 82 micro-H B. 330 micro-H C. 165 micro-H D. 41 micro-H
Show answer — C
Answer: C — Mutual inductance: M = mu_0 N_1 N_2 * A / L, where A = pir^2 = pi(0.05)^2 = 7.854 x 10^-3 m^2. M = (4pi x 10^-7)(500)(10)(7.854 x 10^-3) / 0.30 = (4pi x 10^-7)(5000)(7.854 x 10^-3) / 0.30 = 1.645 x 10^-4 H = 165 micro-H. This mutual inductance means a changing current in the solenoid induces an EMF in the outer coil.
Electrodynamics and Maxwell’s Equations
Q5. Which of Maxwell’s equations describes how a time-varying magnetic field generates an electric field?
A. “Faraday’s law: the line integral of E around a closed loop equals the negative rate of change of magnetic flux”, “Gauss’s law: the flux of E through a closed surface equals the enclosed charge divided by epsilon_0”, “Ampere’s law with Maxwell’s correction: the line integral of B equals mu_0 times the enclosed current plus displacement current”, “Gauss’s law for magnetism: the flux of B through any closed surface is zero”,
Show answer — A
Answer: A — Faraday’s law (integral form: contour integral of E dot dl = -dPhi_B/dt) states that a changing magnetic flux induces a circulating electric field. This is the operating principle behind electric generators, transformers, and inductors. Unlike electrostatic fields, the induced E field is non-conservative.
Q6. A circular loop of radius 0.10 m sits in a uniform magnetic field of 0.50 T perpendicular to the loop plane. If the field drops to zero in 0.050 s, what is the magnitude of the induced EMF?
A. 0.16 V B. 0.63 V C. 0.31 V D. 1.57 V
Show answer — C
Answer: C — By Faraday’s law: |epsilon| = dPhi/dt = d(BA)/dt = A * dB/dt = pir^2 delta_B / delta_t = pi(0.10)^2 0.50 / 0.050 = 3.142 x 10^-2 10 = 0.314 V, rounding to 0.31 V. The rapid collapse of the magnetic field induces a significant EMF in the loop.
Q7. An electromagnetic wave in vacuum has an electric field amplitude of 48 V/m. What is the average intensity (power per unit area) carried by the wave?
A. 6.12 W/m^2 B. 1.53 W/m^2 C. 3.06 W/m^2 D. 12.2 W/m^2
Show answer — C
Answer: C — Average intensity: I = (1/2) epsilon_0 c E_0^2 = 0.5 8.854 x 10^-12 3.00 x 10^8 (48)^2 = 0.5 8.854 x 10^-12 3.00 x 10^8 * 2304 = 3.06 W/m^2. Alternatively, I = E_0B_0/(2mu_0) where B_0 = E_0/c, giving the same result.
Q8. In a plane electromagnetic wave propagating in vacuum, the electric and magnetic field amplitudes are related by E_0 = c*B_0. If E_0 = 120 V/m, what is the magnetic field amplitude and the wavelength when f = 5.0 x 10^14 Hz?
A. B_0 = 4.0 x 10^-7 T and lambda = 600 nm B. B_0 = 4.0 x 10^-7 T and lambda = 300 nm C. B_0 = 8.0 x 10^-7 T and lambda = 600 nm D. B_0 = 2.0 x 10^-7 T and lambda = 600 nm’,
Show answer — A
Answer: A — B_0 = E_0/c = 120/(3.0 x 10^8) = 4.0 x 10^-7 T = 400 nT. Wavelength: lambda = c/f = 3.0 x 10^8 / 5.0 x 10^14 = 6.0 x 10^-7 m = 600 nm (visible orange light). In vacuum, E and B oscillate in phase, are perpendicular to each other and to the propagation direction.
Circuits and Electromagnetic Waves
Q9. A series RLC circuit has R = 100 Ohm, L = 50 mH, and C = 20 micro-F. At what frequency does the circuit resonate?
A. 159 Hz B. 318 Hz C. 79.6 Hz D. 100 Hz
Show answer — A
Answer: A — Resonant frequency: f_0 = 1 / (2pisqrt(LC)) = 1 / (2pisqrt(50 x 10^-3 * 20 x 10^-6)) = 1 / (2pisqrt(1.0 x 10^-6)) = 1 / (2pi1.0 x 10^-3) = 159 Hz. At resonance, the impedance is purely resistive (Z = R = 100 Ohm), the current is maximum, and the reactances cancel: X_L = X_C.
Q10. A light beam travels from air (n = 1.0) into glass (n = 1.5) at an angle of incidence of 30 degrees. What is the angle of refraction inside the glass?
A. 45.0 degrees B. 20.0 degrees C. 19.5 degrees D. 30.0 degrees
Show answer — C
Answer: C — By Snell’s law: n_1sin(theta_1) = n_2sin(theta_2). sin(theta_2) = (n_1/n_2)*sin(theta_1) = (1.0/1.5)*sin(30 degrees) = (2/3)*0.500 = 0.333. theta_2 = arcsin(0.333) = 19.47 degrees, rounding to 19.5 degrees. Light bends toward the normal when entering a denser medium.
Q11. A linearly polarized EM wave (E_0 = 60 V/m) passes through three successive polarizers. The transmission axes are at 0 degrees, 30 degrees, and 60 degrees relative to the initial polarization. What is the intensity after the third polarizer?
A. 4.78 W/m^2 B. 1.19 W/m^2 C. 2.69 W/m^2 D. 0.30 W/m^2
Show answer — C
Answer: C — Apply Malus’s law at each polarizer. Initial intensity: I_0 = E_0^2/(2mu_0c) = 3600/(24pi10^-7310^8) = 4.775 W/m^2. After polarizer 1 (0 degrees): I_1 = I_0cos^2(0) = I_0. After polarizer 2 (30 degrees from I_1 polarization): I_2 = I_0cos^2(30) = I_03/4. After polarizer 3 (30 degrees from I_2 polarization): I_3 = I_0(3/4)cos^2(30) = I_09/16 = 4.775*0.5625 = 2.686 W/m^2 = 2.69 W/m^2.
Common Mistakes
Using cos(θ) instead of cos²(θ) in Malus’s law: Transmitted intensity goes as cos²(θ), not cos(θ). Forgetting the square gives intensity values that are too high.
Confusing inductive and capacitive reactance signs: X_L = ωL is positive; X_C = 1/(ωC) is negative in impedance. Adding them with the wrong sign gives incorrect resonance conditions.
Forgetting that light bends toward the normal when entering a denser medium: Snell’s law gives n₁sin(θ₁) = n₂sin(θ₂). When n₂ > n₁, θ₂ < θ₁ — the refracted ray bends toward the normal, not away from it.
See Also
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.
Advanced Content
This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.
Derivations and Proofs
Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.
Extended Examples
Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.
Research Connections
This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.
Prerequisites
Ensure you have mastered the prerequisite material before attempting this advanced content.