Course: University Physics 212: Electricity and Magnetism | University of Illinois at Urbana-Champaign
Source: Final Assessment review material
Difficulty: Intermediate Prerequisites: Electrostatics and Gauss's Law (Part 1), Electric Potential and Circuits (Part 2), comfort with cross products and integration
Tags: magnetic force, Lorentz force, right-hand rule, magnetic field, solenoid, Ampere's law, Biot-Savart, Faraday's law, electromagnetic induction, motional EMF, Lenz's law, current density, PHYS 212
Magnetism is the second major pillar of PHYS 212. Where electrostatics deals with charges at rest, magnetism arises from charges in motion. Moving charges experience magnetic forces, and currents create magnetic fields. Electromagnetic induction, the final piece, shows that changing magnetic fields produce electric fields (and therefore EMFs), closing the loop between electricity and magnetism. This is the material that leads into electromagnetic waves and Maxwell's equations in more advanced courses.
A moving charge in a magnetic field experiences a force perpendicular to both its velocity and the field (use the right-hand rule). Currents produce magnetic fields described by Ampere's law. Changing magnetic flux through a loop induces an EMF (Faraday's law), and the induced current always opposes the change that created it (Lenz's law).
Magnetic force on a moving charge
F = qv × B. The force is perpendicular to both the velocity and the magnetic field. In simple terms, the magnetic field deflects moving charges sideways without speeding them up or slowing them down.
Right-hand rule
Point your fingers in the direction of v, curl them toward B, and your thumb points in the direction of F (for a positive charge). For a negative charge, the force is in the opposite direction.
Magnetic field of a solenoid
Inside a long solenoid with n turns per unit length carrying current I: B = μ₀nI. The field is uniform and parallel to the axis. Think of a solenoid as the magnetic equivalent of a parallel-plate capacitor: a region of nearly uniform field.
Ampere's law
The line integral of B around a closed loop equals μ₀ times the enclosed current: ∮ B · dl = μ₀I_enc. This is the magnetic analogue of Gauss's law, useful when current distributions have high symmetry.
Current density (J)
Current per unit area: J = I/A for uniform current, or more generally I = ∫J · dA. Measured in A/m². When J varies with position (e.g. J = J₀(r/R) inside a cylinder), you integrate over the cross-section to find total current.
Faraday's law of induction
The induced EMF in a loop equals the negative rate of change of magnetic flux through the loop: EMF = −dΦ_B/dt. In simple terms, a changing magnetic environment creates a voltage.
Motional EMF
When a conductive rod of length L moves with velocity v perpendicular to a uniform field B, the EMF across the rod is: EMF = vBL. This is a special case of Faraday's law.
Lenz's law
The direction of the induced current is always such that it opposes the change in magnetic flux that produced it. This is the negative sign in Faraday's law and is a consequence of conservation of energy.
The magnetic force on a charge q moving with velocity v in a field B is:
F = qv × B
Magnitude: F = qvB sin θ, where θ is the angle between v and B.
Key consequences:
If v is parallel to B (θ = 0° or 180°), the force is zero. The charge travels in a straight line.
If v is perpendicular to B (θ = 90°), the force is maximum: F = qvB. The charge moves in a circle.
The magnetic force never does work (it is always perpendicular to v), so it changes the direction of motion but not the speed.
Example: A proton moves north, B points vertically upward (out of the ground). Using the right-hand rule: fingers point north, curl up (toward B coming out of ground). Wait, let's be precise. v = north, B = up. v × B: point fingers north, curl them upward. The cross product v × B points west. Since the proton is positive, F = qv × B points west. The force on the proton is toward the west.
Actually, let's redo this carefully with coordinates. Let north = ŷ, up = ẑ. Then v × B = ŷ × ẑ = x̂, which is east. Hmm, but the answer needs care. Let's define: north = +ŷ, east = +x̂, up = +ẑ. Then v = vŷ, B = Bẑ. F = q(v × B) = q(vŷ × Bẑ) = qvB(ŷ × ẑ) = qvB x̂. That is east.
So the magnetic force on the proton is toward the east.
Inside a very long solenoid with n turns per unit length and current I:
B = μ₀nI
The field is uniform inside and essentially zero outside.
This result comes from Ampere's law applied to a rectangular loop with one side inside and one side outside the solenoid.
μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space).
Note the distinction: the field of a solenoid depends on n (turns per unit length) and I, not on the solenoid's radius.
Ampere's law: ∮ B · dl = μ₀I_enc
For a long cylindrical conductor of radius R carrying non-uniform current density J = J₀(r/R):
Total current through the conductor: Integrate J over the cross-section.
I = ∫₀ᴿ J · dA = ∫₀ᴿ J₀(r/R) · 2πr dr = (2πJ₀/R) ∫₀ᴿ r² dr = (2πJ₀/R)(R³/3) = 2πJ₀R²/3
Magnetic field at r < R (inside the conductor): Find the enclosed current through a circle of radius r.
I_enc = ∫₀ʳ J₀(r'/R) · 2πr' dr' = (2πJ₀/R) ∫₀ʳ r'² dr' = (2πJ₀/R)(r³/3) = 2πJ₀r³/(3R)
By Ampere's law: B(2πr) = μ₀I_enc = μ₀ · 2πJ₀r³/(3R)
B = μ₀J₀r²/(3R)
The field grows as r² inside the conductor (because the current density increases with r).
EMF = −dΦ_B/dt, where Φ_B = ∫B · dA is the magnetic flux through the loop.
For a flat loop of area A in a uniform field B perpendicular to the loop: Φ_B = BA.
If B changes at a constant rate dB/dt:
EMF = −A(dB/dt)
Magnitude: |EMF| = A|dB/dt|
Example: Circular loop of radius 0.1 m, dB/dt = 2.0 T/s, B perpendicular to the loop.
A = πr² = π(0.1)² = 0.01π m²
|EMF| = (0.01π)(2.0) = 0.02π ≈ 0.0628 V
A conducting rod of length L moving with velocity v perpendicular to a uniform field B generates:
EMF = vBL
This can be understood as charges in the rod experiencing a magnetic force (F = qvB) that separates positive and negative charges to opposite ends, creating a potential difference.
The formula comes directly from Faraday's law: as the rod moves, it sweeps out area at rate dA/dt = Lv, so dΦ/dt = BLv.
Quantity | Formula | Notes |
|---|---|---|
Magnetic force | F = qv × B | |
Solenoid field | B = μ₀nI | n = turns per unit length |
Ampere's law | ∮ B · dl = μ₀I_enc | Useful with cylindrical symmetry |
Total current (non-uniform J) | I = ∫J · dA | Integrate over cross-section |
Faraday's law | EMF = −dΦ_B/dt | Φ_B = ∫B · dA |
Motional EMF | EMF = vBL | Rod perpendicular to B and v |
Permeability of free space | μ₀ = 4π × 10⁻⁷ T·m/A |
Electric generators work on Faraday's law: rotating a coil in a magnetic field produces a changing flux and therefore an EMF. This is how power plants (coal, gas, nuclear, hydro, wind) convert mechanical energy into electrical energy. MRI machines use powerful solenoids to create the uniform magnetic field needed to image the body. Electric motors are essentially generators run in reverse, using current in a magnetic field to produce torque.
Students often think the magnetic force can speed up a charge. It cannot. Because F is always perpendicular to v, the magnetic force changes direction but does zero work. Speed stays constant.
A common error on cross-product direction: forgetting to reverse the force direction for negative charges. The right-hand rule gives the direction for positive charges; for electrons, flip it.
Students sometimes confuse the solenoid formula (B = μ₀nI) with the formula for the field around a long straight wire (B = μ₀I/(2πr)). The solenoid field is uniform inside and depends on n, not on distance from the axis.
In Faraday's law problems, students forget that it is the rate of change of flux that matters, not the flux itself. A large constant flux produces zero EMF.
⚠️ If a charge moves parallel to the magnetic field, the magnetic force is zero. This is tested frequently with simple-sounding questions.
⚠️ Right-hand rule direction problems are worth practising until they are automatic. Set up coordinates explicitly (north, east, up) rather than guessing.
⚠️ The solenoid field formula B = μ₀nI is distinct from the wire formula B = μ₀I/(2πr). Know which applies.
⚠️ Non-uniform current density problems (J = J₀r/R) require integration to find enclosed current. These appear in long-answer sections and carry significant marks.
⚠️ Motional EMF = vBL. Know how to derive it from Faraday's law (area swept per unit time) and from the force on charges in the rod.
⚠️ For Faraday's law calculations, always compute area first (πr² for a circular loop), then multiply by dB/dt.
True or false: The magnetic force on a moving charge can change the charge's speed.
False. The force is always perpendicular to velocity, so it changes direction only.
Fill in the blank: A particle moving parallel to a magnetic field experiences a magnetic force of ______.
Zero.
Fill in the blank: The magnetic field inside a long solenoid with n turns per unit length and current I is ______.
B = μ₀nI
True or false: A large, constant magnetic flux through a loop produces a large EMF.
False. EMF depends on the rate of change of flux, not the flux itself.
Fill in the blank: A rod of length L moving at speed v perpendicular to a field B has a motional EMF of ______.
vBL
Q: A proton moves north in a region where the magnetic field points vertically upward (out of the ground). In which direction is the magnetic force on the proton?
A: East. Using v × B with v = north (ŷ) and B = up (ẑ): ŷ × ẑ = x̂ (east). The proton is positive, so F points east.
Q: A particle with charge +q enters a uniform magnetic field B with velocity v parallel to the field lines. What is the magnitude of the magnetic force?
A: Zero. F = qvB sin θ, and θ = 0° when v is parallel to B.
Q: What is the magnetic field inside a solenoid with n turns per unit length carrying current I?
A: B = μ₀nI. The field is uniform and directed along the solenoid axis.
Q: A conductive rod of length L moves at velocity v perpendicular to a uniform field B. What is the induced EMF?
A: EMF = vBL.
Q: A circular loop of radius 0.1 m sits in a magnetic field that increases at 2.0 T/s, perpendicular to the loop. What is the magnitude of the induced EMF?
A: |EMF| = A(dB/dt) = π(0.1)²(2.0) = 0.02π ≈ 0.0628 V.
Q: An infinitely long cylindrical conductor of radius R carries current density J = J₀(r/R). What is the total current?
A: Integrate: I = ∫₀ᴿ J₀(r/R)(2πr)dr = (2πJ₀/R)(R³/3) = 2πJ₀R²/3.
Q: For the same conductor, what is the magnetic field B at distance r < R from the axis?
A: Find I_enc = 2πJ₀r³/(3R). Apply Ampere's law: B(2πr) = μ₀I_enc, so B = μ₀J₀r²/(3R).
Faraday's law is the bridge between the electric and magnetic worlds. A changing magnetic field induces an electric field, which is the basis for electromagnetic waves (covered in more advanced E&M courses and optics). The concept of inductance, which this exam does not cover heavily, extends the RC circuit ideas from Part 2 into RL and RLC circuits.
Ampere's law has the same mathematical structure as Gauss's law from Part 1. Both relate a field integral over a closed surface (or loop) to an enclosed source (charge or current). Recognising this pattern makes both easier to apply.
Magnetic force, Lorentz force, cross product, right-hand rule, magnetic field, tesla, solenoid, Ampere's law, Amperian loop, Biot-Savart law, current density, non-uniform current, Faraday's law, electromagnetic induction, magnetic flux, weber, motional EMF, Lenz's law, induced EMF, induced current, permeability, μ₀, generator, electric motor, MRI, PHYS 212 UIUC, E&M final review