Unit 3: Magnetic Fields

17–23% of AP Exam · 6 Topics · ~15 Practice Questions

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3.1 Magnets & Magnetic Fields

Every magnet has north and south poles. Like poles repel, opposite attract. Magnetic field lines form closed loops — exit N and enter S outside the magnet, then return from S to N inside. There are no magnetic monopoles (unlike electric charges where single + or − can exist).

The magnetic field \(\vec{B}\) is measured in teslas (T = N/(A·m)). 1 T is very strong; Earth's surface field is ~50 μT. Key properties:

  • \(\vec{B}\) is tangent to field lines; line density ∝ field strength
  • Field lines never start or end (∇·B = 0, Gauss's law for magnetism)
  • The magnetic force does no work (\(\vec{F} \perp \vec{v}\) always)
Exam Tip Because magnetic forces do no work, a pure B-field cannot change a particle's speed — only its direction. This is a classic conceptual question.

3.2 Moving Charges in B-Fields

The magnetic force on a moving charged particle:

$$\vec{F} = q\vec{v} \times \vec{B}, \quad F = |q|vB\sin\theta$$

Right-hand rule (for positive charge): fingers along \(\vec{v}\), curl toward \(\vec{B}\), thumb = \(\vec{F}\) direction. For negative charges, reverse the force direction.

Circular motion in uniform B: When \(\vec{v} \perp \vec{B}\), the magnetic force is the centripetal force:

$$|q|vB = m\frac{v^2}{r} \implies r = \frac{mv}{|q|B}$$

Cyclotron frequency (independent of speed and radius!):

$$\omega = \frac{|q|B}{m}, \quad f = \frac{\omega}{2\pi}, \quad T = \frac{2\pi m}{|q|B}$$

If \(\vec{v}\) has a component parallel to \(\vec{B}\), the motion is helical: circular in the perpendicular plane plus constant drift along the field.

Hall Effect: When a current-carrying conductor sits in a perpendicular B-field, charge carriers are deflected to one side, creating a measurable transverse voltage:

$$V_H = \frac{IB}{nqw} = v_d B w$$

where w is the conductor width, n is charge carrier density, and \(v_d\) is drift velocity. The Hall voltage's sign reveals whether charge carriers are positive or negative — historically crucial for discovering that current in metals is carried by electrons.

Common Mistake The RHR is for positive charges. For electrons, flip the force direction. This is the #1 source of sign errors on the AP exam.
Practice

Q3.1: Proton Circular Motion

A proton (\(m = 1.67\times10^{-27}\ \text{kg}\), \(e = 1.60\times10^{-19}\ \text{C}\)) enters uniform B = 0.50 T at \(v = 2.0\times10^6\ \text{m/s}\) perpendicular to B. Find radius, period, and cyclotron frequency.

Show Solution
\(r = mv/(eB) = (1.67\times10^{-27})(2\times10^6)/(1.6\times10^{-19})(0.5) = 0.0418\ \text{m} = 4.18\ \text{cm}\)
\(\omega_c = eB/m = 4.79\times10^7\ \text{rad/s}\). \(T = 2\pi/\omega_c = 1.31\times10^{-7}\ \text{s}\). \(f_c = 7.63\ \text{MHz}\).
Practice

Q3.2: Velocity Selector

A region has crossed fields: \(\vec{E} = 2000\hat{j}\ \text{V/m}\) and \(\vec{B} = 0.10\hat{k}\ \text{T}\). Find the velocity of particles that pass undeflected through both fields.

Show Solution
For no deflection: \(q\vec{E} + q\vec{v}\times\vec{B} = 0 \implies \vec{E} = -\vec{v}\times\vec{B}\).
With \(\vec{v}\) along \(\hat{i}\): \(2000\hat{j} = -v\hat{i}\times 0.10\hat{k} = 0.10v\hat{j}\). So \(v = 20000\ \text{m/s}\) along \(+\hat{x}\).
Only particles with \(v = E/B = 2\times10^4\ \text{m/s}\) pass straight through, independent of q or m.

3.3 Forces on Current-Carrying Wires

The magnetic force on a straight wire segment:

$$\vec{F} = I\vec{\ell} \times \vec{B}, \quad F = I\ell B\sin\theta$$

Same RHR: fingers along current (\(\vec{\ell}\)), curl toward \(\vec{B}\), thumb = force.

Torque on a current loop in B-field:

$$\vec{\tau} = \vec{\mu} \times \vec{B}, \quad \vec{\mu} = NI\vec{A}$$

where \(\vec{\mu}\) is the magnetic dipole moment: \(N\) turns, current \(I\), area vector \(\vec{A}\) (RHR: curl fingers with current, thumb = direction of \(\vec{A}\)).

The torque tends to align \(\vec{\mu}\) with \(\vec{B}\). Potential energy: \(U = -\vec{\mu}\cdot\vec{B}\). This is the working principle behind electric motors.

Exam Tip \(\vec{\tau} = \vec{\mu}\times\vec{B}\) is the magnetic analog of \(\vec{\tau} = \vec{p}\times\vec{E}\) for electric dipoles. Same form, same PE formula \(U = -\vec{\mu}\cdot\vec{B}\).
Practice

Q3.3: Force on a Wire Segment

A 0.50 m wire carries 3.0 A northward in a uniform 0.20 T B-field pointing east. Find the magnetic force (magnitude and direction).

Show Solution
\(F = I\ell B\sin 90° = (3)(0.5)(0.2) = 0.30\ \text{N}\).
RHR: I north, B east → \(\vec{F}\) is upward, 0.30 N.

3.4 Magnetic Fields from Currents

Every current produces a magnetic field. The simplest and most important result:

Long straight wire:

$$B = \frac{\mu_0 I}{2\pi r}, \quad \mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$$

Direction: RHR — thumb along current, fingers curl in direction of \(\vec{B}\) (concentric circles around the wire).

Force between two parallel wires:

$$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}$$

Parallel currents attract; anti-parallel repel. This is how the ampere is defined.

Circular loop (center): \(B = \mu_0 I/(2R)\). Solenoid (inside): \(B = \mu_0 n I\) where \(n = N/L\) turns per meter. Field is uniform inside, ~zero outside.

Practice

Q3.4: Forces Between Wires

Two parallel wires 3.0 cm apart carry 5.0 A and 8.0 A in the same direction. Find the force per meter on each wire. Is it attractive or repulsive?

Show Solution
\(F/\ell = \mu_0 I_1 I_2/(2\pi d) = (4\pi\times10^{-7})(5)(8)/(2\pi\cdot0.03) = 2.67\times10^{-4}\ \text{N/m}\).
Attractive (parallel currents). The force is equal and opposite on each wire (Newton's 3rd law).

3.5 The Biot-Savart Law

The Biot-Savart law gives the \(d\vec{B}\) contribution from a current element \(Id\vec{\ell}\):

$$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{\ell} \times \hat{r}}{r^2}$$

This is the magnetic analog of \(d\vec{E} = k\frac{dq}{r^2}\hat{r}\) — both inverse-square laws.

Key results via integration:

  • Finite straight wire: \(B = \frac{\mu_0 I}{4\pi r}(\cos\theta_1 - \cos\theta_2)\)
  • Circular loop (on axis): \(B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\)
  • Loop center (x=0): \(B = \mu_0 I/(2R)\)

olen integrate: \(\vec{B} = \int d\vec{B}\). Exploit symmetry to cancel components — the direction comes from \(d\vec{\ell} \times \hat{r}\) at each segment.

Common Mistake Don't forget that \(d\vec{B}\) is a vector. Radial segments often contribute nothing at certain points (when \(d\vec{\ell} \parallel \hat{r}\), the cross product is zero). Always check geometry first.
Practice

Q3.5: Biot-Savart — Quarter Circle

A wire has two straight radial segments and a quarter-circle arc of radius R, carrying current I. Find B at the center of curvature P.

Show Solution
Straight radial segments: Idℓ ∥ r̂, so cross product = 0 → zero contribution at P.
Arc: \(B = \frac{\mu_0 I}{4\pi} \int_0^{\pi/2} \frac{R d\theta}{R^2} = \frac{\mu_0 I}{4\pi R}\cdot\frac{\pi}{2} = \frac{\mu_0 I}{8R}\).
\(B = \mu_0 I/(8R)\), direction ⊥ to page per RHR.

3.6 Ampere's Law

Ampere's Law relates the line integral of \(\vec{B}\) around a closed loop to the enclosed current:

$$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$$

Strategy (mirrors Gauss's Law): (1) Choose Amperian loop exploiting symmetry. (2) B is constant and ∥ or ⟂ on each segment. (3) Find \(I_{\text{enc}}\). (4) Solve for B.

Standard results (must memorize):

  • Long straight wire: \(B = \mu_0 I/(2\pi r)\)
  • Inside wire (uniform J): \(B = \mu_0 I r/(2\pi R^2)\)
  • Ideal solenoid: \(B = \mu_0 n I\) inside, B ≈ 0 outside
  • Toroid: \(B = \mu_0 N I/(2\pi r)\) inside the winding, B = 0 outside

Note: This is the static form. Maxwell will later add the displacement current term \(\mu_0\varepsilon_0 d\Phi_E/dt\) to handle time-varying E-fields.

Critical Ampere's law is always true, but only computationally useful with high symmetry (straight wires, solenoids, toroids). For arbitrary geometries, use Biot-Savart.
Practice

Q3.6: Coaxial Cable

A coaxial cable: inner solid conductor (radius a) carries +I uniformly; outer thin cylindrical shell (radius b > a) carries −I. Find B(r) everywhere.

Show Solution
\(r < a\): \(B(2\pi r) = \mu_0 I(r^2/a^2) \implies B = \mu_0 I r/(2\pi a^2)\).
\(a < r < b\): \(B = \mu_0 I/(2\pi r)\). \(r > b\): \(I_{\text{enc}} = I - I = 0 \implies B = 0\).
No magnetic field outside the cable. This makes coaxial cables self-shielding — they don't interfere with nearby electronics.