Unit 12: Magnetism

AP Physics C: Electricity & Magnetism Review (12.1 – 12.4)

Section 12.1

Magnetic Fields

Magnetic fields \( \vec{B} \) are vector fields produced by moving charges or intrinsic magnetic moments. Unlike electric fields, magnetic field lines always form closed loops (Gauss’s Law for Magnetism: \( \oint \vec{B} \cdot d\vec{A} = 0 \)).

Q1: Which of the following statements about magnetic field lines is always TRUE?

(A) They point toward the North pole of a magnet.
(B) They represent the direction of the force on a stationary electron.
(C) The total magnetic flux through any closed surface is zero.
(D) They can begin and end on magnetic monopoles.

Gauss’s Law for Magnetism states that there are no magnetic monopoles. Therefore, field lines form closed loops, and net flux through any closed surface is zero.
Correct Answer: C

Q2: A magnetic dipole with moment \( \vec{\mu} \) is placed in a uniform magnetic field \( \vec{B} \). The potential energy of the dipole is MINIMIZED when:

(A) \( \vec{\mu} \) is perpendicular to \( \vec{B} \).
(B) \( \vec{\mu} \) is anti-parallel to \( \vec{B} \).
(C) \( \vec{\mu} \) is parallel to \( \vec{B} \).
(D) The field is zero.

Potential energy is \( U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta \). Minimum energy occurs when \( \cos\theta = 1 \), which means \( \theta = 0^\circ \) (parallel).
Correct Answer: C

Q3: At a certain location, the Earth’s magnetic field points horizontally toward the North. A magnetic compass needle will align itself with:

(A) The geographic North pole.
(B) The geographic South pole.
(C) The local magnetic field vector.
(D) The direction of the solar wind.

A compass needle is a small magnetic dipole that experiences a torque \( \vec{\tau} = \vec{\mu} \times \vec{B} \), causing it to align with the local magnetic field lines.
Correct Answer: C

Q4: If the magnetic flux through a flat loop of area \( A \) is \( \Phi_B \), what happens to the flux if the area is doubled and the angle between the field and the normal to the loop is changed from \( 0^\circ \) to \( 60^\circ \)?

(A) Flux remains \( \Phi_B \).
(B) Flux becomes \( 2\Phi_B \).
(C) Flux becomes \( \Phi_B / 2 \).
(D) Flux becomes \( 4\Phi_B \).

\( \Phi_B = BA \cos\theta \). Initial: \( B(A)(1) = BA \).
New: \( B(2A) \cos(60^\circ) = B(2A)(0.5) = BA \).
Correct Answer: A

Q5: Which unit is equivalent to one Tesla (\( 1 \text{ T} \))?

(A) \( \displaystyle\frac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}} \)
(B) \( \displaystyle\frac{\text{V} \cdot \text{m}}{\text{s}} \)
(C) \( \displaystyle\frac{\text{J}}{\text{A} \cdot \text{m}^2} \)
(D) \( \displaystyle\frac{\text{Wb}}{\text{m}} \)

From \( F = qvB \), we have \( B = F / (qv) \).
Units: \( \displaystyle\frac{\text{N}}{\text{C} \cdot (\text{m/s})} = \displaystyle\frac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}} \).
Correct Answer: A

Section 12.2

Magnetism and Moving Charges

The magnetic force on a charge \( q \) moving with velocity \( \vec{v} \) in a field \( \vec{B} \) is \( \vec{F}_M = q(\vec{v} \times \vec{B}) \). This force is always perpendicular to both \( \vec{v} \) and \( \vec{B} \), meaning it does NO work.

Q1: A proton enters a uniform magnetic field pointing into the page. If the proton is moving to the right, in which direction is the initial magnetic force?

(A) Toward the top of the page
(B) Toward the bottom of the page
(C) Into the page
(D) Out of the page

Using the Right-Hand Rule: Fingers toward velocity (right), curl toward field (into page). Thumb points UP (top of page).
Correct Answer: A

Q2: A particle of charge \( q \) and mass \( m \) moves in a circle of radius \( R \) in a uniform magnetic field \( B \). If its kinetic energy is doubled, the new radius is:

(A) \( R / \sqrt{2} \)
(B) \( \sqrt{2} R \)
(C) \( 2R \)
(D) \( 4R \)

\( R = \displaystyle\frac{mv}{qB} \). Since \( K = \displaystyle\frac{1}{2}mv^2 \), we have \( v = \sqrt{\displaystyle\frac{2K}{m}} \).
Thus, \( R \propto v \propto \sqrt{K} \). If \( K \rightarrow 2K \), then \( R \rightarrow \sqrt{2}R \).
Correct Answer: B

Q3: In a velocity selector, an electric field \( \vec{E} \) and a magnetic field \( \vec{B} \) are perpendicular. A particle passes through undeflected if its speed is:

(A) \( EB \)
(B) \( B/E \)
(C) \( E/B \)
(D) \( \sqrt{E/B} \)

Undeflected means \( F_{net} = 0 \). Thus, \( qE = qvB \), which simplifies to \( v = E/B \).
Correct Answer: C

Q4: Which of the following is true regarding the work done by a magnetic force on a moving electron?

(A) It is positive if the electron speeds up.
(B) It is negative if the electron slows down.
(C) It is always zero because the force is perpendicular to displacement.
(D) It depends on the charge-to-mass ratio.

\( dW = \vec{F} \cdot d\vec{s} = (\vec{v} \times \vec{B}) \cdot (\vec{v} dt) \). Since \( (\vec{v} \times \vec{B}) \) is perpendicular to \( \vec{v} \), the dot product is always zero.
Correct Answer: C

Q5: An alpha particle (\( +2e, 4m_p \)) and a proton (\( +e, m_p \)) enter the same magnetic field with the same momentum. The ratio of their orbital radii \( R_\alpha / R_p \) is:

(A) \( 1:1 \)
(B) \( 1:2 \)
(C) \( 2:1 \)
(D) \( 4:1 \)

\( R = \displaystyle\frac{p}{qB} \). If momentum \( p \) and field \( B \) are the same, \( R \propto 1/q \).
\( R_\alpha / R_p = q_p / q_\alpha = e / (2e) = 1/2 \).
Correct Answer: B

Section 12.3

Magnetic Fields of Wires & Biot-Savart Law

The Biot-Savart Law relates a current element \( I d\vec{l} \) to its contribution to the magnetic field: \( d\vec{B} = \displaystyle\frac{\mu_0}{4\pi} \displaystyle\frac{I d\vec{l} \times \hat{r}}{r^2} \).

Q1: What is the magnitude of the magnetic field at the center of a circular wire loop of radius \( R \) carrying current \( I \)?

(A) \( \displaystyle\frac{\mu_0 I}{2R} \)
(B) \( \displaystyle\frac{\mu_0 I}{2\pi R} \)
(C) \( \displaystyle\frac{\mu_0 I}{R} \)
(D) Zero

Integration of Biot-Savart around the loop: \( B = \displaystyle\frac{\mu_0 I}{4\pi R^2} \int dl = \displaystyle\frac{\mu_0 I (2\pi R)}{4\pi R^2} = \displaystyle\frac{\mu_0 I}{2R} \).
Correct Answer: A

Q2: Two long parallel wires carry currents \( I \) in opposite directions. The magnetic field at a point exactly halfway between the wires is:

(A) Zero
(B) Directed along the line connecting the wires
(C) Double the field of a single wire
(D) Half the field of a single wire

Using the Right-Hand Rule, the fields from both wires point in the SAME direction at the midpoint. Thus, they add constructively: \( B_{net} = B_1 + B_2 = 2B_{single} \).
Correct Answer: C

Q3: A straight wire segment of length \( L \) carries current \( I \). At a point on the axis of the wire (beyond its end), the magnetic field is:

(A) Proportional to \( 1/L \)
(B) Proportional to \( L \)
(C) Zero
(D) Infinite

In the cross product \( d\vec{l} \times \hat{r} \), the vectors are parallel for any point on the wire’s axis. Thus, the cross product is zero everywhere.
Correct Answer: C

Q4: The magnetic force per unit length between two parallel wires separated by distance \( d \) is \( F/L = \displaystyle\frac{\mu_0 I_1 I_2}{2\pi d} \). If both currents are doubled and the distance is halved, the force becomes:

(A) \( 2 \times \) original
(B) \( 4 \times \) original
(C) \( 8 \times \) original
(D) \( 16 \times \) original

\( F/L \propto \displaystyle\frac{I_1 I_2}{d} \). New force: \( \displaystyle\frac{(2I_1)(2I_2)}{d/2} = \displaystyle\frac{4}{1/2} = 8 \times \) the original force.
Correct Answer: C

Q5: Which of the following defines the permeability of free space \( \mu_0 \)?

(A) \( 8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2) \)
(B) \( 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)
(C) \( 1.6 \times 10^{-19} \text{ C} \)
(D) \( 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \)

The constant \( \mu_0 \) is exactly \( 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \). Option A is \( \epsilon_0 \), and D is Coulomb’s constant.
Correct Answer: B

Section 12.4

Ampère’s Law

Ampère’s Law relates the line integral of the magnetic field around a closed loop to the enclosed current: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \). It is highly effective for symmetric systems like solenoids and toroids.

Q1: A long straight wire of radius \( R \) carries a uniform current \( I \). How does the magnetic field \( B \) vary for points INSIDE the wire (\( r < R \))?

(A) \( B \propto 1/r \)
(B) \( B \propto r \)
(C) \( B \propto r^2 \)
(D) \( B \) is constant

Using Ampère’s Law: \( B(2\pi r) = \mu_0 I_{enc} \). Since current is uniform, \( I_{enc} = I \left(\displaystyle\frac{\pi r^2}{\pi R^2}\right) \).
Solving for \( B \): \( B = \displaystyle\frac{\mu_0 I r}{2\pi R^2} \). Thus, \( B \propto r \).
Correct Answer: B

Q2: An ideal solenoid of length \( L \) has \( N \) turns and carries current \( I \). If the length is doubled while keeping \( N \) and \( I \) constant, the magnetic field inside:

(A) Doubles
(B) Quadruples
(C) Is halved
(D) Stays the same

Inside a solenoid, \( B = \mu_0 n I = \mu_0 (N/L) I \). If \( L \) doubles, the turn density \( n \) is halved, so the field is halved.
Correct Answer: C

Q3: Which path would be the most appropriate Ampèrian loop for calculating the field inside a toroid?

(A) A circle perpendicular to the toroid’s axis
(B) A circle concentric with the toroid’s center, lying in the plane of the toroid
(C) A rectangle passing through the center of the coils
(D) A sphere enclosing the entire toroid

The field of a toroid is circular and confined within the core. A concentric circular loop allows \( \vec{B} \) to be parallel to \( d\vec{l} \) and constant in magnitude.
Correct Answer: B

Q4: The magnetic field at a distance \( r \) outside an infinite solenoid is:

(A) \( \mu_0 n I / (2\pi r) \)
(B) \( \mu_0 n I \)
(C) \( \mu_0 n I r \)
(D) Zero

For an ideal (infinite) solenoid, the field lines are perfectly contained inside. Outside, any Ampèrian loop would enclose a net current of zero (up and down segments cancel), yielding \( B = 0 \).
Correct Answer: D

Q5: Ampère’s Law, as written here, is technically incomplete because it does not account for:

(A) Changing magnetic fields
(B) Changing electric fields (displacement current)
(C) The presence of dielectric materials
(D) Non-ohmic resistors

In cases with time-varying electric fields (like a charging capacitor), the Maxwell-Ampère Law is required: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 (I + \epsilon_0 \displaystyle\frac{d\Phi_E}{dt}) \).
Correct Answer: B

Magnetism Formulas Recap

Lorentz Force\( \vec{F} = q(\vec{v} \times \vec{B}) \)

Cyclotron Radius\( R = \displaystyle\frac{mv}{qB} \)

Biot-Savart\( d\vec{B} = \displaystyle\frac{\mu_0 I d\vec{l} \times \hat{r}}{4\pi r^2} \)

Ampère’s Law\( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \)

Long Wire B\( B = \displaystyle\frac{\mu_0 I}{2\pi r} \)

Solenoid B\( B = \mu_0 n I \)

AP Physics C Electricity and Magnetism Quick Review: Unit 12 Magnetic Fields and Electromagnetism