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

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

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

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

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

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

\( R = \displaystyle\frac{mv}{qB} \). 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

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

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

\( R = \displaystyle\frac{p}{qB} \). 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

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

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

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

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

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

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

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

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

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

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