AP Physics Practice Questions: Magnetism

AP Physics multiple choice questions on magnetic fields, forces, and electromagnetic induction. Each question includes detailed solutions at the bottom of the page.

1. A proton moves at 2×10⁶ m/s perpendicular to a 0.5 T magnetic field. What is the magnetic force on the proton? (q = 1.6×10⁻¹⁹ C)

   A) 1.6×10⁻¹³ N
   B) 3.2×10⁻¹³ N
   C) 6.4×10⁻¹³ N
   D) 8.0×10⁻¹³ N
   E) 1.6×10⁻¹² N

2. A wire carrying 5 A current is placed in a 0.4 T magnetic field at 30° to the field. If the wire length in the field is 0.2 m, what is the magnetic force on the wire?

   A) 0.2 N
   B) 0.3 N
   C) 0.4 N
   D) 0.5 N
   E) 0.6 N

3. A circular loop of radius 0.1 m carries 2 A current. What is the magnetic field at its center? (μ₀ = 4π×10⁻⁷ T·m/A)

   A) π×10⁻⁶ T
   B) 8π×10⁻⁶ T
   C) 1.26×10⁻⁵ T
   D) 2.51×10⁻⁵ T
   E) 5.02×10⁻⁵ T

4. A square loop of side 0.2 m has resistance 2 Ω. The magnetic field through it changes from 0 to 0.5 T in 0.1 s. What is the average induced current?

   A) 0.01 A
   B) 0.02 A
   C) 0.05 A
   D) 0.10 A
   E) 0.20 A

5. Two parallel wires 0.1 m apart carry currents of 3 A and 5 A in opposite directions. What is the force per unit length between them?

   A) 3×10⁻⁵ N/m attractive
   B) 3×10⁻⁵ N/m repulsive
   C) 6×10⁻⁵ N/m attractive
   D) 6×10⁻⁵ N/m repulsive
   E) 1.5×10⁻⁴ N/m repulsive

Solutions

1. A) 1.6×10⁻¹³ N
   \( F = qvB \sin\theta = 1.6 \times 10^{-19} \times 2 \times 10^6 \times 0.5 \times \sin 90^\circ \)
   \( = 1.6 \times 10^{-13}\, \text{N} \)

2. A) 0.2 N
   \( F = ILB \sin\theta = 5 \times 0.2 \times 0.4 \times \sin 30^\circ \)
   \( = 5 \times 0.2 \times 0.4 \times 0.5 = 0.2\, \text{N} \)

3. C) 1.26×10⁻⁵ T
   \( B = \dfrac{\mu_0 I}{2R} = \dfrac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} \)
   \( = \dfrac{8\pi \times 10^{-7}}{0.2} = 4\pi \times 10^{-6} = 1.26 \times 10^{-5}\, \text{T} \)

4. D) 0.10 A
   Area: \( A = (0.2)^2 = 0.04\,\text{m}^2 \)
   \( \Delta \Phi = B_2 A - B_1 A = 0.5 \times 0.04 - 0 = 0.02\, \text{Wb} \)
   \( \varepsilon = \dfrac{\Delta \Phi}{\Delta t} = \dfrac{0.02}{0.1} = 0.2\,\text{V} \)
   \( I = \dfrac{\varepsilon}{R} = \dfrac{0.2}{2} = 0.10\,\text{A} \)

5. B) 3×10⁻⁵ N/m repulsive
   \( \dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{4\pi \times 10^{-7} \times 3 \times 5}{2\pi \times 0.1} \)
   \( = \dfrac{60\pi \times 10^{-7}}{0.2\pi} = 300 \times 10^{-7} = 3 \times 10^{-5}\, \text{N/m} \)
   Opposite currents repel, so repulsive force