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EXAM 3 PRACTICE
PHYS 112
General Physics II
200-Question Practice Exam
Answers, worked calculations & detailed explanations
200 8 A-D
QUESTIONS CORE AREAS FORMAT
TYPICAL EXAM 3 COVERAGE
Magnetic forces & motion | Fields produced by currents
Faraday-Lenz induction | Inductance & RL circuits
AC & RLC circuits | Electromagnetic waves & radiation
Geometric optics | Interference, diffraction & polarization
Designed for focused practice, calculation fluency and concept review.
PASSPOINTPRO
,PHYS 112 - GENERAL PHYSICS II | EXAM 3 PRACTICE PassPointPro
Exam 3 Practice Overview
This 200-question PHYS 112 practice exam focuses on common later-unit General Physics II material. Questions combine
conceptual reasoning with multi-step calculations. Each item includes four answer choices, the correct answer, and a detailed
explanation immediately below the question.
Coverage Questions
Magnetic Forces & Charged-Particle Motion 25
Magnetic Fields Produced by Currents 25
Electromagnetic Induction & Faraday-Lenz Law 25
Inductance & RL Circuits 25
AC & RLC Circuits 25
Electromagnetic Waves & Radiation 25
Geometric Optics: Reflection, Refraction, Mirrors & Lenses 25
Wave Optics: Interference, Diffraction & Polarization 25
Suggested use: Work each problem without looking at the explanation, then use the rationale to identify the equation, sign
convention, or reasoning step that controls the answer.
Quick Formula Reference
Magnetic force: F = qvB sin(theta) | Wire force: F = ILB sin(theta)
Current fields: B_wire = mu0 I/(2 pi r) | B_solenoid = mu0 nI
Flux: Phi = BA cos(theta) | Faraday: |emf| = N|Delta Phi|/Delta t
Inductors: |emf| = L|dI/dt| | U = (1/2)LI^2 | tau_RL = L/R
AC: X_L = 2 pi fL | X_C = 1/(2 pi fC) | Z = sqrt(R^2 + (X_L-X_C)^2)
EM waves: c = f lambda | E0 = cB0 | I = (1/2)c eps0 E0^2
Optics: n1 sin(theta1) = n2 sin(theta2) | 1/f = 1/do + 1/di
Wave optics: d sin(theta) = m lambda | Malus: I = I0 cos^2(theta)
PHYS 112 | 200-Question Exam 3 Practice Exam Page 2
,PHYS 112 - GENERAL PHYSICS II | EXAM 3 PRACTICE PassPointPro
Magnetic Forces & Charged-Particle Motion
1. A positive particle with charge +2e moves at 3.0e+06 m/s through a 0.25 T magnetic
field at 60 deg to the field. What is the magnetic-force magnitude?
A 4.16e-13 N
B 1.04e-13 N
C 2.08e-13 N
D 8.32e-13 N
Answer: C - 2.08e-13 N
Explanation: Use F = qvB sin(theta). Substituting q = 3.204e-19 C, v = 3.00e+06 m/s, B = 0.25 T, and theta =
60 deg gives F = 2.081e-13 N.
2. A singly charged positive ion of mass 6.692e-27 kg enters a 0.50 T field perpendicular
to the field at 2.00e+06 m/s. What is the radius of its circular path?
A 0.084 m
B 0.334 m
C 1.671 m
D 0.167 m
Answer: D - 0.167 m
Explanation: For perpendicular motion, magnetic force supplies centripetal force: qvB = mv^2/r, so r = mv/(qB).
Substitution gives r = 1.671e-01 m.
3. In a velocity selector, the electric field is 8.0e+04 V/m and the magnetic field is 0.20 T.
What speed passes undeflected?
A 4.00e+05 m/s
B 4.00e+06 m/s
C 8.00e+05 m/s
D 2.00e+05 m/s
Answer: A - 4.00e+05 m/s
Explanation: Undeflected particles satisfy qE = qvB, so v = E/B. Thus v = (8.00e+04)/(0.20) = 4.000e+05 m/s.
4. A straight wire 0.35 m long carries 6 A in a 0.40 T magnetic field. The wire makes 90
deg with the field. What magnetic-force magnitude acts on the wire?
A 3.36 N
B 0.42 N
C 1.68 N
D 0.84 N
Answer: D - 0.84 N
Explanation: Use F = ILB sin(theta). Substitution gives F = (6)(0.35)(0.40)sin(90 deg) = 0.840 N.
PHYS 112 | 200-Question Exam 3 Practice Exam Page 3
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5. A 50-turn current loop carries 0.5 A and has area 0.004 m^2 in a 0.20 T field. If the
loop normal is 90 deg from the field, what is the torque magnitude?
A 0.08 N m
B 0.04 N m
C 0.01 N m
D 0.02 N m
Answer: D - 0.02 N m
Explanation: Torque on a current loop is tau = NIAB sin(theta). Substitution gives tau = 2.000e-02 N m.
6. An electron moves perpendicular to a 0.30 T field in a circle of radius 0.020 m. What is
its speed?
A 1.06e+09 m/s
B 1.06e+10 m/s
C 5.28e+08 m/s
D 2.11e+09 m/s
Answer: A - 1.06e+09 m/s
Explanation: From r = mv/(|q|B), solve v = |q|Br/m. Using electron mass 9.109e-31 kg gives v = 1.055e+09 m/s.
7. A proton moves perpendicular to a uniform 0.20 T field. What is its cyclotron period
(nonrelativistic)?
A 1.64e-07 s
B 3.28e-06 s
C 6.56e-07 s
D 3.28e-07 s
Answer: D - 3.28e-07 s
Explanation: The cyclotron period is T = 2 pi m/(|q|B), independent of speed. Substituting proton mass and
charge gives T = 3.281e-07 s.
8. A charged particle moves exactly parallel to a uniform magnetic field. What magnetic
force acts on it?
A A maximum force qvB.
B A force opposite the velocity.
C Zero magnetic force.
D A force parallel to the field.
Answer: C - Zero magnetic force.
Explanation: Because F = qvB sin(theta), theta = 0 deg gives sin(0) = 0. The magnetic field does not deflect a
particle moving exactly along the field.
PHYS 112 | 200-Question Exam 3 Practice Exam Page 4
EXAM 3 PRACTICE
PHYS 112
General Physics II
200-Question Practice Exam
Answers, worked calculations & detailed explanations
200 8 A-D
QUESTIONS CORE AREAS FORMAT
TYPICAL EXAM 3 COVERAGE
Magnetic forces & motion | Fields produced by currents
Faraday-Lenz induction | Inductance & RL circuits
AC & RLC circuits | Electromagnetic waves & radiation
Geometric optics | Interference, diffraction & polarization
Designed for focused practice, calculation fluency and concept review.
PASSPOINTPRO
,PHYS 112 - GENERAL PHYSICS II | EXAM 3 PRACTICE PassPointPro
Exam 3 Practice Overview
This 200-question PHYS 112 practice exam focuses on common later-unit General Physics II material. Questions combine
conceptual reasoning with multi-step calculations. Each item includes four answer choices, the correct answer, and a detailed
explanation immediately below the question.
Coverage Questions
Magnetic Forces & Charged-Particle Motion 25
Magnetic Fields Produced by Currents 25
Electromagnetic Induction & Faraday-Lenz Law 25
Inductance & RL Circuits 25
AC & RLC Circuits 25
Electromagnetic Waves & Radiation 25
Geometric Optics: Reflection, Refraction, Mirrors & Lenses 25
Wave Optics: Interference, Diffraction & Polarization 25
Suggested use: Work each problem without looking at the explanation, then use the rationale to identify the equation, sign
convention, or reasoning step that controls the answer.
Quick Formula Reference
Magnetic force: F = qvB sin(theta) | Wire force: F = ILB sin(theta)
Current fields: B_wire = mu0 I/(2 pi r) | B_solenoid = mu0 nI
Flux: Phi = BA cos(theta) | Faraday: |emf| = N|Delta Phi|/Delta t
Inductors: |emf| = L|dI/dt| | U = (1/2)LI^2 | tau_RL = L/R
AC: X_L = 2 pi fL | X_C = 1/(2 pi fC) | Z = sqrt(R^2 + (X_L-X_C)^2)
EM waves: c = f lambda | E0 = cB0 | I = (1/2)c eps0 E0^2
Optics: n1 sin(theta1) = n2 sin(theta2) | 1/f = 1/do + 1/di
Wave optics: d sin(theta) = m lambda | Malus: I = I0 cos^2(theta)
PHYS 112 | 200-Question Exam 3 Practice Exam Page 2
,PHYS 112 - GENERAL PHYSICS II | EXAM 3 PRACTICE PassPointPro
Magnetic Forces & Charged-Particle Motion
1. A positive particle with charge +2e moves at 3.0e+06 m/s through a 0.25 T magnetic
field at 60 deg to the field. What is the magnetic-force magnitude?
A 4.16e-13 N
B 1.04e-13 N
C 2.08e-13 N
D 8.32e-13 N
Answer: C - 2.08e-13 N
Explanation: Use F = qvB sin(theta). Substituting q = 3.204e-19 C, v = 3.00e+06 m/s, B = 0.25 T, and theta =
60 deg gives F = 2.081e-13 N.
2. A singly charged positive ion of mass 6.692e-27 kg enters a 0.50 T field perpendicular
to the field at 2.00e+06 m/s. What is the radius of its circular path?
A 0.084 m
B 0.334 m
C 1.671 m
D 0.167 m
Answer: D - 0.167 m
Explanation: For perpendicular motion, magnetic force supplies centripetal force: qvB = mv^2/r, so r = mv/(qB).
Substitution gives r = 1.671e-01 m.
3. In a velocity selector, the electric field is 8.0e+04 V/m and the magnetic field is 0.20 T.
What speed passes undeflected?
A 4.00e+05 m/s
B 4.00e+06 m/s
C 8.00e+05 m/s
D 2.00e+05 m/s
Answer: A - 4.00e+05 m/s
Explanation: Undeflected particles satisfy qE = qvB, so v = E/B. Thus v = (8.00e+04)/(0.20) = 4.000e+05 m/s.
4. A straight wire 0.35 m long carries 6 A in a 0.40 T magnetic field. The wire makes 90
deg with the field. What magnetic-force magnitude acts on the wire?
A 3.36 N
B 0.42 N
C 1.68 N
D 0.84 N
Answer: D - 0.84 N
Explanation: Use F = ILB sin(theta). Substitution gives F = (6)(0.35)(0.40)sin(90 deg) = 0.840 N.
PHYS 112 | 200-Question Exam 3 Practice Exam Page 3
, PHYS 112 - GENERAL PHYSICS II | EXAM 3 PRACTICE PassPointPro
5. A 50-turn current loop carries 0.5 A and has area 0.004 m^2 in a 0.20 T field. If the
loop normal is 90 deg from the field, what is the torque magnitude?
A 0.08 N m
B 0.04 N m
C 0.01 N m
D 0.02 N m
Answer: D - 0.02 N m
Explanation: Torque on a current loop is tau = NIAB sin(theta). Substitution gives tau = 2.000e-02 N m.
6. An electron moves perpendicular to a 0.30 T field in a circle of radius 0.020 m. What is
its speed?
A 1.06e+09 m/s
B 1.06e+10 m/s
C 5.28e+08 m/s
D 2.11e+09 m/s
Answer: A - 1.06e+09 m/s
Explanation: From r = mv/(|q|B), solve v = |q|Br/m. Using electron mass 9.109e-31 kg gives v = 1.055e+09 m/s.
7. A proton moves perpendicular to a uniform 0.20 T field. What is its cyclotron period
(nonrelativistic)?
A 1.64e-07 s
B 3.28e-06 s
C 6.56e-07 s
D 3.28e-07 s
Answer: D - 3.28e-07 s
Explanation: The cyclotron period is T = 2 pi m/(|q|B), independent of speed. Substituting proton mass and
charge gives T = 3.281e-07 s.
8. A charged particle moves exactly parallel to a uniform magnetic field. What magnetic
force acts on it?
A A maximum force qvB.
B A force opposite the velocity.
C Zero magnetic force.
D A force parallel to the field.
Answer: C - Zero magnetic force.
Explanation: Because F = qvB sin(theta), theta = 0 deg gives sin(0) = 0. The magnetic field does not deflect a
particle moving exactly along the field.
PHYS 112 | 200-Question Exam 3 Practice Exam Page 4
