PASSPOINTPRO
PHYS 112
GENERAL PHYSICS II
MAGNETISM & MAGNETIC FORCES
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Magnetic Force • Charged Particles • Current-Carrying Wires • Fields • Flux
, FORMULA REFERENCE
Force on a moving charge: F = |q|vB sin(theta)
Circular path radius: r = mv/(|q|B)
Cyclotron frequency: f = |q|B/(2 pi m)
Velocity selector: v = E/B
Force on a straight wire: F = ILB sin(theta)
Magnetic dipole moment: mu = NIA
Torque on a current loop: tau = mu B sin(theta) = NIAB sin(theta)
Long straight wire: B = mu0 I/(2 pi r)
Long solenoid: B = mu0 nI
Center of N-turn circular coil: B = mu0 NI/(2R)
Parallel wires: F/L = mu0 I1 I2/(2 pi d)
Magnetic flux: Phi_B = BA cos(theta)
Permeability of free space: mu0 = 4 pi x 10^-7 T m/A
Coverage: magnetic force direction and magnitude, circular charged-particle motion, velocity selectors, current-carrying wires, loop torque,
dipole moment, fields from wires and solenoids, parallel-wire forces, circular coils, magnetic flux, and right-hand-rule reasoning.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 2
, 1. A particle with charge magnitude 1 uC moves at 2.5e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 138 mN
B. 206 mN
C. 275 mN
D. 68.8 mN
Answer: A - 138 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (1e-06)(2.5e+05)(0.55)sin(90 degrees) = 0.138 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
2. A particle with charge magnitude 3 uC moves at 2e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 45 degrees with the field. What is the magnitude of the
magnetic force?
A. 117 mN
B. 233 mN
C. 330 mN
D. 467 mN
Answer: B - 233 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2e+05)(0.55)sin(45 degrees) = 0.233 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
3. A particle with charge magnitude 3 uC moves at 2.5e+05 m/s through a 0.18 T magnetic
field. Its velocity makes an angle of 45 degrees with the field. What is the magnitude of the
magnetic force?
A. 47.7 mN
B. 191 mN
C. 95.5 mN
D. 135 mN
Answer: C - 95.5 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2.5e+05)(0.18)sin(45 degrees) = 0.0955 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 3
, 4. A particle with charge magnitude 2 uC moves at 3e+05 m/s through a 0.32 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 288 mN
B. 96 mN
C. 192 mN
D. 384 mN
Answer: C - 192 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (2e-06)(3e+05)(0.32)sin(90 degrees) = 0.192 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
5. A particle with charge magnitude 1 uC moves at 2e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 55 mN
B. 165 mN
C. 220 mN
D. 110 mN
Answer: D - 110 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (1e-06)(2e+05)(0.55)sin(90 degrees) = 0.11 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
6. A particle with charge magnitude 3 uC moves at 2.5e+05 m/s through a 0.32 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 480 mN
B. 360 mN
C. 120 mN
D. 240 mN
Answer: D - 240 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2.5e+05)(0.32)sin(90 degrees) = 0.24 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 4
PHYS 112
GENERAL PHYSICS II
MAGNETISM & MAGNETIC FORCES
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Magnetic Force • Charged Particles • Current-Carrying Wires • Fields • Flux
, FORMULA REFERENCE
Force on a moving charge: F = |q|vB sin(theta)
Circular path radius: r = mv/(|q|B)
Cyclotron frequency: f = |q|B/(2 pi m)
Velocity selector: v = E/B
Force on a straight wire: F = ILB sin(theta)
Magnetic dipole moment: mu = NIA
Torque on a current loop: tau = mu B sin(theta) = NIAB sin(theta)
Long straight wire: B = mu0 I/(2 pi r)
Long solenoid: B = mu0 nI
Center of N-turn circular coil: B = mu0 NI/(2R)
Parallel wires: F/L = mu0 I1 I2/(2 pi d)
Magnetic flux: Phi_B = BA cos(theta)
Permeability of free space: mu0 = 4 pi x 10^-7 T m/A
Coverage: magnetic force direction and magnitude, circular charged-particle motion, velocity selectors, current-carrying wires, loop torque,
dipole moment, fields from wires and solenoids, parallel-wire forces, circular coils, magnetic flux, and right-hand-rule reasoning.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 2
, 1. A particle with charge magnitude 1 uC moves at 2.5e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 138 mN
B. 206 mN
C. 275 mN
D. 68.8 mN
Answer: A - 138 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (1e-06)(2.5e+05)(0.55)sin(90 degrees) = 0.138 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
2. A particle with charge magnitude 3 uC moves at 2e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 45 degrees with the field. What is the magnitude of the
magnetic force?
A. 117 mN
B. 233 mN
C. 330 mN
D. 467 mN
Answer: B - 233 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2e+05)(0.55)sin(45 degrees) = 0.233 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
3. A particle with charge magnitude 3 uC moves at 2.5e+05 m/s through a 0.18 T magnetic
field. Its velocity makes an angle of 45 degrees with the field. What is the magnitude of the
magnetic force?
A. 47.7 mN
B. 191 mN
C. 95.5 mN
D. 135 mN
Answer: C - 95.5 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2.5e+05)(0.18)sin(45 degrees) = 0.0955 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 3
, 4. A particle with charge magnitude 2 uC moves at 3e+05 m/s through a 0.32 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 288 mN
B. 96 mN
C. 192 mN
D. 384 mN
Answer: C - 192 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (2e-06)(3e+05)(0.32)sin(90 degrees) = 0.192 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
5. A particle with charge magnitude 1 uC moves at 2e+05 m/s through a 0.55 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 55 mN
B. 165 mN
C. 220 mN
D. 110 mN
Answer: D - 110 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (1e-06)(2e+05)(0.55)sin(90 degrees) = 0.11 N. Only
the velocity component perpendicular to the field contributes to magnetic force.
6. A particle with charge magnitude 3 uC moves at 2.5e+05 m/s through a 0.32 T magnetic
field. Its velocity makes an angle of 90 degrees with the field. What is the magnitude of the
magnetic force?
A. 480 mN
B. 360 mN
C. 120 mN
D. 240 mN
Answer: D - 240 mN
Explanation: Use F = |q|vB sin(theta). Substituting gives F = (3e-06)(2.5e+05)(0.32)sin(90 degrees) = 0.24 N.
Only the velocity component perpendicular to the field contributes to magnetic force.
PassPointPro | PHYS 112 - Magnetism & Magnetic Forces Page 4
