PASSPOINTPRO
PHYS 112
GENERAL PHYSICS II
ELECTROMAGNETIC INDUCTION
& FARADAY'S LAW
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Flux • Faraday's Law • Lenz's Law • Motional EMF • Inductance • RL Concepts
, FORMULA REFERENCE
Magnetic flux: Phi_B = BA cos(theta)
Faraday's law: emf = -N dPhi_B/dt
Average induced emf magnitude: |emf| = N|Delta Phi_B|/Delta t
Changing field, fixed loop: |emf| = NA|Delta B|cos(theta)/Delta t
Rotating coil: emf = NAB omega sin(omega t); emf_max = NAB omega
Motional emf: emf = BLv (for mutually perpendicular B, L, and v)
Induced current: I = emf/R
Magnetic force on sliding rod: F = BIL
Self-induced emf: |emf_L| = L|Delta I|/Delta t
Inductor energy: U = (1/2)LI^2
Long solenoid inductance: L = mu0 N^2 A/ell
Permeability of free space: mu0 = 4 pi x 10^-7 T m/A
Coverage: magnetic flux, Faraday and Lenz laws, flux changes from changing fields and rotation, generators, motional emf, induced current,
magnetic braking, self-inductance, inductor energy, and solenoid inductance.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 2
, 1. A flat loop of area 0.02 m^2 is in a uniform 0.7 T magnetic field. The field makes an angle of
45 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 14.8 mWb
B. 19.8 mWb
C. 4.95 mWb
D. 9.9 mWb
Answer: D - 9.9 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.7)(0.02)cos(45 degrees) = 0.0099 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
2. A flat loop of area 0.01 m^2 is in a uniform 0.25 T magnetic field. The field makes an angle of
60 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 2.5 mWb
B. 625 uWb
C. 1.88 mWb
D. 1.25 mWb
Answer: D - 1.25 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.25)(0.01)cos(60 degrees) = 0.00125 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
3. A flat loop of area 0.02 m^2 is in a uniform 0.4 T magnetic field. The field makes an angle of
60 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 6 mWb
B. 4 mWb
C. 8 mWb
D. 2 mWb
Answer: B - 4 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.4)(0.02)cos(60 degrees) = 0.004 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 3
, 4. A flat loop of area 0.02 m^2 is in a uniform 0.4 T magnetic field. The field makes an angle of
0 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 8 mWb
B. 12 mWb
C. 16 mWb
D. 4 mWb
Answer: A - 8 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.4)(0.02)cos(0 degrees) = 0.008 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
5. A flat loop of area 0.04 m^2 is in a uniform 0.7 T magnetic field. The field makes an angle of
90 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 8.57 x 10^-19 Wb
B. 2.57 x 10^-18 Wb
C. 1.71 x 10^-18 Wb
D. 3.43 x 10^-18 Wb
Answer: C - 1.71 x 10^-18 Wb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.7)(0.04)cos(90 degrees) = 1.71e-18 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
6. A flat loop of area 0.03 m^2 is in a uniform 0.25 T magnetic field. The field makes an angle of
30 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 3.25 mWb
B. 13 mWb
C. 6.5 mWb
D. 9.74 mWb
Answer: C - 6.5 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.25)(0.03)cos(30 degrees) = 0.0065 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 4
PHYS 112
GENERAL PHYSICS II
ELECTROMAGNETIC INDUCTION
& FARADAY'S LAW
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Flux • Faraday's Law • Lenz's Law • Motional EMF • Inductance • RL Concepts
, FORMULA REFERENCE
Magnetic flux: Phi_B = BA cos(theta)
Faraday's law: emf = -N dPhi_B/dt
Average induced emf magnitude: |emf| = N|Delta Phi_B|/Delta t
Changing field, fixed loop: |emf| = NA|Delta B|cos(theta)/Delta t
Rotating coil: emf = NAB omega sin(omega t); emf_max = NAB omega
Motional emf: emf = BLv (for mutually perpendicular B, L, and v)
Induced current: I = emf/R
Magnetic force on sliding rod: F = BIL
Self-induced emf: |emf_L| = L|Delta I|/Delta t
Inductor energy: U = (1/2)LI^2
Long solenoid inductance: L = mu0 N^2 A/ell
Permeability of free space: mu0 = 4 pi x 10^-7 T m/A
Coverage: magnetic flux, Faraday and Lenz laws, flux changes from changing fields and rotation, generators, motional emf, induced current,
magnetic braking, self-inductance, inductor energy, and solenoid inductance.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 2
, 1. A flat loop of area 0.02 m^2 is in a uniform 0.7 T magnetic field. The field makes an angle of
45 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 14.8 mWb
B. 19.8 mWb
C. 4.95 mWb
D. 9.9 mWb
Answer: D - 9.9 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.7)(0.02)cos(45 degrees) = 0.0099 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
2. A flat loop of area 0.01 m^2 is in a uniform 0.25 T magnetic field. The field makes an angle of
60 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 2.5 mWb
B. 625 uWb
C. 1.88 mWb
D. 1.25 mWb
Answer: D - 1.25 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.25)(0.01)cos(60 degrees) = 0.00125 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
3. A flat loop of area 0.02 m^2 is in a uniform 0.4 T magnetic field. The field makes an angle of
60 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 6 mWb
B. 4 mWb
C. 8 mWb
D. 2 mWb
Answer: B - 4 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.4)(0.02)cos(60 degrees) = 0.004 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 3
, 4. A flat loop of area 0.02 m^2 is in a uniform 0.4 T magnetic field. The field makes an angle of
0 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 8 mWb
B. 12 mWb
C. 16 mWb
D. 4 mWb
Answer: A - 8 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.4)(0.02)cos(0 degrees) = 0.008 Wb. The
angle is measured from the area vector, which is perpendicular to the loop.
5. A flat loop of area 0.04 m^2 is in a uniform 0.7 T magnetic field. The field makes an angle of
90 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 8.57 x 10^-19 Wb
B. 2.57 x 10^-18 Wb
C. 1.71 x 10^-18 Wb
D. 3.43 x 10^-18 Wb
Answer: C - 1.71 x 10^-18 Wb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.7)(0.04)cos(90 degrees) = 1.71e-18 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
6. A flat loop of area 0.03 m^2 is in a uniform 0.25 T magnetic field. The field makes an angle of
30 degrees with the loop's area vector. What is the magnetic flux through the loop?
A. 3.25 mWb
B. 13 mWb
C. 6.5 mWb
D. 9.74 mWb
Answer: C - 6.5 mWb
Explanation: Use magnetic flux Phi_B = BA cos(theta). Thus Phi_B = (0.25)(0.03)cos(30 degrees) = 0.0065 Wb.
The angle is measured from the area vector, which is perpendicular to the loop.
PassPointPro | PHYS 112 - Electromagnetic Induction & Faraday's Law Page 4
