AMP

Chapter-6 Electromagnetic Induction — Online MCQ Test

PHYSICS · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter-6 Electromagnetic Induction with a free chapter-wise online MCQ test. This chapter covers: Faraday's laws - Lenz's law - motional EMF - self induction - mutual induction. AI-generated questions from basic to board-exam level, with instant results and explanations.

Start Exam on Full Site →

Chapter-6 Electromagnetic Induction — Important Questions & Answers

According to Faraday's law of electromagnetic induction, the induced EMF in a coil is equal to:
  • A. The rate of change of magnetic flux through the coil
  • B. The magnetic flux through the coil
  • C. The resistance of the coil multiplied by current
  • D. The number of turns multiplied by magnetic field
Answer: A. The rate of change of magnetic flux through the coil
Faraday's law states that induced EMF = -dΦ/dt, which is the negative rate of change of magnetic flux through the coil.
Which law determines the direction of induced current in a conductor moving through a magnetic field?
  • A. Ohm's law
  • B. Lenz's law
  • C. Ampere's law
  • D. Coulomb's law
Answer: B. Lenz's law
Lenz's law states that the direction of induced current is such that it opposes the change in magnetic flux that produced it.
The SI unit of magnetic flux is:
  • A. Tesla
  • B. Ampere
  • C. Weber
  • D. Henry
Answer: C. Weber
Magnetic flux is measured in Weber (Wb), where 1 Wb = 1 Tesla × meter².
Self-inductance of a coil is the phenomenon where:
  • A. EMF is induced in one coil due to changing current in another coil
  • B. EMF is induced in a coil due to changing current in the same coil
  • C. EMF is induced due to changing magnetic field outside the coil
  • D. Current flows without any EMF being induced
Answer: B. EMF is induced in a coil due to changing current in the same coil
Self-inductance is when a changing current in a coil induces an EMF in the same coil, opposing the change in current.
A rectangular loop is placed in a uniform magnetic field. If the magnetic field is perpendicular to the plane of the loop and remains constant, the induced EMF will be:
  • A. Maximum
  • B. Minimum (zero)
  • C. Increasing with time
  • D. Equal to BL²
Answer: B. Minimum (zero)
Since magnetic flux is constant (uniform field, perpendicular to loop, no change), dΦ/dt = 0, so induced EMF = 0.
A conductor of length L moves with velocity v perpendicular to a magnetic field B. The motional EMF induced is:
  • A. BLv
  • B. B/Lv
  • C. BLv²
  • D. BL/v
Answer: A. BLv
The motional EMF is given by ε = BLv, where the conductor moves perpendicular to the magnetic field.
When a magnet falls through a conducting pipe, its acceleration is less than g because:
  • A. Air resistance opposes motion
  • B. Induced currents in the pipe create a magnetic field opposing the motion
  • C. The magnet's weight decreases
  • D. The pipe exerts a direct magnetic force
Answer: B. Induced currents in the pipe create a magnetic field opposing the motion
As the magnet falls, induced EMF creates currents that produce a magnetic field opposing the change (Lenz's law), creating an upward force.
A rectangular coil ABCD is partially inserted into a magnetic field region. As it moves further into the field, which side experiences maximum motional EMF?
  • A. The side completely in the field
  • B. The side at the boundary (cutting field lines)
  • C. The side parallel to motion
  • D. All sides experience equal EMF
Answer: B. The side at the boundary (cutting field lines)
Only the side crossing the boundary between field and no-field regions cuts magnetic field lines, so it experiences maximum EMF.
A rectangular loop with dimensions a×b moves with velocity v perpendicular to a magnetic field B. The motional EMF developed is Bvb (b is the width perpendicular to motion). Which statement is correct about energy considerations?
  • A. Mechanical work done equals electrical energy dissipated
  • B. Mechanical work is greater than electrical energy dissipated
  • C. Electrical energy exceeds mechanical work input
  • D. No relationship exists between mechanical work and electrical energy
Answer: A. Mechanical work done equals electrical energy dissipated
By energy conservation, mechanical work done against magnetic force equals electrical energy (I²R) dissipated in the loop's resistance.
Two coils A and B have the same self-inductance L and mutual inductance M between them. When connected in series aiding configuration, their combined inductance is L_series. When connected in series opposing configuration, their combined inductance is L_series'. The relationship is:
  • A. L_series + L_series' = 4L
  • B. L_series - L_series' = 4M
  • C. L_series·L_series' = (L² - M²)
  • D. L_series/L_series' = (L + M)/(L - M)
Answer: B. L_series - L_series' = 4M
Series aiding: L_series = L + L + 2M = 2L + 2M; Series opposing: L_series' = 2L - 2M; Difference: L_series - L_series' = 4M.