šŸ›”ļø

Content Protected

Screenshots and recording are not allowed.

Click anywhere or refocus to continue

Self Inductance and Self Induction - UNSOLVED PRACTICE SET

Class 12

Chapter: Electromagnetic Induction | Topic: Self Inductance and Self Induction

Study Material.
Class 12

SELF INDUCTANCE AND SELF INDUCTION - UNSOLVED PRACTICE SET

Topic: Self Inductance and Self Induction

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Self-induction is the phenomenon in which:

  1. A changing current in a coil induces an EMF in a nearby coil
  2. A changing current in a coil induces an EMF in the same coil
  3. A steady current induces EMF
  4. A magnet induces current in a coil

Q2. The self-inductance L of a coil is defined as:

  1. L = NΦ/I
  2. L = I/NΦB
  3. L = Φ/NI
  4. L = NI/Φ

Q3. The SI unit of self-inductance is:

  1. Ohm
  2. Henry
  3. Farad
  4. Weber

Q4. The back EMF in an inductor is given by:

  1. ε = L(dI/dt)
  2. ε = āˆ’L(dI/dt)
  3. ε = L/I
  4. ε = LI

Q5. The self-inductance of a solenoid depends on:

  1. Only the number of turns
  2. Only the current
  3. Geometry, number of turns, and core material
  4. Only the resistance

Q6. When a circuit containing an inductor is switched off, the spark is produced due to:

  1. High resistance
  2. Large back EMF induced by rapid change in current
  3. Capacitance effect
  4. Low voltage

Short Answer Questions

Q7. Define self-inductance. What is its SI unit and dimensional formula?

Q8. Derive the expression for the self-inductance of a long solenoid.

Q9. A coil has self-inductance 0.5 H. Calculate the back EMF when the current changes at 4 A/s.

Q10. Why does a bulb connected in parallel with an inductor glow brighter when the circuit is broken? Explain.

Q11. The self-inductance of a coil is 2 mH. Calculate the energy stored when a current of 3 A flows through it.

Q12. Why does a coil with an iron core have much higher self-inductance than an air-core coil of the same dimensions?

Long Answer Questions

Q13. Derive the expression for the self-inductance of a long solenoid. Show that L = μ₀N²A/l, where N is the number of turns, A is the cross-sectional area, and l is the length. Explain how inserting an iron core increases the inductance.

Q14. Explain the growth and decay of current in an RL circuit when connected to a DC source. Derive the expressions for current as a function of time for both cases. What is the time constant and what does it signify?

Q15. A solenoid has length 50 cm, radius 2 cm, and 1000 turns. It carries a current that increases uniformly from 0 to 3 A in 0.2 s.

(a) Calculate the self-inductance of the solenoid.

(b) Calculate the back EMF during this period.

(c) Calculate the magnetic energy stored when the current reaches 3 A.

(d) If the solenoid has resistance 5 Ī©, calculate the time constant of the RL circuit.

[Given: μ₀ = 4Ļ€ Ɨ 10⁻⁷ T m/A]

Numerical / Application-Based Problems

Q16. In a school physics lab, a student builds an electromagnet using a solenoid with the following specifications: length 40 cm, diameter 4 cm, 2000 turns, iron core with relative permeability μ_r = 800.

(a) Calculate the self-inductance of the solenoid with and without the iron core.

(b) The solenoid is connected to a 12 V battery with internal resistance 1 Ī©. Calculate the time constant and the final steady current.

(c) Calculate the energy stored in the magnetic field at steady state.

(d) When the switch is opened, a spark jumps across the gap. Estimate the voltage of the spark if the current drops to zero in 1 ms.

(e) Explain why a diode is often placed across relay coils to prevent spark damage.

[Given: μ₀ = 4Ļ€ Ɨ 10⁻⁷ T m/A]

Q17. An RL circuit consists of an inductor L = 2 H (resistance 5 Ī©) and a resistor R = 15 Ī© connected in series to a 20 V DC source.

(a) Calculate the time constant of the circuit.

(b) Write the expression for current as a function of time after the switch is closed.

(c) Calculate the current at t = 0, t = Ļ„, t = 2Ļ„, and t = 5Ļ„.

(d) Calculate the voltage across the inductor and resistor at t = Ļ„.

(e) After steady state is reached, the switch is opened. Write the expression for current decay and calculate the time for current to fall to 10% of its maximum value.

(f) Plot graphs of current versus time for both growth and decay.

Q18. A fluorescent tube light uses a choke (inductor) to limit current. The choke has L = 2 H and resistance 20 Ī©. The tube operates at 220 V, 50 Hz and requires 0.4 A.

(a) Calculate the inductive reactance of the choke at 50 Hz.

(b) Calculate the total impedance of the choke.

(c) Calculate the voltage across the choke when the tube is operating.

(d) Explain why the choke is needed — why can't a simple resistor be used?

(e) Calculate the power factor of the choke and the power dissipated in it.

(f) When the starter switch opens, the collapsing magnetic field in the choke generates a high voltage pulse. Estimate this voltage if the current drops from 0.4 A to 0 in 1 ms.


Total: 30 Marks | Time: 40 mins

Explore more topics in Electromagnetic Induction