Test: Self Inductance - 2

Concept:

• Self-inductance: The property of the current-carrying coil that opposes or resists the change of current flowing through itself is called self-inductance.

It is calculated by:

where N is the number of turns, ϕ is the magnetic flux, i is current, and L is the self-inductance.

Calculation:
The self-inductance of a coil is calculated by

So the correct answer is option 3.

Let two inductors are L₁ and L₂ and their mutual inductance is M.
When both inductors are in additive connection

When both inductors are in subtractive connection

Subtracting equation (i) and (ii)
100 = 4M
M = 100/4 = 25 µH
Mutual inductance will 25 µH.

Concept:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –
  
  Where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid and A = Area of the solenoid.

Explanation:

We know self-inductance,

Hence the correct answer is N²

Concept:
Self-Induction:

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –
  

  Where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid, the resistance of the coil (R) = 4Ω, and A = Area of the solenoid.
  Induced e.m.f can be given as,
  

Calculation
Given -
di/dt = -2 Amp/sec and L = 5 H (Since the current decreases at a rate of 2 Amp/sec).
The emf induced across the inductor is,

The correct answer is option 2) i.e. 2.5 H.

Concept:

• Self-inductance: Self-inductance is the phenomenon where a current-carrying coil resists the change in the current flowing through it.
  • ​The resistance to the change in current is due to the induced emf in the coil. 
  • The direction of the induced emf is opposite to the applied voltage if the current is increasing and the direction of the induced emf is in the same direction as the applied voltage if the current is decreasing.
  • The unit of inductance is Henry (H).

The induced emf is related to the change in current as follows:

 

Where e is the induced emf, L is the self-inductance of the coil, and di/dt is the rate of change of current in the coil.

Calculation:
Given that:
Change in current, di = (0 - 4) A
Time interval, dt = 0.1 s
Induced emf, e = 100 V

Self-inductance, L = 2.5 H​​​​​

Concept:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –
  
  Where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid, the resistance of the coil (R) = 4Ω, and A = Area of the solenoid.

Induced e.m.f can be given as

Where,
V_L = induced voltage in volts
N = self-inductance of the coil
dI/dt= rate of change of current in ampere/second

Calculation:
Given I₁ = 4A, I₂ = 2A, dt = 0.05 sec, V_L = 8 volt
⇒ dI = I₂ - I₁ = (2 - 4) = -2A

From equation 1,

Hence, option 3 is correct.

Explanation:
Solenoid: 

• The solenoid is a long cylindrical coil of wire consisting of a large number of turns bound together very tightly.

Self-Induction:

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.

Self-inductance of a solenoid is given by:

where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = cross-sectional area of the solenoid.

Thus we can say that

Concept:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.

Induced e.m.f can be given as:

Where V_L = induced voltage in volts, L = self-inductance of the coil, dI/dt = rate of change of current in ampere/second.

Calculation:
Given:
I₁ = +2 Amp, I₂ = -2 Amp, dt = 0.05 sec, VL = 8 volt
⇒ dI = I₂ - I₁ = (-2 - 2) = -4 Amp
From equation 1,

Concept:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –
  
  Where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.

Induced e.m.f can be given as
Calculation:
Given:
L = 1 H, di / dt = 1 A/s

Here negative sigh indicated the nature of opposing.
∴ The opposing emf X = 1 V

Concept:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a solenoid is given by –
  
  Where μ_o = Absolute permeability, N = Number of turns, l = length of the solenoid, the resistance of the coil (R) = 4Ω, and A = Area of the solenoid.
• Induced e.m.f can be given as
  Where ,
  V_L = induced voltage in volts
  N = self-inductance of the coil
  dI/dt = rate of change of current in ampere/second

Calculation:
Given I₁ = 4A, I₂ = 2A, dt = 0.05 sec, VL = 8 volt
⇒ dI = I₂ - I₁ = (2 - 4) = -2A
From equation 1,

Hence, option 1 is correct.