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When the current changes from +2A to 2A in 0.05 second, an emf of 8V is induced in the coil. The coefficient of self inductance of the coil is
Concept:
Induced e.m.f can be given as:
Where V_{L} = induced voltage in volts, L = selfinductance of the coil, dl/dt = rate of change of current in ampere/second.
Calculation:
Given:
I_{1} = +2 Amp, I_{2} = 2 Amp, dt = 0.05 sec, VL = 8 volt
⇒ dI = I_{2}  I_{1} = (2  2) = 4 Amp
From equation 1,
An emf of 5 V is produced by self  inductance, when the current changes at a steady rate from 3 A to 2 A in 1 ms. The value of self  inductance is
Concept:
Where E is the electromotive force or the potential, t is the time elapsed, L is the selfinductance of the coil and ΔI is the change in the electric current of the coil.
Calculation:
Given: E = 5 V, t = 1 ms = 0.001 s, ΔI = (2A  3A) =  1A
According to Faraday's Law,
If we apply law of conversion of energy to electromagnetic induction, electrical energy induced in a conductor comes from
Concept:
Explanation:
This principle is used in electric generators.
So option 3 is correct.
When number of turns per unit length of a solenoid is doubled, its self inductance becomes:
Concept:
Selfinductance 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:
μ_{o} ,I, A are constants. So, we can say that
L = k N^{2}  (2)
k is constant, N is the number of turns
L is directly proportional to the square of a number of turns.
If number of turns becomes N' = N2, then inductance
L' = k N'^{2}
⇒ L' = k (2N)^{2} = k 4 N^{2} = 4 K N^{2}
⇒ L' = 4 L
So, the inductance is increased by 4 times.
Hence the correct option is 4 times.
When the length of the solenoid is doubled without any change in the number of turns and the area of the coil.Then its selfinductance will
Concept:
Where μ_{o} = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.
Explanation
Given  l_{2} = 2l_{1}
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance:
Concept:
Where L is inductance of the inductor and I is current flowing.
Explanation:
Given that:
Magnetic potential energy (U) = 25 mJ = 25 x 10^{3} J
Current (I) = 60 mA = 60 x 103A
Use the formula:
If we increase the current in an inductor, self inductance of the inductor will __________.
Concept:
Explanation:
Concept:
Mutual Induction:
Explanation:
Dependence of mutual inductance –
The coefficient of selfinductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of selfinductance will become nearly
Concept:
SelfInduction:
Where N = number of turns, A = area of crosssection, l = length of the solenoid and μ_{0} = absolute permeability
Calculation:
Give  L_{1} = 0.18 mH = 0.18 x 10^{3} H , μ_{1} = μ_{₀}μ, μ_{2} = μ_{₀} and relative permeability (μ) = 900
The selfinductance of a solenoid is given by:
Here N, A and l is constant
∴ L ∝ μ
If the magnitude of the applied potential is decreased in the coil, then the direction of the induced current and the current due to applied potential will be
Concept:
Selfinduction:
Lenz's Law:
where N = number of loops and dϕ = Change in magnetic flux
The above equation is given by Faraday's law, but the negative sign is a result of Lenz's law.
Explanation:
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