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In a coil of N turns, a current of i produces a magnetic flux of ϕ. Find the self inductance of this coil.
Concept:
It is calculated by:
where N is the number of turns, ϕ is the magnetic flux, i is current, and L is the selfinductance.
Calculation:
The selfinductance of a coil is calculated by
So the correct answer is option 3.
Two inductors when connected in series with additive connection, their equivalent inductance is 400 µH and when same inductors connected in series with subtractive connection, their equivalent inductance is 300 µH. Find their mutual inductance.
Let two inductors are L_{1} and L_{2} 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.
If N is the number of turns in a coil, the value of self inductance varies as
Concept:
SelfInduction
Explanation:
We know selfinductance,
Hence the correct answer is N^{2}
A current in a coil of inductance 5 H decreases at a rate of 2 Amp/sec. The induced emf is
Concept:
SelfInduction:
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,
Current in a coil changes from 4 A to zero in 0.1 second and the emf induced is 100 V. The self inductance of the coil is
The correct answer is option 2) i.e. 2.5 H.
Concept:
The induced emf is related to the change in current as follows:
Where e is the induced emf, L is the selfinductance 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
Selfinductance, L = 2.5 H
In a coil, the current changes from 4A to 2A in 0.05 sec. If the induced e.m.f. is 8 volt, then selfinductance of the coil is:
Concept:
SelfInduction
Induced e.m.f can be given as
Where,
V_{L} = induced voltage in volts
N = selfinductance of the coil
dI/dt= rate of change of current in ampere/second
Calculation:
Given I_{1} = 4A, I_{2} = 2A, dt = 0.05 sec, V_{L} = 8 volt
⇒ dI = I_{2}  I_{1} = (2  4) = 2A
From equation 1,
Hence, option 3 is correct.
The self inductance of a long solenoid of length l is proportional to
Explanation:
Solenoid:
SelfInduction:
Selfinductance of a solenoid is given by:
where μ_{o} = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = crosssectional area of the solenoid.
Thus we can say that
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:
SelfInduction
Induced e.m.f can be given as:
Where V_{L} = induced voltage in volts, L = selfinductance of the coil, dI/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,
A circuit possesses an inductance of 1 H when a current through coil is changing uniformly at the rate of 1 A/s inducing an opposing emf of 'X' volts in it.
What is the value of 'X'?
Concept:
SelfInduction
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
In a coil current changes from 2A to 4A in 0.05 second. If the average induced e.m.f. is 8 volt, then coefficient of self inductance is:
Concept:
SelfInduction
Calculation:
Given I_{1} = 4A, I_{2} = 2A, dt = 0.05 sec, VL = 8 volt
⇒ dI = I_{2}  I_{1} = (2  4) = 2A
From equation 1,
Hence, option 1 is correct.
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