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The current in a self-inductance L=40 mH is to be increased uniformly from 1A to 11A in 4 ms. The emf indued in the inductor during the process is
Options
(a) 100 V
(b) 0.4 V
(c) 40 V
(d) 440 V
Correct Answer:
100 V
Explanation:
Consider the inductor of inductance L.
The current flowing through the inductor is i.
Now, we can write ɸ = Li
where, ɸ is magnetic flux linked with the inductor dɸ / dt = L (di / dt)
Given, L = 40 mH
dt = change in time = t₂ – t₁ = 4 ms = Δi
So, dɸ / dt = Δɸ / dt = L (Δi / dt) = (40 mH) [10 /4 ms] = 10 × 10 = 100 —-(i)
According to Faraday’s law of electromagnetic induction, Emf induced, e = -(dɸ / dt)
|e| = Δɸ / dt = 100 V [from equation (i)]
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Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Assuming the sun to have a spherical outer surface of radius r
- An electric dipole is placed at an angle 30°with an electric field intensity 2×10⁵N/C.
- The dimensional formula for Young’s modulus is
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Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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