A. 2V
B. 0.02V
C. 0.2V
D. Zero
Answer:- B. 0.02V
Right option is B)
Specified, L=2×10−3H
Δt=0.1sec
ΔI=1amp.
We identify that,
∴ε=−ΔtLΔI
=−2×10−3×0.11=−2×10−2
=−0.02V
If in a coil with 2×10−3H coefficient of self-inductance, current flows for 0.1s is increased at a constant rate by 1A,then the magnitude of induced emf is:
by
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