An emf of 5V is produced by an inductor due to self inductance, when the current changes at a steady rate from 3A to 2A in 1ms. The value of self inductance of inductor is
A
Zero
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B
0.5H
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C
50H
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D
5 mH
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Solution
The correct option is D5 mH The emf produced across an inductor is, ε=−Ldidt...(i) didt=rate of change of current through inductor. ⇒didt=2−31×10−3=−1000As−1 Substituting in Eq (i), 5=−L×(−1000) ∴L=5×10−3H=5 mH