The average emf induced in a coil is 2 V when current is changed in 0.4 s
(A) from 5 A to 2 A and the self-inductance of the coil is 0.266 mH
(B) from 4 A to 4 A in the opposite direction, the self-inductance of the coil is 0.10 mH
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Both (A) and (B) are incorrect
The average emf induced in a coil by a changing current is set by its self-inductance.
\( \displaystyle \varepsilon = L\,\frac{|\Delta I|}{\Delta t} \;\Rightarrow\; L = \frac{\varepsilon\,\Delta t}{|\Delta I|} \)
(A) The current changes from 5 A to 2 A, so \( |\Delta I| = 3\,\mathrm{A} \).
\( \displaystyle L = \frac{2 \times 0.4}{3} = 0.267\,\mathrm{H} \)
The statement gives \( 0.266\,\mathrm{mH} \), which is 1000 times smaller — Incorrect.
(B) The current reverses from \( +4\,\mathrm{A} \) to \( -4\,\mathrm{A} \), so \( |\Delta I| = 8\,\mathrm{A} \).
\( \displaystyle L = \frac{2 \times 0.4}{8} = 0.10\,\mathrm{H} \)
The statement gives \( 0.10\,\mathrm{mH} \), again 1000 times smaller — Incorrect.
Both statements are incorrect as printed, which is Option (4).
Note: It seems that there is a mistake in the original NTA paper, which prints “mH” in both statements when “H” is meant, since the numbers 0.266 and 0.10 are the values in henry. If the unit were henry, both statements would be correct and the answer would be Option (2). As printed (mH), Option (4) is the answer.