Match List I with List II.
| LIST I | LIST II | ||
|---|---|---|---|
| A. | NBA\( \omega\sin\omega t \) | I. | Blv |
| B. | Joule's loss of heat | II. | Mutual inductance |
| C. | Motional emf | III. | \( B^{2}l^{2}v^{2}/R \) |
| D. | \( \mu_{\mathrm{r}}\mu_0 n_1 n_2 \pi r^{2} l \) | IV. | A.C generator |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-III, C-I, D-II
Each expression in List I is identified from the electromagnetic induction result it represents.
(A) NBA\( \omega\sin\omega t \) — this matches (IV) A.C generator. A coil of N turns and area A rotating with angular speed \( \omega \) in a field B has an induced emf \( \varepsilon = NBA\omega\sin\omega t \), which is the emf of an a.c. generator.
(B) Joule's loss of heat — this matches (III) \( B^{2}l^{2}v^{2}/R \). For a rod of length l moving with speed v, the motional emf is \( Blv \) and the current is \( Blv/R \), so the power lost as heat is \( \frac{\varepsilon^{2}}{R} = \frac{B^{2}l^{2}v^{2}}{R} \).
(C) Motional emf — this matches (I) Blv. The emf induced across a conductor of length l moving with speed v perpendicular to a field B is \( \varepsilon = Blv \).
(D) \( \mu_{\mathrm{r}}\mu_0 n_1 n_2 \pi r^{2} l \) — this matches (II) Mutual inductance. It is the mutual inductance of two long coaxial solenoids of length l, with the inner one of radius r, wound on a core of relative permeability \( \mu_{\mathrm{r}} \), having \( n_1 \) and \( n_2 \) turns per unit length.