A copper rod of length L is moving with a uniform velocity v parallel to a long straight wire carrying a current I. The rod itself is perpendicular to the wire with its ends at distances X and Y from it. The emf induced in the whole rod will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\mu_0 Iv}{2\pi}\left[\log_e Y - \log_e X\right] \)
Note: The given answer is as per the official NTA Answer Key. However, the question can have more than one correct answer, i.e. both Option (1) and Option (3), as explained below.
The field varies along the rod, so integrate the motional emf of small elements. At distance r from the wire,
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \)
B, v and the rod are mutually perpendicular, so an element dr has emf
\( \displaystyle d\varepsilon = Bv\,dr = \frac{\mu_0 I v}{2\pi}\,\frac{dr}{r} \)
Integrating from the nearer to the farther end gives a positive emf, but the question does not say which end is nearer the wire.
Option (1) \( \displaystyle\frac{\mu_0 Iv}{2\pi}\left[\log_e Y - \log_e X\right] \) — Correct if X is the nearer end (Y > X), probably NTA’s reading.
\( \displaystyle \varepsilon = \frac{\mu_0 I v}{2\pi}\int_{X}^{Y}\frac{dr}{r} = \frac{\mu_0 I v}{2\pi}\left[\log_e Y - \log_e X\right] \)
Option (3) \( \displaystyle\frac{\mu_0 Iv}{2\pi}\left[\log_e X - \log_e Y\right] \) — Correct if X is the farther end (X > Y).
\( \displaystyle \varepsilon = \frac{\mu_0 I v}{2\pi}\int_{Y}^{X}\frac{dr}{r} = \frac{\mu_0 I v}{2\pi}\left[\log_e X - \log_e Y\right] \)
Option (2) \( \displaystyle\frac{\mu_0 I}{2\pi v}\left[\log_e Y - \log_e X\right] \) — Incorrect. The motional emf must be proportional to v, not to 1/v.
Option (4) \( \displaystyle\frac{\mu_0 Iv}{2}\left[\log_e X - \log_e Y\right] \) — Incorrect. The factor \( 2\pi \) from the field of a straight wire is missing.