A \( 20\,\mathrm{m} \) long horizontal straight wire extending from east to west falls with a speed of \( 10\,\mathrm{m/s} \), at right angle to the horizontal component of the earth’s magnetic field, \( 3\times10^{-5}\,\mathrm{Wb\,m^{-2}} \). The instantaneous value of the emf induced in the wire is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6.0\,\mathrm{mV} \)
The falling wire cuts the field lines, and the motional emf is \( \varepsilon = Blv \) when the wire, its velocity and the field are mutually perpendicular. Here the wire is east-west, it moves vertically, and the horizontal component of the earth’s field points north-south, so all three are perpendicular.
\( \varepsilon = B_H l v \)
\( \varepsilon = 3\times10^{-5}\,\mathrm{Wb\,m^{-2}}\times20\,\mathrm{m}\times10\,\mathrm{m/s} = 6\times10^{-3}\,\mathrm{V} = 6.0\,\mathrm{mV} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
