When a metal rod of length L placed normal to a uniform magnetic field B is moved at a velocity of v perpendicular to the field, the induced emf across its ends is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( BLv \)
This is motional emf. Each free electron in the moving rod feels the magnetic force \( qvB \) along the rod, and charges pile up at the ends until the electric field they create balances this force.
\( \displaystyle qE = qvB \;\Rightarrow\; E = vB \)
The potential difference across the length \( L \) is then
\( \displaystyle \varepsilon = E\,L = B\,L\,v \)
This holds when the rod, its velocity and the field are mutually perpendicular, as here.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Summary – “… 4. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its ends is
\( \displaystyle \varepsilon = Bl\,v \)”