A conducting rod of length \( 2 \) m moves at a velocity of \( 1.5\,\mathrm{m\,s^{-1}} \) in a uniform magnetic field \( 5\times10^{-5} \) T, perpendicular to its length. The induced emf produced in the rod is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.15 \) mV
A rod of length l moving with velocity v perpendicular to a uniform field B has a motional emf between its ends, \( e = Blv \), when B, l and v are mutually perpendicular.
\( \displaystyle e = Blv \)
\( \displaystyle e = (5\times10^{-5}\,\mathrm{T})\times(2\,\mathrm{m})\times(1.5\,\mathrm{m\,s^{-1}}) \)
\( \displaystyle e = 1.5\times10^{-4}\,\mathrm{V} \)
\( 1.5\times10^{-4}\,\mathrm{V} = 0.15\times10^{-3}\,\mathrm{V} = 0.15\,\mathrm{mV} \)