A rectangular loop of wire, 10 cm by 4 cm, has a small gap cut in it. The loop lies normal to a uniform magnetic field of 0.5 T. It is pulled out of the field region at 2 cm s−1, with the velocity perpendicular to its 10 cm sides. While the loop is leaving the field, the emf across the gap is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1\ \mathrm{mV}\)
While the loop leaves, one 10 cm side is still inside the field and moves across it. The area of the loop inside the field falls at the rate \( lv \). So the emf is the motional emf \( Blv \), where \( l \) is the side normal to the velocity. The gap keeps the circuit open, so this emf appears across the gap.
\( B=0.5\ \mathrm{T} \)
\( l=10\ \mathrm{cm}=0.10\ \mathrm{m} \)
\( v=2\ \mathrm{cm\,s^{-1}}=0.02\ \mathrm{m\,s^{-1}} \)
\( \begin{aligned} \varepsilon &= Blv \\ &= 0.5\times0.10\times0.02 \\ &= 1\times10^{-3}\ \mathrm{V} \\ &= 1\ \mathrm{mV} \end{aligned} \)
NCERT (2026-27):
Motional Electromotive Force – “Since x is changing with time, the rate of change of flux ΦB will induce an emf given by:”
\( \displaystyle \varepsilon=-\frac{d\Phi_B}{dt}=-\frac{d}{dt}(Blx)=-Bl\frac{dx}{dt}= \)\( Blv \)\( \qquad (6.5) \)