A straight metal rod of length 40 cm moves with a constant velocity in a uniform magnetic field of 0.25 T. The velocity is perpendicular to the rod and to the field. An emf of 0.3 V appears between the ends of the rod. The speed of the rod is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\ \mathrm{m\,s^{-1}}\)
A straight conductor moving at right angles to a uniform field has a motional emf across its ends. Its value is \( \varepsilon=Blv \). Here the velocity is normal to both the rod and the field, so this relation applies directly.
\( l=40\ \mathrm{cm}=0.4\ \mathrm{m} \)
\( B=0.25\ \mathrm{T} \)
\( \varepsilon=0.3\ \mathrm{V} \)
\( \begin{aligned} v &= \frac{\varepsilon}{Bl} \\ &= \frac{0.3}{0.25\times0.4} \\ &= 3\ \mathrm{m\,s^{-1}} \end{aligned} \)
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:”
\( \displaystyle \varepsilon=-\frac{d\Phi_B}{dt}=-\frac{d}{dt}(Blx)=-Bl\frac{dx}{dt}= \)\( Blv \)\( \qquad (6.5) \)
“where we have used dx/dt = –v which is the speed of the conductor PQ. The induced emf Blv is called motional emf. …”