A metal rod slides at \(5\ \mathrm{m\,s^{-1}}\) along two parallel rails 0.5 m apart. A uniform magnetic field of 0.4 T acts normal to the plane of the rails. The rod has a resistance of \(0.2\ \Omega\), and the rails are joined at one end by a \(0.8\ \Omega\) resistor. Neglecting the resistance of the rails, the current in the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1\ \mathrm{A}\)
The moving rod acts like a cell of emf \( \varepsilon=Blv \). Its own resistance \( r \) works like the internal resistance of the cell. It is in series with the resistor \( R \), so the current is \( I=\varepsilon/(R+r) \).
\( \begin{aligned} \varepsilon &= Blv \\ &= 0.4\times0.5\times5 \\ &= 1\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} I &= \frac{\varepsilon}{R+r} \\ &= \frac{1}{0.8+0.2} \\ &= 1\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Motional Electromotive Force – “The induced emf Blv is called motional emf. …”
Summary – “… If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”