Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = \( 180.0\,\mathrm{m\Omega} \). Assume the field to be uniform.
Suppose K is open and the rod is moved with a speed of \( 12\,\mathrm{cm\,s^{-1}} \) in the direction. The magnitude of the induced emf will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 9.0 \times 10^{-3}\,\mathrm{V} \)
A rod moving through a magnetic field has a motional emf across its ends even when the circuit is open; with K open only the current is zero. The rod, the field and the velocity are mutually perpendicular, so
\( \displaystyle \varepsilon = B\,v\,l \)
\( l = 15\,\mathrm{cm} = 0.15\,\mathrm{m} \)
\( v = 12\,\mathrm{cm\,s^{-1}} = 0.12\,\mathrm{m\,s^{-1}} \)
\( \displaystyle \varepsilon = 0.50 \times 0.12 \times 0.15 = 9.0\times10^{-3}\,\mathrm{V} \)
The resistance of the loop does not enter: it decides the current when K is closed, not the emf.