Appeared in: 2025 (3 sets)
A metal rod of length $50\text{ cm}$ is held vertically and moved with a velocity of $10\text{ m/s}$ towards east. The horizontal component of the Earth's magnetic field at the place is $0\cdot4\text{ G}$. The emf induced across the ends of the rod is :
(A) $0\cdot1\text{ mV}$
(B) $0\cdot2\text{ mV}$
(C) $0\cdot8\text{ mV}$
(D) $1\cdot6\text{ mV}$
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Official Evaluation Steps & EL10 Insight
Correct Option: (B) $0\cdot2\text{ mV}$
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by the formula:
$$e = Bvl$$Given data in S.I. units:
- Length of the rod, $l = 50\text{ cm} = 0\cdot5\text{ m}$
- Velocity of the rod, $v = 10\text{ m/s}$
- Horizontal component of Earth's magnetic field, $B = 0\cdot4\text{ G} = 0\cdot4 \times 10^{-4}\text{ T}$
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
Substituting the values into the formula:
$$e = (0\cdot4 \times 10^{-4}\text{ T}) \times (10\text{ m/s}) \times (0\cdot5\text{ m})$$ $$e = 0\cdot4 \times 10^{-4} \times 5$$ $$e = 2\cdot0 \times 10^{-4}\text{ V}$$Step 3 (Final Concluded Answer): [0.25 Mark]
Converting the result into millivolts ($\text{mV}$):
$$e = 0\cdot2 \times 10^{-3}\text{ V} = 0\cdot2\text{ mV}$$Thus, the emf induced across the ends of the rod is $0\cdot2\text{ mV}$.
💡 EL10 Concept Insight: Imagine the rod is a "net" catching magnetic field lines as it moves; the faster it moves or the longer it is, the more "magnetic energy" it sweeps up, which gets converted into the electrical voltage we measure.