A rod of length 0.5 m moves with a velocity of 3 m/s in a perpendicular magnetic field of strength 2 T. Find the emf induced across its ends.
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The induced emf in a moving conductor in a magnetic field is given by \( \mathcal{E} = B \cdot v \cdot L \), where \( B \) is the magnetic field strength, \( v \) is the velocity of the rod, and \( L \) is the length of the rod.
The induced emf \( \mathcal{E} \) in a moving conductor in a magnetic field is given by:
\[
\mathcal{E} = B \cdot v \cdot L
\]
Where:
- \( B \) is the magnetic field strength,
- \( v \) is the velocity of the rod,
- \( L \) is the length of the rod.
Substitute the given values:
- \( B = 2 \, \text{T} \),
- \( v = 3 \, \text{m/s} \),
- \( L = 0.5 \, \text{m} \).
Thus, the induced emf is:
\[
\mathcal{E} = 2 \times 3 \times 0.5 = 3 \, \text{V}
\]
Therefore, the correct answer is:
\[
\boxed{(C) \, 3 \, \text{V}}
\]
