An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when moves with a velocity of:
- 20 ms-1
- 2 ms-1
- 0.5 ms-1
- 3.2 ms-1
The Correct Option is B
Solution and Explanation
The induced emf (\( \epsilon \)) in a moving conductor is given by: \[ \epsilon = B \cdot l \cdot v, \] where: - \( B = 0.4 \, \text{T} \) (magnetic field), - \( l = 10 \, \text{cm} = 0.1 \, \text{m} \) (length of the rod), - \( v \) is the velocity. \bigskip Rearrange the formula to solve for \( v \): \[ v = \frac{\epsilon}{B \cdot l}. \] Substitute the known values: \[ v = \frac{0.08}{0.4 \cdot 0.1}. \] Simplify: \[ v = \frac{0.08}{0.04} = 2 \, \text{ms}^{-1}. \]
Final Answer: The velocity of the rod is: \[ \boxed{2 \, \text{ms}^{-1}}. \]
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