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 move with a velocity of:

We need to find the velocity of a metal rod that produces an EMF of $$\varepsilon = 0.08$$ V when moving perpendicular to a uniform magnetic field $$B = 0.4$$ T. The rod has length $$l = 10$$ cm $$= 0.1$$ m.

When a conducting rod moves with velocity $$v$$ perpendicular to a magnetic field, the free charges experience a Lorentz force $$F = qvB$$ that drives them to opposite ends, creating a potential difference. The motional EMF is given by:

$$\varepsilon = Blv$$

Rearranging to solve for $$v$$ and substituting:

$$v = \frac{\varepsilon}{Bl} = \frac{0.08}{0.4 \times 0.1} = \frac{0.08}{0.04} = 2 \text{ m/s}$$

Hence, the correct answer is Option 4.