F = BW (5.00 x 10⁻⁵) (2.0x106) (7.80 x 103 m/s)

The problem asks to calculate the induced electric field ($E$) using the formula $E = BV$.
From the image, we identify the following values:

• Magnetic field strength, $B = 5.00 \times 10^{-5} \text{ T}$ (assuming this is the intended magnetic field strength, as it's a typical value for such problems).
• Velocity of the conductor, $V = 7.80 \times 10^3 \text{ m/s}$ (indicated by the units and consistent with previous problems). The value $2.0 \times 10^4$ is not used in this calculation, assuming it is extraneous or for a different part of a problem not shown.

$$E = BV$$ $$E = (5.00 \times 10^{-5} \text{ T}) \times (7.80 \times 10^3 \text{ m/s})$$

$$E = (5.00 \times 7.80) \times 10^{-5+3} \text{ V/m}$$ $$E = 39.0 \times 10^{-2} \text{ V/m}$$ $$E = 0.390 \text{ V/m}$$

The induced electric field is $\text{0.390 V/m}$.