Perpendicular to 85.0 x 10⁻⁵ T magnetic field E = BIV E = BW(5.00x10⁻⁵)(2.0x104) 7.80 x 103 m/s induce electric field the work done on the conductor in moving unit charge from one

The problem describes a conductor moving perpendicular to a magnetic field.

• Magnetic field strength, $B = 85.00 \times 10^{-5} \text{ T}$
• Velocity of the conductor, $V = 7.80 \times 10^3 \text{ m/s}$

The student's notes mention "induce electric field" and the formula for motional electromotive force (EMF), $\epsilon = BlV$. Since the length ($L$) of the conductor is not provided, we will calculate the induced electric field ($E$), which is the EMF per unit length.

When a conductor moves with velocity $V$ perpendicular to a magnetic field $B$, the induced electric field $E$ within the conductor is given by:

Substitute the given values of $B$ and $V$ into the formula:

The induced electric field is $\text{6.63 V/m}$.