Answer
a. $0.675V$.
b. End b.
c. $E=2.25V/m$ from b to a.
d. End b.
e. i. 0.
e ii. 0.
Work Step by Step
a. $\epsilon=vBLsin\phi$, where the angle is between the velocity and the magnetic field.
$$\epsilon=vBL=(5.00m/s)(0.450T)(0.300m)sin90^{\circ}=0.675V$$
b. Positive charges are pushed toward b. That end is at the higher electric potential.
c. $E=\frac{V}{L}=\frac{0.675V}{0.300m}=2.25V/m$. The direction of the electric field is from high potential to low potential, b to a.
d. Positive charges are pushed toward b. That end is at the higher electric potential and has an excess of positive charge.
e. i. Here, we assume L = 0, so there is no emf.
e ii. The emf is zero because the rod moves parallel to the magnetic field.
