## University Physics with Modern Physics (14th Edition)

a. $0.675V$. b. End b. c. $E=2.25V/m$ from b to a. d. End b. e. i. 0. e ii. 0.
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.