Part A In the figure, a conducting rod of length L = 29.0 cm moves in a magnetic field B of magnitude 0.490 T directed into the plane of the figure. The rod moves with speed v 4.00 m/s in the direction shown. (Figure 1) When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge? Figure 1 of 1 a b Submit Previous Answers Correct The force on a charge q moving with velocity v in a magnetic field B is given by The magnitude of the force is given by F = qu B. F = |q|vB sin where q is the magnitude of the charge, v and B are the magnitudes of the velocity and magnetic field and is the angle measured from the direction of to the direction of B. When v is perpendicular to B, 6 = 90 and sin = 1. You can use the right-hand rule to determine the direction in which the electrons in the rod are forced to move. Part B What is the direction of the electric field in the rod? from a to b from b to a Submit Previous Answers Correct Keep in mind that the electric field exists due to the separation of the positive and negative charges caused by the magnetic force. Also, recall that electric field lines start on positive charges and end on negative charges. Part C When the charges in the rod are in equilibrium, what is the magnitude E of the electric field within the rod? Express your answer in volts per meter to at least three significant figures. View Available Hint(s) E = Submit c. ? V/m Figure a X V Part D Which point, a or b, has a higher potential? View Available Hint(s) a b Points a and b have the same potential. Submit Previous Answers Part E Correct What is the magnitude Vb of the potential difference between the ends of the rod? Express your answer in volts to at least three significant figures. View Available Hint(s) Vba = Submit Part F 1 of 1 ? What is the magnitude of the motional emf induced in the rod? Express your answer in volts to at least three significant figures. View Available Hint(s) E = Submit - Provide Feedback ? V