(b) The solution in this case is the same except that the cross-product v B and the bar are at θ ¼ 90α ¼ 60 to each other.
The induced emf is found in two steps. First, we find the vector product E ¼ v B. This gives the electric field intensity
in the direction perpendicular to v and B. However, we are only interested in that component of the electric field
intensity that is parallel to the bar. Thus, the second step is to calculate the projection of E on the bar: