Chapter 30 - Induction and Inductance - Problems - Page 897: 18b

$\mathscr{E} = 56.8~V$

In part (a), we found that $A = h^2$, where $h$ is the height of the bar above the vertex. We can find the magnitude of the emf: $\mathscr{E} = \frac{d\Phi}{dt}$ $\mathscr{E} = B~\frac{dA}{dt}$ $\mathscr{E} = B~\frac{dA}{dh}~\frac{dh}{dt}$ $\mathscr{E} = B~2h~v$ $\mathscr{E} = (2)(0.350~T)(15.6~m)(5.20~m/s)$ $\mathscr{E} = 56.8~V$