Electromotive force and Faraday's law
You left out some units, hence the confusion. Here's some very hand-wavy math that gets us to the right answer. Faraday's law states that (for a single loop) the EMF, \$ \mathcal{E} \$, is equal to the magnetic flux per unit time:  $$\mathcal{E} = -{{d\Phi_B} \over dt} \ $$ Magnetic flux, \$\Phi_B\$ is the magnetic field through an open surface. Magnetic field is in units of Tesla, mightiest of units: $$ \mathrm{T} 
= \dfrac{\mathrm{V}\cdot{\mathrm{s}}}{\mathrm{m}^2} $$ and a surface has the units \$\mathrm{m}^2\$. Putting that all together: 
$$ \mathcal{E} 
= \dfrac{\dfrac{\mathrm{-V}\cdot{\mathrm{s}}}{\mathrm{m}^2}\cdot\mathrm{m}^2}{\mathrm{s}} $$

Note that this math is not too valid, since we're dealing with time varying vectors. However, for exactly perpendicular, simple, instantaneous cases, it suites the purpose. 
So, with this little bit of dimensional analysis we see that EMF has units of Volts.

$$ \mathcal{E} = \mathrm{-V} $$