Answer
6.32 volts.
Work Step by Step
Consider the coil to be made up of a pure resistance and a pure inductance, connected in series.
There is a voltage drop across the resistor, given by Ohm’s law.
$$V_R=IR$$
There is a voltage drop across the inductance due to the changing current, given by equation 21–9. The current is increasing, so the induced emf opposes the increasing current. It is like a little battery that opposes the direction of the current, trying to reduce it.
$$V_L=\epsilon=-L\frac{\Delta I}{\Delta t}$$
The coil consists of a pure resistance and a pure inductance, connected in series, so the potential difference across the coil is the sum of these two potential drops.
$$V_{coil}=IR-L\frac{\Delta I}{\Delta t}=(3.00A)(2.25\Omega) -(0.112H)(3.80A/s)=6.32V$$
