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Step-by-Step Solution
Step 1: Identify the given information
• The current flowing through a purely inductive circuit is given by
$I(t) = 5 \,\sin(49\,\pi t - 30^\circ).$
• The inductance of the circuit is $L = 30\,\mathrm{mH} = 30 \times 10^{-3}\,\mathrm{H}.$
Step 2: Recall the voltage-current relationship for an inductor
For an inductor, the voltage $V(t)$ is related to the current $I(t)$ by the expression
$$
V(t) = L \frac{dI(t)}{dt}.
$$
In an AC scenario, this can also be expressed as
$$
V(t) = \omega L \, I_0 \,\sin\!\bigl(\omega t + \phi + 90^\circ\bigr),
$$
where $I_0$ is the amplitude of the current, $\omega$ is the angular frequency, and the additional $90^\circ$ (or $\frac{\pi}{2}$ in radians) accounts for the fact that voltage leads the current by $90^\circ$ in a purely inductive circuit.
Step 3: Determine the angular frequency $\omega$
The given current has the form:
$$
I(t) = 5 \,\sin(49\,\pi \,t - 30^\circ).
$$
Here, the coefficient of $t$ inside the sine function ($49\,\pi$) represents the angular frequency $\omega$. Therefore,
$$
\omega = 49\,\pi.
$$
Step 4: Write the general form for voltage across the inductor
Since $I(t)$ is in the sine form, the voltage leads by $90^\circ$:
$$
V(t) = I_0 \,\omega\,L \,\sin(\omega t - 30^\circ + 90^\circ).
$$
Substituting $I_0 = 5,$ $\omega = 49\,\pi,$ and $L = 30 \times 10^{-3}\,\mathrm{H}$ gives:
$$
V(t) = 5 \times 49\,\pi \times 30 \times 10^{-3} \,\sin\bigl(49\,\pi\,t - 30^\circ + 90^\circ\bigr).
$$
Step 5: Simplify the expression
First, calculate the numerical factor:
$$
49\,\pi \times 30 \times 10^{-3} = 49 \times \pi \times 0.03.
$$
Since $\pi \approx \frac{22}{7},$
$$
49 \times \frac{22}{7} = 49 \times \frac{22}{7} = 7 \times 22 = 154,
$$
so
$$
49\,\pi \approx 154 \quad \text{and} \quad 154 \times 0.03 = 4.62.
$$
Then multiply by $5$:
$$
5 \times 4.62 = 23.1.
$$
Also, $-30^\circ + 90^\circ = 60^\circ.$
Thus,
$$
V(t) = 23.1 \,\sin(49\,\pi\,t + 60^\circ).
$$
Final Answer
The voltage across the inductor is
$$
\boxed{23.1 \,\sin\bigl(49\,\pi\,t + 60^\circ\bigr).}
$$
