A sinusoidal voltage V(t) = 100 sin(500t) is applied across a pure inductance of L = 0.02H. The current through the coil is:

We are given a sinusoidal voltage $$ V(t) = 100 \sin(500t) $$ applied across a pure inductance of $$ L = 0.02 \, \text{H} $$. We need to find the current through the coil.

For a pure inductor, the relationship between voltage and current is given by:

$$ V(t) = L \frac{di}{dt} $$

Substituting the given values:

$$ 100 \sin(500t) = 0.02 \cdot \frac{di}{dt} $$

Now, solve for $$ \frac{di}{dt} $$:

$$ \frac{di}{dt} = \frac{100 \sin(500t)}{0.02} $$

Calculate the numerical value:

$$ \frac{100}{0.02} = 5000 $$

So,

$$ \frac{di}{dt} = 5000 \sin(500t) $$

To find the current $$ i(t) $$, integrate both sides with respect to $$ t $$:

$$ i(t) = \int 5000 \sin(500t) \, dt $$

The integral of $$ \sin(500t) $$ is $$ -\frac{\cos(500t)}{500} $$, because the derivative of $$ \cos(500t) $$ is $$ -500 \sin(500t) $$, so the antiderivative of $$ \sin(500t) $$ is $$ -\frac{\cos(500t)}{500} $$. Therefore:

$$ i(t) = 5000 \cdot \left( -\frac{\cos(500t)}{500} \right) + C $$

where $$ C $$ is the constant of integration.

Simplify the expression:

$$ i(t) = -\frac{5000}{500} \cos(500t) + C $$

$$ i(t) = -10 \cos(500t) + C $$

In AC circuits with pure inductance, the average current over a full cycle is zero, and there is no DC component. Therefore, we can assume the constant $$ C = 0 $$. Additionally, the current in an inductor lags the voltage by $$ 90^\circ $$ (or $$ \frac{\pi}{2} $$ radians). Since the voltage is a sine function, the current should be a negative cosine function, which matches the form above without a constant term.

Thus, the current is:

$$ i(t) = -10 \cos(500t) $$

Comparing with the options:

A. $$ 10 \cos(500t) $$

B. $$ -10 \cos(500t) $$

C. $$ 10 \sin(500t) $$

D. $$ -10 \sin(500t) $$

Our solution matches option B.

Hence, the correct answer is Option B.